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Psycholinguistics/A comparison of language processing in Chinese and English

Introduction
Language is central to societal life. It is a window on cognition, and may be a window on (a partial explanation of) cultural differences. To understand how the brain processes language, it can be useful to contrast distinct languages. In the current work, Chinese is contrasted with English

English Grammar System
To understand how Chinese language processing is different from English, it is useful to review the basic structures of English. English, as well as other Indo-European languages (e.g., French, German), is an alphabetic language. The basic structural unit is the phoneme (the smallest unit of sound). Phonemes are combined to make morphemes, which are the basic units of meaning (the smallest semantic unit). A morpheme may be one or more phonemes. A word may be a single morpheme (e.g., cat) or a combination of morphemes (e.g., cat + s = cats). Some morphemes, such as suffixes (e.g., the “s” in “cats”) and prefixes (e.g., “un” in “undo”), cannot stand alone as words. Critically, in English, one can understand the meaning of a word by identifying and understanding the component morphemes, and then combining the meanings of those morphemes. Essentially, the meaning of morphemes is invariant (though there are context-dependent exceptions) and morphemes add to create meaning of the word. In spoken English, the phonemes are combined to create syllables. which are typically, a consonant-vowel-consonant combination (i.e., three phonemes), though there are variations. Phonemes, morphemes and syllables are related by not equated (Jay, 2002). For example, the word “preconception” has three morphemes (“pre” + “concept” + “ion”), but four syllables (“pre” + “con” + “cept” + “ion”) and 12 phonemes (“p” + “r” + “iy” + “k” + “ah” + “n” + “s” + eh + p + “sh” + “an” + “n”).

In English, words are combined into meaningful sentences using syntax rules (Ashcraft & Klin, 2006). The comprehension of a sentence is based on the meanings of the words, but the identification, and ultimately the specific meaning, of words is aided by the syntactical rules. Different categories of words fulfill different functions (e.g., nouns, verbs, objects, prepositions, adverbs, adjectives, conjunctions). Most words belong to a single category though there are exceptions (e.g., impact, project) particularly for verbs and nouns (note that the pronunciation typically changes when a word changes categories; project as a verb is pronounced differently than project as a noun). More typically, words are altered by suffixes or prefixes when changing categories (e.g., noun and verb forms of a word typically differ). Syntax constrains the sequential word order (word categories) by defining aspects like word order, subject-verb agreement (Liu, Li, Shu, Zhang & Chen, 2010). For example, noun or adjectives follow articles (e.g., “the car”, “a red car”), clauses have a typical subject-verb-object construction (SVO, e.g., “the red car drove into the garage”), and objects follow prepositions (e.g., “into the garage”). Syntax also constrains subject-verb agreement and tense (e.g., “he runs”, “they run” “they ran”). Some Indo-European languages such as French that are structurally similar to English also have gender (Liu,et al 2010; Zhang, Yu & Boland).

To understand spoken English, one must parse the sequence of syllables into morphemes and then into words (and sentences), using both semantics and syntax. Semantics is collective meaning of all the words, while syntax is essentially grammar. The difficulty for the comprehension of spoken English is the identification of breaks between words (and sentences). That is, even though it appears effortless for listeners, the breaks between words are not always obvious in the acoustic signal. In written English, the phoneme is essentially associated with a particular letter string, while words are identified by breaks and sentences are clearly marked. Hence, the breaks between words (and sentences) are obvious. However, morphemes (or even the phonemes) within a word are not necessarily obvious (Jay, 2002).

In English, syntax and semantics can be processed independently (Liu., et al., 2010), even though both are “required” for interpretation. For example, it is easy to construct sentences that follow the syntactical rules of English (e.g., the water breathes furiously) without having meaning. It is also possible to construct sentences that have meaning while violating syntactical rules (e.g., the boys is gone). Parents can understand the speech of children even if it is ungrammatical. The absence of proper syntax or semantics will make an utterance more difficult to understand or more ambiguous, but not impossible (Liu., et al, 2010).

For the comparison of English comprehension to Chinese comprehension, there are a number of other important issues. Firstly, there is a loose hierarchical nesting of meanings particularly for nouns and verbs (e.g, a ginger cat is a type of cat, which is a type of animal, which is a type of animate object; running and walking are types of motion, which is a type of animation). Secondly, synonyms are different words with the same meanings. They can be used to change the word without changing the meaning (e.g., “he taught the student/pupil”), although syntax might be affected (e.g., “he does that everyday” vs “he does that day after day”; “there is one dog” vs “there is a single dog”). Synonyms can be used to generate errors (e.g., “the music was dampened/muted” but not “the cloth was dampened/muted”). Homonyms are words with the same sound but different meanings. They are not detectable (applicable) in speech but are easily identified by both syntax and semantics in written work (e.g., “their” “there” and “they’re” are not interchangeable syntactically or semantically). Like synonyms, homonyms can be used to generate errors, but only in written work. Thirdly, semantics and/or syntax can constrain word category and/or word order (Liu., et al, 2010). In that constraint, he prior context of a word may be highly constraining (e.g., “he put the key into the____”), moderately constraining (e.g., “the sky is ____”), or minimally constraining (e.g., “he wanted a ____”), though the definition of the degree of constraint may be somewhat subjective. Those constraints may be semantic or syntactic or both. Finally, one must also remember that there are numerous exceptions to every rule, be that morphological or syntactical, and that colloquial speech (common, everyday speech) does not follow all the rules (Storkel.,et al, 2004). In English, the rules should be seen as strong guidelines rather than as "rules" (other alphabetic languages are more rigorous).

Chinese Grammar System
Unlike English, Chinese is an orthographic language. In an orthographic language, each morpheme is only used "in isolation". Each morpheme has a stand-alone meaning (Newman, 2011). Each morpheme is unbound. In English, one has the morphemes "dog", "toy", "escutcheon" and "s" and one may combine “dog” with “s” to get “dogs”, and “toy” with “s” to get “toys”. Even if one were to be unfamiliar with the meaning of escutcheon (the bit of metal or wood that surrounds a key hole), one could surmise that escutcheons is the plural form. Note that “s” cannot be used as a stand-alone word. It can only be used in combination. It must be bound to another morpheme. On the other hand, “dog” and “toy” may be unbound. English has a number of bound morphemes like “s” (plural), “ed” (past tense), “ex” or “un” (inverting the meaning of the following morpheme) “er” (person who), or “il/ile” (pertaining to, or capability). Many prefixes or suffixes are bound morphemes, but note that suffixes and prefixes are not limited to bound morphemes – some can also function as unbound morphemes (e.g., “ate” or “ant” or “an”). Bound morphemes are still morphemes – they have a fixed meaning that is used to modify the other morpheme of the word. For example, “able” in “portable” means more or less the same as the noun “able” and it adds that meaning to any word it modifies. However, there are numerous exceptions that can be confusing (particularly during the acquisition of English). For example, “ate” in “mitigate” (where “ate” means to “make”) is quite different from the verb “ate” (past tense of eat). These exceptions are a consequence of the historical development of English. In addition, one must be mindful of the fact that combining morphemes into new words might (or might not) affect the syntax (grammar). For example, “there is an ant” vs “they are ants” (contrasted with “he walked” “they walked”).

Chinese does not work this way. In Chinese, all orthographies are unbound. There are no bound morphemes in Chinese. Each orthography has its single sound. For example, the character for people/person is 人 (rén); one cannot say how many morphemes, phonemes and syllables it has. Although Chinese developed Pinyin (is a pronunciation system to unify pronunciation by using letters) around 50 years ago, it is impossible to recognize anything like morphemes phonemes and syllables by looking at characters, and Pinyin is generally not used for writing. Thus, one does not modify the meaning of an orthography by adding a anything like a bound morpheme. Instead, one denotes different conditions (e.g., plural, past tense) by adding additional unbound orthographies to the sentence. For example, to denote the past, one adds additional words. To be specific about the present tense, one adds additional words. To denote the plural, one adds words. Note that English uses the same tactic for some situations. For example, the future tense uses additional unbound morphemes in an auxiliary plus verb structure (e.g., “he will run”). The structures of Chinese past, present and future tense are all analogous to the structure of the English future tense. This orthographic structure affects both the spoken and written versions of the language.

Chinese is complicated by the fact that there are many different spoken dialects of Chinese (e.g., Mandarin and Cantonese) which are not generally intelligible to each other. However, there is only one written form for all. Those who speak different dialects use completely different pronunciations for the same written symbols. Mandarin is the official dialect (the dialect based on the Northern (Beijing) style pronunciation) of the Chinese language, and its use is enforced throughout the education system. Hence it is useful to discuss the written form before discussing the spoken form.

In written Chinese, the most basic characters have developed from pictures (Pictographs).These characters “look like” the object represented (e.g., sun, moon tiger). See Figure 1. Other characters (Ideographs) represent abstract concepts, often in a meaningful way (e.g., 一 one, 二 two) but sometime not so (e.g., 四 four). See Figure 2.In these, the most basic element of the construction for each character is called the Stroke (See example 1). The strokes are classified as horizontal, vertical, rise, hook, dot and so on (Xue, Yue, Cao & Wang, 2003). The basic strokes are combined to make basic characters (e.g., 一 one, 木 wood, 水 water, etc). The basic characters are elaborated upon to make more complex characters (e.g., 林woods, 冰 ice, etc). However, the individual strokes have no intrinsic meaning, and do not associate with particular sounds (they are not the same as phonemes or letters in English). Thus, strokes cannot add up to give pronunciations and meanings to characters.

At times, one can have a “logical” combination of simpler characters to create a more complex concept. This is called a logical aggregate. For example, east (東) is the representation of the sun rising over trees. Some are shown in Figure 3, Logical Aggregates. However, most of the characters in the Chinese language are not logical-aggregates.

Alternatively, aggregates may be “semantically-phonetically based” (see later comments on the spoken language). One component of the character is a simpler character used for its broad semantic meaning to establish a semantic context (Xue., et al, 2003). To this, a second layer is added which may be based on an irrelevant syllable (e.g.,榜 means list of names. It is 木 “wood”+ 旁“side”, see more in Example 2) (Xin Hua Zi Dian, 2005). The semantic phonetic aggregates represent 90 - 95% of all symbols used in Chinese (Xue., et al, 2003). Some are shown in Figure 4:Semantic-phonetic Aggregates.

For those who only know English, there is a critical issue for the interpretation of semantic-phonetic aggregates or logical aggregates as “words” or concepts. (Xue, Yue, Cao & Wang, 2003). Generally one cannot add the components of a character to create a more complex concept even if it is a semantic-phonetic aggregate or a logical aggregate. That is, one could not know that the combination of person (the semantic root) with the irrelevant phonetic component “ban” produces “biased”. No one would reasonably guess that the characters woman and child sum to represent “good”, that “pig” plus “roof” is home (家), while “cow” plus “roof” is prison (牢). On the other hand, the fact that sun plus moon is “bright” (明) does seem reasonable. The character for steam (汽) is not the combination of the symbols for water and heat.

Chinese characters have evolved throughout Chinese history. However, most people would not be able to figure out the meaning of most of the characters even if they knew that history. Note that the same is true for English. Each English word has its origins, but most English speakers do not know those origins, and knowledge of the origin does not necessarily convey an understanding of its current meaning.

Generally, the interpretation of aggregate characters -- even the interpretation of pictographs or ideographs -- is not obvious. They must be memorized. Even if the meanings of the simpler pictographs and ideographs are memorized, the meaning of aggregate characters is not obvious. Turned around, one cannot decompose a Chinese character to understand its meaning. For example, a character may seem to contain the symbol for one (一) or two (二), but that does not mean the character refers to one or two. It is almost impossible to parse a character into components that can be added to create meaning. This is in stark contrast to English where one can determine (to a large part) the meaning of a word by simply understanding its component morphemes (which are dependent on its phonemes) (Xue,et al, 2003). This English decomposition process has no analogue in Chinese. One simply memorizes the meanings of all symbols.

There is a further complication. Some complex concepts are the combination of two (or more) simpler characters in very close proximity (e.g., 木 means “wood”, 林 is combined by two木 which means “woodlot”). Finally, different characters may be combined to make more complex concepts. Note that the previous symbol for woods is two copies of the symbol for wood written in very close proximity (closer than two distinct symbols). The combination of two characters to create a single concept is the most common, but a single concept may be represented by from one to four characters. This structure seems analogous to the notion of morphemic addition in English, but one cannot assume that the total meaning is anything like the sum of the two (or more) individual meanings: sometimes it is (e.g., 天真 means naive. 天 means sky, 真 means true.), sometimes it is not (e.g., 学 means study. 习means study. Ï̊ means practice) (XinHua dictionary, 2005).

For those who write only in English, sentence structure in Chinese is paradoxically simpler and yet more complicated. In Chinese, semantic information is far more important than syntactical information for comprehension (Liu., et al, 2010; Zhang.,et al, 2010). The context of a statement is essential to the proper interpretation of the statement. For example, 我去学校 (I go to school) may mean “I am going to school” or “I went to school” or “I will go to school” The reader must use context to resolve the ambiguity. In addition, as noted above, a single concept may be denoted by several adjacent characters. Chinese is a type of analytic language. The term “analytic language” refers to a language in which all the words have separate meanings from each other. In other words, each concept is structured by a graphic unit with its own meaning (a graphic unit may be several characters). To express different meanings in Chinese, one simply uses different combinations of characters. One does not modify the structure of a single character to alter its meanings. For example, 木 means “wood”, 树means “tree”, 林 means and lots of trees together, 森 has three distinct meanings as luxuriant growth of trees; obscure dark, and severe. Note that the same symbol may have different meaning based on context. Similarly, the SYMBOL in 树林 means “woodlot”, while the same SYMBOL in 森林 means “forest”.

Syntax in Chinese sentences is limited (Liu., et al, 2010). The structure of a sentence typically follows the usual subject-verb-object that is common in English, but it is not so constrained as in English. Additional words are inserted to qualify or change the meaning, but the insertion point is, generally, not as constrained as English. For example, time indicators are placed before or after the subjects. One can say “yesterday, he went to work” or “he, yesterday, went to work”. Verb modifiers (such as the location of activity) are placed before the verb. Prepositional phrases that modify the verb are placed between the verb and location. Objects are typically placed after the verb, but can be before the verb, or even before the subject (Xue., et al, 2003).

As noted, context is important for the interpretation of tense. Tense is not indicated by the verb. The standard verb is all tenses (past, present, future) simultaneously. One does not modify a verb to change tense as is done in English (e.g., “run”, “ran”). Instead, if it is necessary to be precise (if context is insufficient), one adds a (single) character modifier. To be precise about the future, one simple adds a single character as in 将( 我将去学校“I will go to school”). To be precise about the past, one uses 过 ( 我去过学校“I went to school”). To be precise about the present one adds two characters that sum to a single meaning as 现在 (我现在去学校“I am going to school now”). Note that the two characters 现在 as a single whole word means “right now”, but the one character 现 means “unveil”, “now”, and “ready-made” (depending on context) and the one character 在 can be a adverb or preposition (depends on situations use as different function) that and it means “exist”, “consist”, or simple emphasis (XinHua dictionary, 2005). Hence, in Chinese, the notion of word agreement does not exist, and the comprehension heavily depends on contexts. Similarly, in Chinese, there is no notion of singular vs plural in the construction of basic sentences. There is no issue of subject verb agreement (Liu., et al, 2010). In contrast to English, Chinese requires the reader to infer the proper meaning of a particular sentence from its context.

In Chinese, one cannot count on word order to define the meaning of a sentence or the role of the word in a sentence. The object might be first or last. This implies that the during reading, any prior context is going to have strong effect on the interpretation of a sentence -- much stronger than in English where syntax will establish the meaning sentence. Said another way, in English a sentence may “obviously” conflict with its prior context, and a word in a sentence could conflict with the prior sentence (e.g., a noun where a verb is required). As such, it would stand out as “obviously wrong” or “inconsistent”. This is less likely in Chinese because the structure is more fluid and because the interpretation of the current sentence requires context. That is, in Chinese, even if a word or sentence “makes no sense” it is the job of the reader to “make it make sense”. It also implies that a reader of Chinese must read the entire sentence before meaning can be established. In contrast, in English, one can build meaning word by word. These points are particular issues for sentences read in isolation (without context).

When considering spoken Chinese, there is another layer of complication. As noted previously, there are many different dialects of Chinese, all based on the same written form. The Chinese characters, unlike English phonemes, do not provide any clues to their pronunciation. Thus, the match between the character and the pronunciations is completely done through memory. Each dialect has its own memorized repertoire. On the other hand, most people in China must learn Mandarin – it is the language of the educational system (Xue., et al, 2003). In addition, one needs to note that there are many characters that have very different meanings, but the same sound (these are homophones in English) in Mandarin. For example, 植物 zhí wù (plant) and 职务 (position) sound the same despite very different meaning. Thus, in spoken Chinese, context is probably even more important than it is for written Chinese. In Chinese, it is particularly important for listeners to classify phrases with the same sounds but different meanings. For example: in 这里有一株植物 (There is a plant) 我的职务是老师(My position is a teacher), the bolded characters sound exactly same. Considerable effort in the educational system is devoted to the learning of such distinctions.

Spoken Chinese also has four different tones that help in the identification of the concept, It is important to pronounce words in a clear and correct tone. For example, 织物 zhī wù(wool), it has same basic sound as zhí wù 职务 and 植物, but has a slightly different tone. These are the most difficult part of spoken Chinese for children to learn.

Generally, in verbal and written Chinese, there is a one morpheme to one character to one sound correspondence. However, all of these correspondences are memorized. In addition, the written form does not provide any clues to pronunciation and the verbal form does not provide any clues to the writing. Thus, in verbal and written Chinese, people need to know the meanings of the words (职务 and 植物) and the context of the words to determine the meanings of sentences.

Unlike English, one cannot determine the spelling of a verbal word by “sounding it out” and one cannot determine the meaning of a written word by “sounding it out”. In truth, there is a group of characters called xíng shēng zì (形声字) which do actually give clues to pronunciation. For example, 肢 (limb/limbs), the pronunciation is “zhī”. The pronunciation of the right part “之” is “zhī”. However, the number of these characters is very limited. In addition, to be able to pronounce 肢Chinese speakers have to learn Ö§ first, which is a process of memorization.

A storty of 36 Chinese Characters

Event-Related Brain Potentials (ERPs) in English
To study syntax and semantic process in brain activities, Event related brain potentials (ERPs) are always used. Typically, electrodes are placed on the skull and then used to record the electrical potential -- over some time span -- at that point. This is called a waveform. The recording is usually tied to some specific behavioral event (the “event related” part of ERP), which is usually the presentation of a visual or auditory signal. In particular, ERPs record the timing of specific brain activities and “the summed electrical activity of apical dendrites of synchronously activated clusters of pyramidal neurons within the cortex” (Newman, Pancheva, Ozawa, Neville & Ullman, 2001). Different waveforms are identified by both location (at the skull) and timing (post stimulus). A number of waveforms have well established functional correlates. For example, the N400 is a negative waveform that peaks approximately 400ms after stimulus onset. The peak will be largest over center-parietal electrode sites (Newman, 2001).N400 is usually associated with semantic integration (Liu., et al, 2010). The P600 is a positive wave form that occurs 600ms after the stimulus onset, and is thought to be associated with syntactic violations, ambiguity and syntactic complexity. The peaks will be at the largest at 600 and 800ms over center-parietal electrode sites. (Newman., et al, 2001; Liu., et al 2010). It is often evoked when a stimulus needs to be retrospectively reinterpreted. Some waveforms are labeled by location rather than time. For example, the ELAN (early left anteriour negativity) and LAN (left anterior negativity) are negative waveforms that appear within 300 ms of stimulus onset and are associated with, respectively, word category identification and syntactical errors (Liu., et al, 2010). In the view of some (e.g., Friederici, Gunter, Hahne & Mauth, 2003) they are similar to the P600.

Research into English Semantic and Syntactic Processes
The research into the processing of English, as well as other Indo-European languages, is very mature in Western countries. In this type of study, psychologists usually design four conditions to compare similarities and differences between semantic and syntactic processes. These conditions are the correct condition (has correct sentences), the semantically incorrect condition (has sentences with semantic violations), the syntactically incorrect condition (has sentences with syntactic violations) and the combined incorrect condition (has sentences with both semantic and syntactic violations). Bornkessel and Schleswksy (2006) suggested a model, called the extended Argument Dependency Model (eADM), to explain how semantic and syntax processing happens in English and other Indo-European languages. They suggested that in sentence comprehension, there is a a primacy of syntax over semantics. Hence, ELAN is indexed first by syntactic process. The primacy of syntax is argued in two ways. First of all, in English and other Indo-European languages, syntactic information (e.g., word category and word agreement) provides very important clues for sentence comprehension. Secondly, semantic integration will be blocked if primary syntactic processing fails. This model was supported by the finding of a syntax-related ELAN-600 effect, and the absence of the semantic-related N400 effect.

Kuperberg (2007) suggested that there are at least two neural routes towards language comprehension. The first route is called semantic memory-based stream. In sentences, it is for analyzing semantic factors, and semantic relationships of words within content (e.g., word agreements). Moreover, these relationships are compared within prior lexical semantic memory (Kuperberg, 2007; Zhang., et al, 2010). This has suggested the N400 event-related potential is sensitive to semantic computations. The second route of the stream is named combinatorial, and it is contrasted semantic memory-based stream. This stream is sensitive to morpho-syntactic and semantic–thematic constraints. The P600 is representative of this processing. Generally, this model is supported by the finding that sentences with both syntax and semantic violations produce both a ELAN-600 and a N400 (Liu., et al, 2010). Kuperberg also suggested that morphological languages (e.g., Chinese) may have less syntactocentric and more dynamic processing (Kuperberg, 2007). In an auditory study of semantic and syntactic process, Friderici, Gunter, Hahne & Mauth (2003) noted that N400 was found in the semantic-violated condition. In the syntactically incorrect condition, the study showed that left lateralized fronto-temporal P600 (150-600 sm) and LAN were found. Both P600 and LANs are sensitive to syntactic processes.

Research into Chinese Semantic and Syntactic Processes
Since Chinese has a more limited grammatical morphology (i.e., minimal syntax), it is useful for the investigation of the ELAN and LAN as well as the P600. It provides a cross-language test of syntactical and semantic effects (Liu., et al, 2010). So far, there are very few studies on this topic, and all the studies have suggested that semantic processing in Chinese does not need license from syntax (Liu, et al., 2010; Ye., et al., 2006; Ye, Zhan & Zhou, 2007; Zhang, Yu & Boland, 2010). For example, Ye et al. (2007) designed a research study by using “ba” sentence structure. The “Bǎ” sentence structure is a SVO ordered sentence. “Ba” is a functional and particle verb, which leads the main verb (Ye., et al; Zhang., et al) and defines the actions that the verb has on objects (Xin Hua Zi Dian, 2005). For example, in 设计师做新衣， 把布裁了（ Shì jì shi-(tailor) zuò-(make) xīn yī-(new clothing), bǎ bù- (material) cái(-cut) le, the bǎ serves to define the action of cutting the cloth (Ye., et al, 2006). Ye et al designed ba sentences which were correct, semantically incorrect, syntactically incorrect and combined incorrect. The semantic violations appeared in around the 150-400 ms time window, which is prior to the classical N400 which occurs in the 300-500 ms time window. This observation suggested that N400 in auditory comprehension in Chinese is not quite the same as auditory comprehension in English. In the early window (150-250 ms), semantic and syntactic processes appeared to be independent. However, in the later window (250-400 ms), they started to interact. The broadly negative distribution appeared during N400 latency range, and it was observed in all the conditions. However, P600 was not observed.

Liu., et al (2010), designed another experiment to study semantics and syntactic processing by using idioms. Idioms have been very popular in Chinese daily conversation for thousands of years, and many idioms are grounded in historical events, as well as in cultural and traditional values. Although idioms have a long history and a constrained structure, they have a similar syntactic structure to modern Chinese. Idioms are usually consist of four characters. For example, the idiom, 叶公好龙 （yè gōng hào lóng） came from a story, “Mr. Ye likes dragons”. In the story, Mr. Ye liked dragons, and because he liked dragons, he put paintings of dragons everywhere in his place. However, when a real dragon came in a friendly visit to his place, Mr Ye was scared to death. This idiom is used to describe someone who claims that they like something, but are actually afraid of it. Using such idioms, Liu et al. designed a correct condition, a syntactic violation condition, a semantic violation condition and combined syntactic and semantic violation condition. Participants were presented with all four conditions visually. The results suggested that no ELAN and LAN was observed in combined violation condition. The N400 was broadly found in all there conditions, although the syntactic violation was found to produce a smaller N400 than the semantic violation. A left-hemisphere dominant P600 was found in all three violation conditions. Their results were consistent with Yu and Zhang (2008) in that an N400 was found in the syntactically incorrect condition, semantically incorrect condition and combined incorrect condition, but no ELAN was found.

In Experiment 1 of another study, Zhang, Yu and Boland (2010) also used “bá” sentences. In syntactically incorrect condition, they used the adverb “hěn” （很）, which corresponds to the word very in English. It is followed by a noun. In Chinese, “hěn” （很） can be bound with adjective or verb, but not with a noun. In the semantically incorrect condition, they used a wrong verb to replace the verb in the correct condition. They used both incorrect syntactic and semantic in the combined condition. Their results suggested that N400 was found in all three incorrect conditions, and that the combined condition had the bigger N400. Furthermore, the results suggested that in syntactic violations, a left-lateralized negativity was found in 300-500 ms time window, followed by a late posterior positivity in the 500-1,000 ms time window (P600). The P600 effect was not produced by verb replacement (Zhang., et al, 2010). In summary, in Chinese sentence comprehension, semantics processes do not need syntax in both verbal and visual presentation.

Summary
In summary, English and Chinese represent distinct classes of language and therefore language processing. Chinese is a morphological language whereas English is an alphabetic/phonetic language. Chinese is much more dependent on semantic integration than English and Chinese is much less dependent on syntax than English. The learning processes of the two languages are also different. The learning of Chinese requires a great deal of memorization, and the visual characters are memorized separately from the auditory symbols. In Chinese ERPs studies, N400 is largely found, the P600 might be found, and LEAN and LAN are not found. All are found in English. However, in contrast to English, there are only a few studies in Chinese. Thus, for future directions, psychologists can, for example, research garden path sentences and one would predict that a P600 may be observed.

Characters story of Chinese Zodiac Animals

Reference
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Friederici, A., Gunter, T., Hahne, A., & Mauth, K (2003). The relative timing of syntactic and semantic processes in sentence comprehension. COGNITIVE NEUROSCIENCE AND NEUROPSYCHOLOGY. Vol 15 No 1 19 January 2004.

Kuperberg., G (2007). Neural mechanisms of language comprehension: Challenges to syntax. Brain Reseach. 1146 (2007) 23-49.

Liu, Y., Li, P., Shu, H., Zhang. Q.,& Chen.,L (2010). Structure and meaning in Chinese: an ERP of idioms. Journal of Neurolinguistics. 23 (2010), 615-630.

Newman, A., Pancheva, R., Ozawa, K., Neville, H., & Ullman, M (2001). An Event-Related fMRI study of syntactic and semantic violations. Journal of Psychology Research. 30(3), 339-361.

Ye, Z., Zhan, W., & Zhou., X (2007). The semantic processing of syntactic structure in sentence comprension: an ERP study. Brain Reseach. 1142 (2010), 135-145

Ye, Z., Zhan, W., & Zhou., X (2005). Semantic and syntactic processing in Chinese sentence comprehension: Evidence from event-related potential. Brain Reseach.

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Learning Exercise
1) For English speakers learning Chinese as an adult, what are the hardest things for them to learn or understand?

2) For Chinese speakers learning English as an adult, what are the hardest things for them to learn or understand?

3) In Chinese, what is an idiom and what are the rules/guidelines that govern its use?
 * What the rules of idioms in Chinese?
 * For bonus points, which Star Trek episode approximates this linguistic system?

4) The movie “Hero” told a real story about the Zhan Guo dynasty (see Figure 1). The state Qin was very powerful and strong and wanted to rule all other existing six states. Because his state was not powerful enough to resist, the king of Yan asked Jinke (the hero in the movie) to kill the king of the Qin state. To maintain the peace, the king of Yan offered to give some cities to the king of Qin. However, it was a trick. The plan was to assassinate the king of Qin. The hero, Jinke, was supposed to deliver a scroll map to Qin. The map was supposed to identify the cities that would be surrendered. However, in the rolled up map there was a knife. As the map was unrolled, Jinke was supposed to grab the knife and kill the king of Qin. However, he failed the mission. In Chinese, the idiom 图穷匕见 is a reference to this history with 图 (map), 穷 (the end), 匕 (dagger) 见 (occur). Translated, it states "at the end of the map, the dagger occurs". It is interpreted to mean that "at the end of the event, the true event occurs". Given the following four scenerios, choice the one that would be the best application for this idiom?
 * A: Every four years, Americans vote for their president.
 * B: During the second World War, before the D-Day invasion at Normandy, the British dumped a dead body with fake information about the invasion to deceive Germany. The Gemans were never sure of the correct invasion site until it actually happened.
 * C. Criminals have been know to use high wage jobs to entice women from another country to come to Canada. Initially, the criminals are very nice to the women. However, once the women arrive in Canada, the criminals become very agressive and force the women are forced to work in prostitution and pornography.
 * D. Criminals have been know to use high wage jobs to entice women from another country to come to Canada. Initially, the criminals are very nice to the women. However, once the women arrive in Canada, the criminals become very agressive and force the women are forced to work in prostitution and pornography. Evemtually, the criminals are arrested, and the women regain their freedom.

5) In Chinese, what is a semantic conjunction for characters, and why is difficult for English speakers to learn.

6) Explain the “Ba” structure in Chinese, and contrast it with its functional equivalent in English.

7) Unlike English, Chinese writing does not give clear clues of pronunciation. Thus, those who can speak and write Chinese must remember the pronunciation and the written form of a morpheme separately. With reference to Figure 2, please indicate the most likely processing routes for a native Chinese person listening to Chinese and for a native Chinese person reading Chinese. Also with reference to Figure 2, please indicate the most likely processing routes for a native English person listening to English and for a native English person reading English. Discuss any differences.

Links
http://en.wikipedia.org/wiki/Chinese_character

http://en.wikipedia.org/wiki/Chinese_grammar


text:
User:Egm6341.s11.team4.KTH/HW2

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

Find
2-12-1) Construct 2-12-2) Plot

Solution
2-12-1) Plot

Matlab Code:

2-12-2) Plot Matlab Code:

Find
Give details to obtain

Solution
Let

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

title: User:Egm6341.s11.team4.KTH/HW2


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

= Problem 11 - Derivation of simple Simpson's rule = From (meeting 10 page 4)

Given
From the class note p.8-3

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

I(f_n)=\int_a^b \! f_n(x) \, \mathrm{d}x $$ $$ ,where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f_n(x)=P_n(x)=\sum_{i=0}^n l_{i,n}(x)f(x_i) $$ $$ Let $$\displaystyle n=2.$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_2(x)=\sum_{i=0}^{n=2} l_{i,2}(x)f(x_i) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-3)
 * }
 * $$\displaystyle
 * $$\displaystyle

\begin{align} x_0&=a \\ x_1&=\frac{a+b}{2} \\ x_2&=b \end{align} $$

Find
Derive simple Simpson's rule using Eq.(11-3).
 * Simple Simpson's rule:
 * $$\displaystyle

I_2=\frac{h}{3}[f_0+4f_1+f_2] $$ ,where $$\displaystyle h=\frac{b-a}{2} $$

Solution
First, Lagrange interpolation functions need to be computed using the definition shown below.
 * {| style="width:100%" border="0" align="left"

l_{i,n}(x)=\prod_{j=0}^n \frac{x-x_j}{x_i-x_j} $$ $$ So, the following steps can be done.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} l_{0,2}(x)&=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}=\frac{1}{(x_0-x_1)(x_0-x_2)}\left[x^2-(x_1+x_2)x+x_1 x_2\right]\\ l_{1,2}(x)&=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}=\frac{1}{(x_1-x_0)(x_1-x_2)}\left[x^2-(x_0+x_2)x+x_0 x_2\right]\\ l_{2,2}(x)&=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}=\frac{1}{(x_2-x_0)(x_2-x_1)}\left[x^2-(x_0+x_1)x+x_0 x_1\right] \end{align} $$ $$ Also, Eq (11-3) can be rewritten in the following form.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_2(x)=l_{0,2}f(x_0)+l_{1,2}f(x_1)+l_{2,2}f(x_2) $$ $$ Also, the integral equation (11-1) can be rewritten as follows.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} I(f_n)&=\int_{x_0}^{x_2} \! P_2(x) \, \mathrm{d}x \\ &=\int_{x_0}^{x_2} \! l_{0,2}f(x_0)+l_{1,2}f(x_1)+l_{2,2}f(x_2) \, \mathrm{d}x \\ &=f(x_0) \int_{x_0}^{x_2} \! l_{0,2} \, \mathrm{d}x +f(x_1) \int_{x_0}^{x_2} \! l_{1,2} \, \mathrm{d}x +f(x_2) \int_{x_0}^{x_2} \! l_{2,2} \, \mathrm{d}x \end{align} $$ $$ Each integral terms in Eq.11-7 can be obtained through the following steps.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-7)
 * }
 * }
 * $$\displaystyle

\begin{align} \int_{x_0}^{x_2} \! l_{0,2} \, \mathrm{d}x &=\frac{1}{(x_0-x_1)(x_0-x_2)} \int_{x_0}^{x_2} \! \left[x^2-(x_1+x_2)x+x_1 x_2\right] \, \mathrm{d}x\\ &=\frac{1}{(x_0-x_1)(x_0-x_2)} \left[\frac{1}{3}x^3-\frac{(x_1+x_2)}{2}x^2+x_1 x_2 x \right]_{x_0}^{x_2}\\ &=\frac{1}{(x_0-x_1)(x_0-x_2)} \left[\frac{1}{3}(x_2^3-x_0^3)-\frac{(x_1+x_2)}{2}(x_2^2-x_0^2)+x_1 x_2(x_2-x_0) \right] \\ &=\frac{(x_2-x_0)}{(x_0-x_1)(x_0-x_2)} \left[\frac{(x_2^2+x_2 x_0+x_0^2)}{3}-\frac{(x_1+x_2)(x_2+x_0)}{2}+x_1 x_2 \right] \\ & \mbox{at this point, we should take advantage of the fact that } x_1=\frac{x_2+x_0}{2} \\ &=\frac{(-1)}{(x_0-x_1)} \left[\frac{(x_2^2+x_2 x_0+x_0^2)}{3}-(x_1+x_2)x_1+x_1 x_2 \right] \\ &=\frac{(-1)}{(x_0-x_1)} \left[\frac{x_2^2+x_2 x_0+x_0^2-3x_1^2}{3} \right] \\ & \mbox{substitute } x_1=\frac{x_2+x_0}{2} \mbox{ into the above equation}\\ &=\frac{(-2)}{(x_0-x_2)} \left[\frac{4(x_2^2+x_2 x_0+x_0^2)-3(x_0^2+2x_0 x_2+x_2^2)}{12} \right] \\ &=\frac{(-2)}{(x_0-x_2)} \left[\frac{x_2^2-2 x_2 x_0+x_0^2}{12} \right] \\ &=\frac{(-2)}{(x_0-x_2)} \left[\frac{(x_2-x_0)^2}{12} \right] \\ &=\frac{(x_2-x_0)}{6} \end{align} $$
 * $$\displaystyle

\begin{align} \int_{x_0}^{x_2} \! l_{1,2} \, \mathrm{d}x &=\frac{1}{(x_1-x_0)(x_1-x_2)} \int_{x_0}^{x_2} \! \left[x^2-(x_0+x_2)x+x_0 x_2\right] \, \mathrm{d}x\\ &=\frac{1}{(x_1-x_0)(x_1-x_2)} \left[\frac{1}{3}x^3-\frac{(x_0+x_2)}{2}x^2+x_0 x_2 x \right]_{x_0}^{x_2}\\ &=\frac{1}{(x_1-x_0)(x_1-x_2)} \left[\frac{1}{3}(x_2^3-x_0^3)-\frac{(x_0+x_2)}{2}(x_2^2-x_0^2)+x_0 x_2(x_2-x_0) \right] \\ &=\frac{(x_2-x_0)}{(x_1-x_0)(x_1-x_2)} \left[\frac{(x_2^2+x_2 x_0+x_0^2)}{3}-\frac{(x_2+x_0)^2}{2}+x_0 x_2 \right] \\ &=\frac{(x_2-x_0)}{(x_1-x_0)(x_1-x_2)} \left[\frac{(2x_2^2+2x_2 x_0+2x_0^2)}{6}-\frac{3x_2^2+6x_2 x_0+3x_0^2}{6}+\frac{6x_0 x_2}{6} \right] \\ &=\frac{(x_2-x_0)}{(x_1-x_0)(x_1-x_2)} \left[\frac{-x_2^2+2x_2 x_0-x_0^2}{6} \right] \\ & \mbox{substitute } x_1=\frac{x_2+x_0}{2} \mbox{ into the above equation}\\ &=\frac{(x_2-x_0)}{(\frac{x_2-x_0}{2})(\frac{x_0-x_2}{2})} \left[\frac{-(x_2-x_0)^2}{6} \right] \\ &=\frac{4}{6}(x_2-x_0) \end{align} $$
 * $$\displaystyle

\begin{align} \int_{x_0}^{x_2} \! l_{2,2} \, \mathrm{d}x &=\frac{1}{(x_2-x_0)(x_2-x_1)} \int_{x_0}^{x_2} \! \left[x^2-(x_0+x_1)x+x_0 x_1\right] \, \mathrm{d}x\\ &=\frac{1}{(x_2-x_0)(x_2-x_1)} \left[\frac{1}{3}x^3-\frac{(x_0+x_1)}{2}x^2+x_0 x_1 x \right]_{x_0}^{x_2}\\ &=\frac{1}{(x_2-x_0)(x_2-x_1)} \left[\frac{1}{3}(x_2^3-x_0^3)-\frac{(x_0+x_1)}{2}(x_2^2-x_0^2)+x_0 x_1(x_2-x_0) \right] \\ &=\frac{(x_2-x_0)}{(x_2-x_0)(x_2-x_1)} \left[\frac{(x_2^2+x_2 x_0+x_0^2)}{3}-\frac{(x_0+x_1)(x_2+x_0)}{2}+x_0 x_1 \right] \\ & \mbox{at this point, we should take advantage of the fact that } x_1=\frac{x_2+x_0}{2} \\ &=\frac{1}{(x_2-x_1)} \left[\frac{(x_2^2+x_2 x_0+x_0^2)}{3}-(x_0+x_1)x_1+x_0 x_1 \right] \\ &=\frac{1}{(x_2-x_1)} \left[\frac{x_2^2+x_2 x_0+x_0^2-3x_1^2}{3} \right] \\ & \mbox{substitute } x_1=\frac{x_2+x_0}{2} \mbox{ into the above equation}\\ &=\frac{2}{(x_2-x_0)} \left[\frac{4(x_2^2+x_2 x_0+x_0^2)-3(x_0^2+2x_0 x_2+x_2^2)}{12} \right] \\ &=\frac{2}{(x_2-x_0)} \left[\frac{x_2^2-2 x_2 x_0+x_0^2}{12} \right] \\ &=\frac{2}{(x_2-x_0)} \left[\frac{(x_2-x_0)^2}{12} \right] \\ &=\frac{(x_2-x_0)}{6} \end{align} $$ The result from the integrations above shows the followings.
 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x_2} \! l_{0,2} \, \mathrm{d}x = \frac{(x_2-x_0)}{6} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x_2} \! l_{1,2} \, \mathrm{d}x =\frac{4}{6}(x_2-x_0) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x_2} \! l_{2,2} \, \mathrm{d}x = \frac{(x_2-x_0)}{6} $$ $$ We know that $$\displaystyle x_2=b$$, $$\displaystyle x_0=a$$. In addition, the term $$\displaystyle h $$ in the definition of simple Simpson's rule is defined as $$\displaystyle h=\frac{b-a}{2}$$. As a result, the following can be concluded.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11-10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{align} I(f_n)&=f(x_0) \int_{x_0}^{x_2} \! l_{0,2} \, \mathrm{d}x +f(x_1) \int_{x_0}^{x_2} \! l_{1,2} \, \mathrm{d}x +f(x_2) \int_{x_0}^{x_2} \! l_{2,2} \, \mathrm{d}x \\ &=\frac{(x_2-x_0)}{6}f(x_0)+\frac{4(x_2-x_0)}{6}f(x_1)+\frac{(x_2-x_0)}{6}f(x_2) \\ &=\frac{(b-a)}{6}f(x_0)+\frac{4(b-a)}{6}f(x_1)+\frac{(b-a)}{6}f(x_2) \\ &=\frac{h}{3}f(x_0)+\frac{4h}{3}f(x_1)+\frac{h}{3}f(x_2) \end{align} $$
 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |
 * }
 * }
 * }

= Problem 12 - Newton-Cotes method = From (meeting 10 page 4)

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

Find

 * 1) Construct $$\displaystyle f_n(x)$$ defined in Eq.12-1 for $$\displaystyle n=1,2,4,8,16$$.
 * 2) Plot $$\displaystyle f, f_{n}$$ for $$\displaystyle n=1,2,4,8,16$$.
 * 3) Compute $$\displaystyle I(f_n)=\int_a^b \! f_n(x) \, \mathrm{d}x$$,(n=1,2,4,8,16) and compare to $$\displaystyle I$$ from WA.
 * 4) For $$\displaystyle n=5$$, plot $$\displaystyle l_{0},l_{1},l_{2}$$. How will $$\displaystyle l_{3},l_{4},l_{5}$$ look?

Solution
1. Construct $$\displaystyle f_n(x) $$  CASE 1 :  $$\displaystyle n=1 $$
 * $$\displaystyle x_0=-1,\quad x_1=1 $$
 * $$\displaystyle

f_1(x) = P_1(x)=\sum_{i=0}^{1} l_{i,1}(x)f(x_i)=l_{0,1}(x)f(x_0)+l_{1,1}f(x_1) $$
 * , where
 * $$\displaystyle

\begin{align} l_{0,1}(x)&=\frac{(x-x_1)}{(x_0-x_1)}=\frac{x-1}{-2}, \quad f(x_0)=f(-1)=1-e^{-1} \\ l_{1,1}(x)&=\frac{(x-x_0)}{(x_1-x_0)}=\frac{x+1}{2}, \quad f(x_1)=f(1)=e-1 \end{align} $$  CASE 2 :  $$\displaystyle n=2 $$
 * $$\displaystyle x_0=-1,\quad x_1=0,\quad x_2=1 $$
 * $$\displaystyle

f_2(x) = P_2(x)=\sum_{i=0}^{2} l_{i,2}(x)f(x_i)=l_{0,2}(x)f(x_0)+l_{1,2}f(x_1)+l_{2,2}f(x_2) $$
 * , where
 * $$\displaystyle

\begin{align} l_{0,2}(x)&=\frac{(x-x_1)}{(x_0-x_1)}\frac{(x-x_2)}{(x_0-x_2)}=\frac{x(x-1)}{2}, \quad f(x_0)=f(-1)=1-e^{-1} \\ l_{1,2}(x)&=\frac{(x-x_0)}{(x_1-x_0)}\frac{(x-x_2)}{(x_1-x_2)}=\frac{(x+1)(x-1)}{-1}, \quad f(x_1)=f(0)=1 \\ l_{2,2}(x)&=\frac{(x-x_0)}{(x_2-x_0)}\frac{(x-x_1)}{(x_2-x_1)}=\frac{(x+1)x}{2}, \quad f(x_1)=f(1)=e+1 \end{align} $$  CASE 3 :  $$\displaystyle n=4 $$
 * $$\displaystyle

x_0=-1,\quad x_1=-\frac{1}{2}, \quad x_2=0, \quad x_3=\frac{1}{2}, \quad x_4=1 $$
 * $$\displaystyle

f_4(x)=P_4(x)=\sum_{i=0}^{4}l_{i,4}(x)f(x_{i})=l_{0,4}(x)f(x_{0})+l_{1,4}(x)f(x_{1})+l_{2,4}(x)f(x_{2})+l_{3,4}(x)f(x_{3})+l_{4,4}(x)f(x_{4}) $$
 * , where
 * $$\displaystyle

\begin{align} l_{0,4}&=\prod_{j=0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x+\frac{1}{2})(x-0)(x-\frac{1}{2})(x-1)}{(-1+\frac{1}{2})(-1-0)(-1-\frac{1}{2})(-1-1)}, \quad f(x_{0})=f(-1)=1-e^{-1} \\ l_{1,4}&=\prod_{j=0}^{4}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x+1)(x-0)(x-\frac{1}{2})(x-1)}{(-\frac{1}{2}+1)(-\frac{1}{2}-0)(-\frac{1}{2}-\frac{1}{2})(-\frac{1}{2}-1)}, \quad f(x_{1})=f(-\frac{1}{2})=\frac{e^{-\frac{1}{2}}-1}{-\frac{1}{2}} \\ l_{2,4}&=\prod_{j=0}^{4}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x+1)(x+\frac{1}{2})(x-\frac{1}{2})(x-1)}{(0+1)(0+\frac{1}{2})(0-\frac{1}{2})(0-1)}, \quad f(x_{2})=f(0)=1 \\ l_{3,4}&=\prod_{j=0}^{4}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x+1)(x+\frac{1}{2})(x-0)(x-1)}{(\frac{1}{2}+1)(\frac{1}{2}+\frac{1}{2})(\frac{1}{2}-0)(-\frac{1}{2}-1)}, \quad f(x_{3})=f(\frac{1}{2})=\frac{e^{\frac{1}{2}}-1}{\frac{1}{2}} \\ l_{4,4}&=\prod_{j=0}^{4}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x+1)(x+\frac{1}{2})(x-0)(x-\frac{1}{2})}{(-1+1)(-1+\frac{1}{2})(-1-0)(-1-\frac{1}{2})}, \quad f(x_{4})=f(1)=e^{1}-1 \\ \end{align} $$  CASE 4 :  $$\displaystyle n=8 $$
 * $$\displaystyle

x_0=-1,\quad x_1=-\frac{3}{4},\quad x_2=-\frac{1}{2},\quad x_3=-\frac{1}{4},\quad x_4=0,\quad x_5=\frac{1}{4},\quad x_6=\frac{1}{2},\quad x_7=\frac{3}{4}, \quad x_8=1 $$
 * $$\displaystyle

\begin{align} f_8(x)=P_8(x)=\sum_{i=0}^{8}l_{i,8}(x)f(x_{i})=&l_{0,8}(x)f(x_{0})+l_{1,8}(x)f(x_{1})+l_{2,8}(x)f(x_{2})+l_{3,8}(x)f(x_{3})+l_{4,8}(x)f(x_{4})+l_{5,8}(x)f(x_{5})+l_{6,8}(x)f(x_{6}) \\ &+l_{7,8}(x)f(x_{7})+l_{8,8}(x)f(x_{8}) \end{align} $$
 * , where
 * $$\displaystyle

\begin{align} l_{0,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{4})(x-\frac{1}{2})(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(-1+\frac{3}{4})(-1+\frac{1}{2})(-1+\frac{1}{4})(-1-0)(-1-\frac{1}{4})(-1-\frac{1}{2})(-1-\frac{3}{4})(-1-1)}, \quad f(x_{0})=f(-1)=1-e^{-1} \\ l_{1,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x+1)(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{4})(x-\frac{1}{2})(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(-\frac{3}{4}+1)(-\frac{3}{4}+\frac{1}{2})(-\frac{3}{4}+\frac{1}{4})(-\frac{3}{4}-0)(-\frac{3}{4}-\frac{1}{4})(-\frac{3}{4}-\frac{1}{2})(-\frac{3}{4}-\frac{3}{4})(-\frac{3}{4}-1)}, \quad f(x_{1})=f(-\frac{3}{4})==\frac{e^{-\frac{3}{4}}-1}{-\frac{3}{4}} \\ l_{2,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{4})(x-0)(x-\frac{1}{4})(x-\frac{1}{2})(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(-\frac{1}{2}+1)(-\frac{1}{2}+\frac{3}{4})(-\frac{1}{2}+\frac{1}{4})(-\frac{1}{2}-0)(-\frac{1}{2}-\frac{1}{4})(-\frac{1}{2}-\frac{1}{2})(-\frac{1}{2}-\frac{3}{4})(-\frac{1}{2}-1)}, \quad f(x_{2})=f(-\frac{1}{2})=\frac{e^{-\frac{1}{2}}-1}{-\frac{1}{2}} \\ l_{3,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x-0)(x-\frac{1}{4})(x-\frac{1}{2})(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(-\frac{1}{4}+1)(-\frac{1}{4}+\frac{3}{4})(-\frac{1}{4}+\frac{1}{2})(-\frac{1}{4}-0)(-\frac{1}{4}-\frac{1}{4})(-\frac{1}{4}-\frac{1}{2})(-\frac{1}{4}-\frac{3}{4})(-\frac{1}{4}-1)}, \quad f(x_{2})=f(-\frac{1}{4})=\frac{e^{-\frac{1}{4}}-1}{-\frac{1}{4}} \\ l_{4,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-\frac{1}{4})(x-\frac{1}{2})(x-\frac{3}{4})(x-1)}{(0+1)(0+\frac{3}{4})(0+\frac{1}{2})(0+\frac{1}{4})(0-\frac{1}{4})(0-\frac{1}{2})(0-\frac{3}{4})(0-1)}, \quad f(x_{4})=f(0)=1 \\ l_{5,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{5}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{2})(x-\frac{3}{4})(x-1)}{(\frac{1}{4}+1)(\frac{1}{4}+\frac{3}{4})(\frac{1}{4}+\frac{1}{2})(\frac{1}{4}+\frac{1}{4})(\frac{1}{4}-0)(\frac{1}{4}-\frac{1}{2})(\frac{1}{4}-\frac{3}{4})(\frac{1}{4}-1)}, \quad f(x_{5})=f(\frac{1}{4})=\frac{e^{\frac{1}{4}}-1}{\frac{1}{4}} \\ l_{6,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{6}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(\frac{1}{2}+1)(\frac{1}{2}+\frac{3}{4})(\frac{1}{2}+\frac{1}{2})(\frac{1}{2}+\frac{1}{4})(\frac{1}{2}-0)(\frac{1}{2}-\frac{1}{4})(\frac{1}{2}-\frac{3}{4})(\frac{1}{2}-1)}, \quad f(x_{6})=f(\frac{1}{2})=\frac{e^{\frac{1}{2}}-1}{\frac{1}{2}} \\ l_{7,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{7}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{2})(x-\frac{1}{4})(x-1)}{(\frac{3}{4}+1)(\frac{3}{4}+\frac{3}{4})(\frac{3}{4}+\frac{1}{2})(\frac{3}{4}+\frac{1}{4})(\frac{3}{4}-0)(\frac{3}{4}-\frac{1}{4})(\frac{3}{4}-\frac{1}{2})(\frac{3}{4}-1)}, \quad f(x_{7})=f(\frac{3}{4})=\frac{e^{\frac{3}{4}}-1}{\frac{3}{4}} \\ l_{8,8}&=\prod_{j=0}^{8}\frac{x-x_{j}}{x_{8}-x_{j}}=\frac{(x+1)(x+\frac{3}{4})(x+\frac{1}{2})(x+\frac{1}{4})(x-0)(x-\frac{1}{2})(x-\frac{1}{4})(x-\frac{3}{4})}{(1+1)(1+\frac{3}{4})(1+\frac{1}{2})(1+\frac{1}{4})(1-0)(1-\frac{1}{4})(1-\frac{1}{2})(1-\frac{3}{4})}, \quad f(x_{8})=f(1)=\frac{e-1}{1} \\ \end{align} $$  CASE 5 :  $$\displaystyle n=16 $$
 * $$\displaystyle

\begin{align} x_0&=-1,\quad x_1=-\frac{7}{8},\quad x_2=-\frac{3}{4},\quad x_3=-\frac{5}{8},\quad x_4=-\frac{1}{2},\quad x_5=-\frac{3}{8},\quad x_6=-\frac{1}{4},\quad x_7=-\frac{1}{8}, \quad x_8=0 \\ x_9&=\frac{1}{8},\quad x_{10}=\frac{1}{4},\quad x_{11}=\frac{3}{8},\quad x_{12}=\frac{1}{2},\quad x_{13}=\frac{5}{8},\quad x_{14}=\frac{3}{4},\quad x_{15}=\frac{7}{8},\quad x_{16}=1 \end{align} $$
 * $$\displaystyle

\begin{align} f_{16}(x)=P_{16}(x)=&\sum_{i=0}^{16}l_{i,16}(x)f(x_{i})=l_{0,16}(x)f(x_{0})+l_{1,16}(x)f(x_{1})+l_{2,16}(x)f(x_{2})+l_{3,16}(x)f(x_{3})+l_{4,16}(x)f(x_{4})+l_{5,16}(x)f(x_{5}) \\ &+l_{6,16}(x)f(x_{6})+l_{7,16}(x)f(x_{7})+l_{8,16}(x)f(x_{8})+l_{9,16}(x)f(x_{9})+l_{10,16}(x)f(x_{10})+l_{11,16}(x)f(x_{11})+l_{12,16}(x)f(x_{12}) \\ &+l_{13,16}(x)f(x_{13})+l_{14,16}(x)f(x_{14})+l_{15,16}(x)f(x_{15})+l_{16,16}(x)f(x_{16}) \end{align} $$
 * , where
 * $$\displaystyle

\begin{align} l_{0,16}&=\prod_{j=0}^{16}\frac{x-x_{j}}{x_{0}-x_{j}}, \quad f(x_{0})=f(-1)=\frac{e^{-1}-1}{-1} \\ l_{1,16}&=\prod_{j=0}^{16}\frac{x-x_{j}}{x_{1}-x_{j}}, \quad f(x_{1})=f(-\frac{7}{8})=\frac{e^{-\frac{7}{8}}-1}{-\frac{7}{8}}\\ & \cdots \\ l_{16,16}&=\prod_{j=0}^{16}\frac{x-x_{j}}{x_{16}-x_{j}}, \quad f(x_{16})=f(1)=\frac{e-1}{1} \end{align} $$ 2. Plot $$\displaystyle f_n(x), f(x)$$ for n=1,2,4,8,16. Plot: Matlab code:
 * {| style="width:100%" border="0" align="left"

3. Compute $$\displaystyle I(f_n)=\int_a^b \! f_n(x) \, \mathrm{d}x$$,(n=1,2,4,8,16) and compare to $$\displaystyle I$$
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

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

4 For $$\displaystyle n=5$$, plot $$\displaystyle l_{0},l_{1},l_{2}$$. How will $$\displaystyle l_{3},l_{4},l_{5}$$ look? Plot: Matlab code:
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

= Problem 14 - Transformation of variable in integration = From (meeting 11 page 2)
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Given

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

I(f(x))=\int_a^b \! f(x) \, \mathrm{d}x = \int_{-1}^{+1} \! \overline{f}(y) \, \mathrm{d}y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (14-1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

f(x)=f(x(y))=\overline{f}(y) $$, $$\displaystyle x = x(y)$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (14-2)
 * }
 * }

Find
Give details to obtain Eq.14-1 using the given information.

Solution
First of all, set up the followings.
 * $$\displaystyle x(-1)=a$$, $$\displaystyle x(+1)=b$$

With the setup above, the following can be concluded.
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{align} \int_{-1}^{+1} \! \overline{f}(y) \, \mathrm{d}y &= \int_{x(-1)}^{x(+1)} \! f(x(y)) \, \mathrm{d}x \\ &= \int_{a}^{b} \! f(x) \, \mathrm{d}x \\ &= I(f(x)) \end{align} $$
 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" align="center" |
 * }
 * }
 * }

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


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User:Callicoma

Marghanita da Cruz

http://ramin.com.au/marg/resume.html

title: User:Callicoma


text:
Template:IRC channel

#

title: Template:IRC channel


text:
File:HW2 P12 A.png

= Summary=

title: File:HW2 P12 A.png


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File:HW2 P12 B.png

= Summary=

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File:HW2 P12 C.png

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User:Egm6341.s11.team2.zou/HW3

=Problem 3.1 Compare Taylor series approximation with Lagrangian Interpolation=

Problem Statement
Use two methods to approximate the function $$\displaystyle f(x)=\sin x$$,

1) Taylor series expaned with respect to $$\displaystyle \hat{x}=\frac{3\pi }{8}$$ to the order $$\displaystyle n$$,

2) Lagrangian interpolation to the order 4

Compare the error by two methods at $$\displaystyle x=\frac{7\pi }{8}$$, find $$\displaystyle n$$ such that the error of Taylor series expansion is less than 4th order Lagrangian interpolation. And then give the maximum value of Lagrangian interpolation error.

Solution
Taylor series expansion

The Taylor expansion to $$\displaystyle f(x)=\sin x$$ is

Lagrangian interpolation

Use equidistance discretion, the 4th order Lagrangian interpolation requires 5 nodes: While the Lagrangian Interpolation evaluated at $$\displaystyle x=\frac{7\pi }{8}$$ is,

where,

Plug them back into the Eq 1.4 we have,

Compare the two methods

The exact value of $$\displaystyle f(\frac{7\pi }{8})$$ is,

Then we have the error of Lagrangian interpolation,

To find the order n which makes the Taylor series produce small error, we need to try different order.

From the calculation above, the 6th order of Taylor expansion can give better result than Lagrangian interpolation.

The maximum value of Lagrangian interpolation error

By Lagrangian interpolation error theorem,

Thus the maximum error is contained by $$\displaystyle \left| {{q}_{5}}(t) \right|$$. If $$\displaystyle t$$ is confined within $$\displaystyle (0,\pi )$$, the maximal error is given by

=Problem 3.8: Deriving the Error in Simple Simpson's=

Given: The Convergence of Numerical Integration
The convergence of numerical integration when applied to the error associated with the Simple Simpson's formula as seen in [[media:nm1.s11.mtg17.djvu|Lecture 17]] is given by

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

$$ \displaystyle $$    (8.1)
 * style="width:98%" |
 * style="width:98%" |
 * E_n| \leq \frac{M_{n+1}}{(n+1)!} \int_a^b{|q_{n+1}(x)|dx}
 *
 * }

Where $$ \displaystyle M_{n+1}(x):= $$ be defined as,

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

$$ \displaystyle M_{n+1}(x):=\underset{a\le x\le b}{\mathop{\max }}\,f^{(n+1)}(x)$$ (8.2)
 * style="width:98%" |
 * style="width:98%" |
 *
 * }

Derive: The Error in the Simple Simpson's Rule
Derive the error $$ \displaystyle |E_2| $$ for the simple Simpson's rule such that $$ \displaystyle n=2 $$ and where

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

$$\displaystyle n=2 \quad \implies \quad q_3(x)=(x-x_0)(x-x_1)(x-x_2)=(x-a)(x-\frac{a+b}{2})(x-b) $$     (8.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Then go on to show that

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

$$ \displaystyle (8.4)
 * style="width:98%" |
 * style="width:98%" |
 * E_2| \leq \frac{M_{3}}{3!} \int_a^b{|q_{3}(x)|dx} = \frac{192}{{M}_{3}}$$
 *
 * }

Solution
To find the bounded error for $$\displaystyle \left| {{E}_{2}} \right|$$ one can simply expand the polynominal in the integrand of (8.4) and integrate, as shown here.

Since function $$\displaystyle g(x)=\left| (x-a)(x-\frac{a+b}{2})(x-b) \right|$$ is symmetric with respect to $$\displaystyle x=\frac{a+b}{2}$$(from the fact $$\displaystyle g(\frac{a+b}{2}+t)=g(\frac{a+b}{2}-t)$$), then we use this property to eliminate the absolute value sign.

Then we have ,

The length of the panel is $$\displaystyle h=\frac{b-a}{2}$$, so we can write the solution like,

title: User:Egm6341.s11.team2.zou/HW3


text: TESOL/Emotions

title: TESOL/Emotions


text:
User talk:Anz2~enwikiversity

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User talk:SpikeToronto

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title: User talk:SpikeToronto


text: File:Hw2.12-6 Xia.jpg

title: File:Hw2.12-6 Xia.jpg


text:
University of Canberra/RCC2010/Participants

 * 1) Leighblackall 03:04, 16 July 2010 (UTC)
 * 2) Robinmcconnell 07:48, 16 July 2010 (UTC) UCNISS
 * 3) James Neill, Lecturer, Centre for Applied Psychology, University of Canberra - until 2.30pm
 * 4) Margaret O'Connell - CIT
 * 5) Kerrie Doyle, Lecturer, Nursing & Midwifery, Faculty of Health, University of Canberra
 * 6) Margaret Robson - CIT
 * 7) Robert Fitzgerald - Dean - Education (Research), Faculty of Education, University of Canberra
 * 8) Michael de Percy - Broadbanding the Nation
 * 9) Dr Bob Jansen - Cancer Council of Australia I am a private consultant to the Cancer Council and am interested in using wiki technology to foster collaboration between people at a distance
 * 10) Paul Harvey - Taxonomy Research & Information Network informatics
 * 11) Alice Winter-Irving - Cancer Council Australia
 * 12) Christine Vuletich - Cancer Council Australia
 * 13) Marina Lobastov - ANU
 * Graeme Bartlett a Wikipedia editor
 * 1) Joelle Vandermensbrugghe, Research Services Office, UC
 * 2) Marghanita da Cruz, Ramin Communications Pty Ltd, Local Community Engagement 2.0 (will be arriving in the afternoon)
 * 3) Kirsty Sharp, Tasmanian Polytechnic Wiki Projects
 * 4) Andrew Garrett, Wikimedia Foundation
 * 5) Tony Krone - University of Canberra
 * 6) Mark Spain - UC Organisation Development
 * 7) Peter Shanks - Skills Tasmania
 * 8) Ben Rattray UCNISS
 * 9) Rose Grozdanic - Australian Flexible Learning Framework
 * 10) Nerida Hart - Australian River Restoration Centre
 * 11) Sue Hollands - ACODE, TLC, University of Canberra
 * 12) Lauren Kane- Educational Developer, Australian National University
 * 13) Lulu Turner, Business &Government/HDR, University of Canberra
 * 14) Leonard Low - University of Canberra, Teaching & Learning Centre
 * 15) Lubna Alam - UC
 * 16) Alan Arnold - UC Teaching & Learning Centre


text:
BCP/Eupomatia laurina

In Bloom Annandale, 11am, 6pm 23 November 2010, about 13.15 24 November 2010 and 15.30 4 December 2010 http://ramin.com.au/travel/eupomatia-laurina-annandale-23nov2010.shtml

title: BCP/Eupomatia laurina


text:
BCP/Callicoma serratifolia

In Bloom in Annandale in October http://ramin.com.au/annandale/eco-annandale-2010-october-walk.shtml

title: BCP/Callicoma serratifolia


text: File:Hw2.12-7 Xia.jpg

title: File:Hw2.12-7 Xia.jpg


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User:Eml5526.s11.team6.tupsakhare/hwk2

=Problem 2.5=

Given
We have Eq. (2) from page 7-2 of mtg 7 ,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

where  i =  1, 2 ,..., n

Find
Show that $$ \underline{w} .\underline{P}(\underline{v})=0 $$ ,

where $$  \underline{w}= \sum_{i=1}^{n}\alpha_{i} \underline{b} _{i}$$ ,

and $$ \forall (\alpha _{1},\alpha _{2},...,\alpha _{n})\in R^{n} $$

Solution
We have,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

With i = 1 to n, we can then write -

$$ \underline{b}_{1}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . ..

. . . . ..

. . . . ..

. . . . ..

. . . . ..

$$ \underline{b} _{n}.\underline{P}(\underline{v})=0 $$

Now consider a collection of arbitrary numbers $$  (  \alpha _{1}, \alpha _{2} ,..., \alpha _{n}  ) $$ i.e.   \alpha _{i}      with i = 1 to n

Now successively multiplying above n equations, on both sides by corresponding  \alpha _{i}, we get the following new set of n equations -

With i = 1 to n, we can then write -

$$ \alpha _{1}. \underline{b}_{1} .\underline{P}(\underline{v})=0 $$

$$ \alpha _{2}. \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \alpha _{3}. \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . ..

. . . . ..

. . . . ..

. . . . ..

. . . . ..

$$ \alpha _{n}. \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

Adding LHS and RHS respectively for above n equations, we get the following -

$$ \alpha _{1} \underline{b}_{1}.\underline{P}(\underline{v})+ $$ $$ \alpha _{2} \underline{b}_{2}.\underline{P}(\underline{v})+...+ $$ $$ \alpha _{n} \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

which in index notation can be expressed as the following:

$$ \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i} .\underline{P}(\underline{v})=0 $$

But as per previous declaration in 'Find' tab, we have stated -

$$ \underline{w} = \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i}$$ ,

Hence we can now say that,

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

This shows that required equation (2) of page 7- 2 mtg 7 can be used to get equation (2) of page 8-2 mtg 7

= Problem 2.6 Linearly Independent Basis Function.=

Find
1) Construct Γ5x5(F) and observe the proportionality of Γ.

2) Find det Γ(F)

3) Conclude whether F is orthogonal basis i.e. $$\Gamma _{ij}=\left \langle b_{i},b_{j} \right \rangle=\delta _{ij}$$

Given
$$F=\left \{ 1,cos(\omega x),cos(2\omega x),sin(\omega x),sin(2\omega x) \right \} , \Omega =[0,T], i=1,2$$

Solution
We have from reference of Mtg 10_Page 2

1) $$ \mathbf{\Gamma} _{5x5}=\int_{\Omega }{}b_{i}\cdot b_{j}\;dx=\int_{0}^{T}b_{i}\cdot b_{j}\;dx $$

Constructing a matrix of bi ˑ bj yields:

$$ \Rightarrow \begin{bmatrix} 1\cdot 1 & cos(\omega x)\cdot 1 & cos(2\omega x)\cdot 1 & sin(\omega x)\cdot 1 & sin(2\omega x)\cdot 1\\ 1\cdot cos(\omega x) & cos(\omega x)\cdot cos(\omega x) & cos(2\omega x)\cdot cos(\omega x) & sin(\omega x)\cdot cos(\omega x) & sin(2\omega x)\cdot cos(\omega x)\\ 1\cdot cos(2\omega x) & cos(\omega x)\cdot cos(2\omega x) & cos(2\omega x)\cdot cos(2\omega x) & sin(\omega x)\cdot cos(2\omega x) & sin(2\omega x)\cdot cos(2\omega x)\\ 1\cdot sin(\omega x) & cos(\omega x)\cdot sin(\omega x) & cos(2\omega x)\cdot sin(\omega x) & sin(\omega x)\cdot sin(\omega x) & sin(2\omega x)\cdot sin(\omega x)\\ 1\cdot sin(2\omega x) & cos(\omega x)\cdot sin(2\omega x) & cos(2\omega x)\cdot sin(2\omega x) & sin(\omega x)\cdot sin(2\omega x) & sin(2\omega x)\cdot sin(2\omega x) \end{bmatrix} $$

Multiplying out the above matrix we integrate the resulting terms within limits from 0 to T to get the $$ \mathbf{\Gamma} _{5x5} $$ :

$$\Rightarrow \begin{bmatrix} x & sin(\omega x) & sin(\omega x)cos(\omega x) & -cos(\omega x) & -\frac{1}{2}cos(2\omega x)\\ sin(\omega x) & \frac{1}{2}(x+sin(\omega x)cos(\omega x)) & \frac{1}{6}(3sin(\omega x)+sin(3\omega x)) & -\frac{1}{2}cos^2(\omega x) & -\frac{2}{3}cos^3(\omega x)\\ sin(\omega x)cos(\omega x) & \frac{1}{6}(3sin(\omega x)+sin(3\omega x)) & \frac{1}{8}(4x+sin(4\omega x)) & \frac{1}{6}(3cos(\omega x)-cos(3\omega x)) & -\frac{1}{8}cos(4\omega x))\\ -cos(\omega x)) & -\frac{1}{2}cos^2(\omega x) & \frac{1}{6}(3cos(\omega x)-cos(3\omega x)) & \frac{1}{2}(x-sin(\omega x)cos(\omega x)) & \frac{2}{3}sin^3(\omega x)\\ -\frac{1}{2}cos(2\omega x) & -\frac{2}{3}cos^3(\omega x) & -\frac{1}{8}cos(4\omega x)) & \frac{2}{3}sin^3(\omega x) & \frac{1}{8}(4x-sin(4\omega x)) \end{bmatrix} $$

By definition of Angular Frequency

$$2\pi =\omega T\rightarrow \omega=\frac{2\pi}{T}$$

Above is the resulting matrix to be integrated within limits of 0 to T.

And now integrating the results using a table of integrals yielded the following terms for the final integrated matrix.

$$\begin{array}{l} \int_0^T {1 \cdot 1dx = x} |_0^T = T \\ \int_0^T {1 \cdot \cos (\omega x)dx = \frac{1}{\omega }\sin (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \cos (2\omega x)dx = \frac{1}\sin (2\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (\omega x)dx = - \frac{1}{\omega }\cos (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (2\omega x)dx = - \frac{1}\cos (2\omega x)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \cos (\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \int_0^T {\cos (\omega x) \cdot \cos (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \cos (2\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\cos (2\omega x) \cdot \sin (\omega x)dx = \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\sin (\omega x) \cdot \sin (\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\sin (\omega x) \cdot \sin (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\sin (2\omega x) \cdot \sin (2\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \end{array}$$
 * }

Above integrals results were verified with Wolframalpha.

The $$ \mathbf{\Gamma} _{5x5} $$ matrix is thus :

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

 * }
 * }

We note that this Gram matrix is a diagonal matrix.

2 ) Determinant of the above gram matrix $$ \mathbf{\Gamma} _{5x5} $$   :

Det $$ \mathbf{\Gamma} _{5x5} $$ = $$ \frac{T^5}{16} $$ < Shawn, please put the matlab figure here for calculation of determinant >

We note that the determinant is non- zero and hence we can say that the given functions form a family of Linearly Independent Basis Functions

3) We see that the $$ \mathbf{\Gamma} _{5x5} $$ derived is not equivalent to The Kronecker Delta.

Hence, we can conclude that $$ \mathbf{\Gamma} _{5x5} $$ is  NOT an ORTHOGONAL Basis.

title: User:Eml5526.s11.team6.tupsakhare/hwk2


text:
Pauli matrices

 * The following is modified from Pauli matrices.

In physics, the Pauli matrices are a set of 2 × 2 complex Hermitian and unitary matrices. Usually indicated by the Greek letter "sigma" (σ), they are occasionally denoted with a "tau" (τ) when used in connection with isospin symmetries. They are:

\sigma_1 = \sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} $$

\sigma_2 = \sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix} $$

\sigma_3 = \sigma_z = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} $$

The name refers to Wolfgang Pauli.

The real (hence also, complex) subalgebra generated by the σi (that is, the set of real or complex linear combinations of all the elements which can be built up as products of Pauli matrices) is the full set M2(C) of complex 2×2 matrices. The σi can also be seen as generating the real Clifford algebra of the real quadratic form with signature (3,0): this shows that this Clifford algebra Cℓ3,0(R) is isomorphic to M2(C), with the Pauli matrices providing an explicit isomorphism. (In particular, the Pauli matrices define a faithful representation of the real Clifford algebra Cℓ3,0(R) on the complex vector space C2 of dimension 2.)

Algebraic properties

\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = -i\sigma_1 \sigma_2 \sigma_3 = \begin{pmatrix} 1&0\\0&1\end{pmatrix} = I$$ where I is the identity matrix, i.e. the matrices are involutory.

 * The determinants and traces of the Pauli matrices are:

 * $$\begin{matrix}

\det (\sigma_i) &=& -1 & \\[1ex] \operatorname{Tr} (\sigma_i) &=& 0 & \quad \hbox{for}\ i = 1, 2, 3. \end{matrix}$$

From above we can deduce that the eigenvalues of each σi are ±1.

 * Together with the identity matrix I (which is sometimes written as σ0), the Pauli matrices form an orthogonal basis, in the sense of Hilbert-Schmidt, for the real Hilbert space of 2 × 2 complex Hermitian matrices, or the complex Hilbert space of all 2 × 2 matrices.

Pauli vector
The Pauli vector is defined by
 * $$\vec{\sigma} = \sigma_1 \hat{x} + \sigma_2 \hat{y} + \sigma_3 \hat{z} \,$$

and provides a mapping mechanism from a vector basis to a Pauli matrix basis as follows

\begin{align} \vec{a} \cdot \vec{\sigma} &= (a_i \hat{x}_i) \cdot (\sigma_j \hat{x}_j ) \\ &= a_i \sigma_j \hat{x}_i \cdot \hat{x}_j \\ &= a_i \sigma_i \end{align} $$

(summation over indices implied). Note that in this vector dotted with Pauli vector operation the Pauli matrices are treated in a scalar like fashion, commuting with the vector basis elements.

Commutation relations
The Pauli matrices obey the following commutation and anticommutation relations:

 * $$\begin{matrix}

[\sigma_a, \sigma_b]    &=& 2 i \varepsilon_{a b c}\,\sigma_c \\[1ex] \{\sigma_a, \sigma_b\} &=& 2 \delta_{a b} \cdot I\\ \end{matrix}$$

where $$\varepsilon_{abc}$$ is the Levi-Civita symbol, $$\delta_{ab}$$ is the Kronecker delta, and I is the identity matrix.

The above two relations are equivalent to:

 * $$\sigma_a \sigma_b = \delta_{ab} \cdot I + i \sum_c \varepsilon_{abc} \sigma_c \,$$.

For example,
 * $$\begin{matrix}

\sigma_1\sigma_2 &=& i\sigma_3,\\ \sigma_2\sigma_3 &=& i\sigma_1,\\ \sigma_2\sigma_1 &=& -i\sigma_3,\\ \sigma_1\sigma_1 &=& I.\\ \end{matrix}$$

and the summary equation for the commutation relations can be used to prove
 * $$(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) \, I + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} ) \quad \quad \quad \quad (1) \,$$
 * (as long as the vectors a and b commute with the pauli matrices)

as well as
 * $$e^{i (\vec{a} \cdot \vec{\sigma})} = \cos{a} + i (\hat{n} \cdot \vec{\sigma}) \sin{a} \quad \quad \quad \quad \quad \quad (2) \,$$

for $$\vec{a} = a \hat{n} $$.

Proof of (1)

 * $$(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) \,$$
 * $$ = a_i \sigma_i b_j \sigma_j \,$$
 * $$ = a_i b_j \sigma_i \sigma_j \,$$
 * $$ = a_i b_j \left( \delta_{ij} \cdot I+ i \varepsilon_{ijk} \sigma_k \right) \,$$
 * $$ = a_i b_j \delta_{ij} \cdot I+ i \sigma_k \varepsilon_{ijk} a_i b_j \,$$
 * $$ = ( \vec{a} \cdot \vec{b} ) \cdot I+ i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = a_i b_j \left( \delta_{ij} \cdot I+ i \varepsilon_{ijk} \sigma_k \right) \,$$
 * $$ = a_i b_j \delta_{ij} \cdot I+ i \sigma_k \varepsilon_{ijk} a_i b_j \,$$
 * $$ = ( \vec{a} \cdot \vec{b} ) \cdot I+ i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = ( \vec{a} \cdot \vec{b} ) \cdot I+ i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = ( \vec{a} \cdot \vec{b} ) \cdot I+ i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }

Proof of (2) First notice that for even powers
 * $$(\hat{n} \cdot \vec{\sigma})^{2n} = I \,$$

but for odd powers
 * $$(\hat{n} \cdot \vec{\sigma})^{2n+1} = \hat{n} \cdot \vec{\sigma} \,$$

Combine these two facts with the knowledge of the relation of the exponential to sine and cosine:

Which, when we use $$x = a (\hat{n} \cdot \vec{\sigma}) \,$$ gives us
 * $$e^{ix} \,$$
 * $$= \sum_{n=0}^\infty{\frac{i^n x^n}{n!}} \,$$
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n x^{2n}}{(2n)!}} + i\sum_{n=0}^\infty{\frac{(-1)^n x^{2n+1}}{(2n+1)!}} \,$$
 * }
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n x^{2n}}{(2n)!}} + i\sum_{n=0}^\infty{\frac{(-1)^n x^{2n+1}}{(2n+1)!}} \,$$
 * }
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n (a\hat{n}\cdot \vec{\sigma})^{2n}}{(2n)!}} + i\sum_{n=0}^\infty{\frac{(-1)^n (a\hat{n}\cdot \vec{\sigma})^{2n+1}}{(2n+1)!}} \,$$
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n a^{2n}}{(2n)!}} + i (\hat{n}\cdot \vec{\sigma}) \sum_{n=0}^\infty{\frac{(-1)^n a^{2n+1}}{(2n+1)!}} \,$$

The sum on the left is cosine, and the sum on the right is sine so finally,
 * $$e^{i a(\hat{n} \cdot \vec{\sigma})} = \cos{a} + i (\hat{n} \cdot \vec{\sigma}) \sin{a} \,$$

Completeness relation
An alternative notation that is commonly used for the Pauli matrices is to write the vector index $$i$$ in the superscript, and the matrix indices as subscripts, so that the element in row $$\alpha$$ and column $$\beta$$ of the $$i$$th Pauli matrix is $$\sigma^i_{\alpha\beta}$$.

In this notation, the completeness relation for the Pauli matrices can be written
 * $$\vec{\sigma}\cdot\vec{\sigma}=\sum_i \sigma^i_{\alpha\beta}\sigma^i_{\gamma\delta} = 2 \delta_{\alpha\delta} \delta_{\beta\gamma} - \delta_{\alpha\beta}\delta_{\gamma\delta}.\,$$

Proof The fact that the Pauli matrices, along with the identity matrix $$I$$, form an orthogonal basis for the complex Hilbert space of all 2 × 2 matrices mean that we can express any matrix $$M$$ as
 * $$M = c \mathbf{I} + \sum_i a_i \sigma^i$$

where $$c$$ is a complex number, and $$a$$ is a 3-component complex vector. It is straightforward to show, using the properties listed above, that
 * $$\mathrm{tr}\, \sigma^i\sigma^j = 2\delta_{ij}$$

where $$\mathrm{tr}$$ denotes the trace, and hence that $$c=\frac{1}{2}\mathrm{tr}\,M$$ and $$a_i = \frac{1}{2}\mathrm{tr}\,\sigma^i M$$. This therefore gives
 * $$2M = I \mathrm{tr}\, M + \sum_i \sigma^i \mathrm{tr}\, \sigma^i M$$

which can be rewritten in terms of matrix indices as
 * $$2M_{\alpha\beta} = \delta_{\alpha\beta} M_{\gamma\gamma} + \sum_i \sigma^i_{\alpha\beta} \sigma^i_{\gamma\delta} M_{\delta\gamma}$$

where summation is implied over the repeated indices $$\gamma$$ and $$\delta$$. Since this is true for any choice of the matrix $$M$$, the completeness relation follows as stated above.

Relation with the permutation operator
Let $$P_{ij} $$ be the permutation (transposition, actually) between two spins $$ \sigma_i $$ and $$ \sigma_j $$ living in the tensor product space $$ \mathbb{C}^2 \otimes \mathbb{C}^2 $$, $$ P_{ij}|\sigma_i \sigma_j\rangle = |\sigma_j \sigma_i\rangle $$. This operator can be written as $$ P_{ij} = \frac{1}{2}(\vec{\sigma}_i\cdot\vec{\sigma}_j+1)$$, as the reader can easily verify.

SU(2)
The matrix group SU(2) is a Lie group, and its Lie algebra is the set of the anti-Hermitian 2×2 matrices with trace 0. Direct calculation shows that the Lie algebra su(2) is the 3 dimensional real algebra spanned by the set {$$i \sigma_j$$}. In symbols,

 * $$\; \operatorname{su}(2) = \operatorname{span} \{ i \sigma_1, i \sigma_2, i \sigma_3 \}.$$

As a result, $$i \sigma_j$$s can be seen as infinitesimal generators of SU(2).

A Cartan decomposition of SU(2)
This relationship between the Pauli matrices and SU(2) can be explored further, as can be seen from the following simple example. We can write

 * $$\; \operatorname{su}(2) = \operatorname{span} \{i \sigma_2\} \oplus \operatorname{span} \{ i \sigma_1, i \sigma_3\}.$$

We put

 * $$\; \mathfrak{k} = \operatorname{span} \{i \sigma_3\},$$

and

 * $$\; \mathfrak{p} = \operatorname{span} \{ i \sigma_1, i \sigma_2\}.$$

Using the algebraic identities listed in the previous section, it can be verified that $$\mathfrak{k}$$ and $$\mathfrak{p}$$ form a Cartan pair of the Lie algebra SU(2). Furthermore,

 * $$\; \mathfrak{a} = \operatorname{span} \{ i \sigma_2\}$$

is a maximal abelian subalgebra of $$\mathfrak{p}$$. Now, a version of Cartan decomposition states that any element U in the Lie group SU(2) can be expressed in the form

 * $$U = e^{k_1} e^a e^{k_2}\,\!$$ where $$k_1, k_2 \in \mathfrak{k}$$ and $$a \in \mathfrak{a}.$$

In other words, any unitary U of determinant 1 is of the form

 * $$U = e^{i \alpha \sigma_3} e^{i \beta \sigma_2} e^{i \gamma \sigma_3}\,\!$$

for some real numbers α, β, and γ.

Extending to unitary matrices gives that any unitary 2 × 2 U is of the form

 * $$U = e^{i \delta} e^{i \alpha \sigma_3} e^{i \beta \sigma_2} e^{i \gamma \sigma_3}\,\!$$

where the additional parameter δ is also real (also compare with Leonhardt 2010, eq 5.22, pg. 99)

SO(3)
The Lie algebra su(2) is isomorphic to the Lie algebra so(3), which corresponds to the Lie group SO(3), the group of rotations in three-dimensional space. In other words, one can say that $$i \sigma_j$$'s are a realization (and, in fact, the lowest-dimensional realization) of infinitesimal rotations in three-dimensional space. However, even though su(2) and so(3) are isomorphic as Lie algebras, SU(2) and SO(3) are not isomorphic as Lie groups. SU(2) is actually a double cover of SO(3), meaning that there is a two-to-one homomorphism from SU(2) to SO(3).

Quaternions
The real linear span of $$\{I, i\sigma_1, i\sigma_2, i\sigma_3\}\,$$ is isomorphic to the real algebra of quaternions H. The isomorphism from H to this set is given by the following map (notice that Pauli matrices are in reversed order):

1 \mapsto I, \quad i \mapsto i \sigma_3, \quad j \mapsto i \sigma_2, \quad k \mapsto i \sigma_1. $$

As the quaternions of unit norm is group-isomorphic to SU(2), this gives yet another way of describing SU(2) via the Pauli matrices. The two-to-one homomorphism from SU(2) to SO(3) can also be explicitly given in terms of the Pauli matrices in this formulation.

Quaternions form a division algebra—every non-zero element has an inverse—whereas Pauli matrices do not. For a quaternionic version of the algebra generated by Pauli matrices see biquaternions, which is a venerable algebra of eight real dimensions.

Quantum mechanics

 * In quantum mechanics, each Pauli matrix represents an observable describing the spin of a spin ½ particle in the three spatial directions. Also, as an immediate consequence of the Cartan decomposition mentioned above, $$i \sigma_j$$ are the generators of rotation acting on non-relativistic particles with spin ½. The state of the particles are represented as two-component spinors. An interesting property of spin ½ particles is that they must be rotated by an angle of 4$$\pi$$ in order to return to their original configuration. This is due to the two-to-one correspondence between SU(2) and SO(3) mentioned above, and the fact that, although one visualizes spin up/down as the north/south pole on the 2-sphere S2, they are actually represented by orthogonal vectors in the two dimensional complex Hilbert space.

 * For a spin $1⁄2$ particle, the spin operator is given by $$\mathbf{J} =\frac\hbar2\boldsymbol{\sigma}$$. The Pauli matrices can be generalized to describe higher spin systems in three spatial dimensions. The spin matrices for spin 1 and spin $$\textstyle\frac{3}{2}$$ are given below:

$$j=1$$:

J_x = \frac\hbar\sqrt{2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix} $$

J_y = \frac\hbar\sqrt{2} \begin{pmatrix} 0&-i&0\\ i&0&-i\\ 0&i&0 \end{pmatrix} $$

J_z = \hbar \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix} $$

$$j=\textstyle\frac{3}{2}$$:

J_x = \frac\hbar2 \begin{pmatrix} 0&\sqrt{3}&0&0\\ \sqrt{3}&0&2&0\\ 0&2&0&\sqrt{3}\\ 0&0&\sqrt{3}&0 \end{pmatrix} $$

J_y = \frac\hbar2 \begin{pmatrix} 0&-i\sqrt{3}&0&0\\ i\sqrt{3}&0&-2i&0\\ 0&2i&0&-i\sqrt{3}\\ 0&0&i\sqrt{3}&0 \end{pmatrix} $$

J_z = \frac\hbar2 \begin{pmatrix} 3&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-3 \end{pmatrix}. $$

 * Also useful in the quantum mechanics of multiparticle systems, the general Pauli group Gn is defined to consist of all n-fold tensor products of Pauli matrices.

 * The fact that any 2 × 2 complex Hermitian matrices can be expressed in terms of the identity matrix and the Pauli matrices also leads to the Bloch sphere representation of 2 × 2 mixed states (2 × 2 positive semidefinite matrices with trace 1). This can be seen by simply first writing a Hermitian matrix as a real linear combination of {σ0, σ1, σ2, σ3} then impose the positive semidefinite and trace 1 assumptions.

Quantum information

 * In quantum information, single-qubit quantum gates are 2 × 2 unitary matrices. The Pauli matrices are some of the most important single-qubit operations. In that context, the Cartan decomposition given above is called the Z-Y decomposition of a single-qubit gate. Choosing a different Cartan pair gives a similar X-Y decomposition of a single-qubit gate.

title: Pauli matrices


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title: File:2-12-2.jpg


text:
User:Eml5526.s11.team7.snook/HW2

Problem Description
Show:

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$$ \displaystyle \int \frac{x^2}{1+x}dx=\frac{x^2}{2}-x+\log(1+x)+k $$ (2.4.1)
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Given
Prove the following through integration by parts:

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$$ \displaystyle \int log(x)dx=xlog(x)-x $$ (2.4.2)
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Solution
Using a similar derivation as p.48 of "A First Course in Finite Elements."

Looking first at using the product rule for two functions:
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$$ \displaystyle d(uv)=udv+vdu $$ (2.4.3) Rearranging Eq. 2.4.3 gives:
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$$ \displaystyle udv=d(uv)-vdu$$ (2.4.4)
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Integrating Eq. 2.4.4:
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$$ \displaystyle \int udv=\int d(uv)-\int vdu$$ (2.4.5) Note the first term on the right hand side of Eq. 2.4.5 becomes the following from the fundamental theorem of calculus:
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$$ \displaystyle \int d(uv)=uv$$ (2.4.6) Plugging Eq. 2.4.6 into Eq. 2.4.5 gives the following familiar equation used for integration by parts:
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$$ \displaystyle \int udv=(uv)-\int vdu$$ (2.4.6) $$\displaystyle \text{Returning to Eq. 2.4.2 using } u=log(x) \text{, } du=\frac{1}{x}\text{, } dv=dx\text{ and } v=x$$:
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$$ \displaystyle \int log(x)dx= xlog(x)-\int\frac{1}{x}xdx$$ (2.4.7) Which yields:
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$$ \displaystyle \int log(x)dx= xlog(x)-x$$ (2.4.8) Which is in agreement with Eq. 2.4.2.
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Given
Prove the following through integration by parts:

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$$ \displaystyle \int xlog(x)dx=\frac{1}{2}x^2[log(x)-\frac{1}{2}] $$ (2.4.9)
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Solution
Following from Eq. 2.4.6:

$$\displaystyle u=log(x), du=\frac{1}{x}, dv=xdx,\text{ and } v=\frac{1}{2}x^2$$

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$$ \displaystyle \rightarrow\int xlog(x)dx=\frac{x^2}{2}log(x)-\int \frac{x^2}{2}\frac{1}{x}dx=\frac{x^2}{2}log(x)-\frac{x^2}{4} $$ (2.4.10)
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Factoring out $$\displaystyle\frac{x^2}{2}$$ gives:

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$$ \displaystyle \int xlog(x)dx=\frac{1}{2}x^2[log(x)-\frac{1}{2}] $$ (2.4.11)
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Which agrees with Eq 2.4.9.

Given
Find the following:

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$$ \displaystyle \int \frac{x^2}{1+bx}dx $$ (2.4.12)
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Solution
Using integration by parts, so following Eq 2.4.6:

$$\displaystyle u=x^2, du=2xdx, dv=\frac{1}{1+bx},\text{ and } v=\frac{1}{b}log(1+bx)$$

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$$ \displaystyle\rightarrow \int \frac{x^2}{1+bx}dx=\frac{x^2}{b}log(1+bx)-\int \frac{2x}{b}log(1+bx)dx=$$
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$$ \frac{x^2}{b}log(1+bx)-\frac{2}{b}\int xlog(1+bx)dx $$ (2.4.13)
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Looking at the far right integral of Eq. 2.4.13 and using substitution of variables for easier integration:

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$$ \displaystyle z=1+bx $$ (2.4.14)
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Rearrangement yields:

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$$ \displaystyle x=\frac{z-1}{b} $$ (2.4.15)
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Lastly, taking the derivative of 2.4.15:

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$$ \displaystyle dx=\frac{1}{b}dz $$ (2.4.16)
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Subsituting 2.4.14-2.4.16 into the far right integral of 2.4.13 yields (omitting the constant coefficients at this time):

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$$ \displaystyle \int\frac{z-1}{b} log(z)\frac{1}{b}dz=\frac{1}{b^2}[\int zlog(z)dz-\int log(z)dz] $$ (2.4.17)
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Notice the first integral on the right hand side is the same as part two, Eq. 2.4.10 and the second integral on the right hand side is the same as part one, Eq. 2.4.8. Using these equations on Eq. 2.4.17:

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$$ \displaystyle \int\frac{1}{b^2}[\int zlog(z)dz-\int log(z)dz]=\frac{1}{b^2}[\frac{z^2log(z)}{2}-\frac{z^2}{4}-zlog(z)+z] $$ (2.4.18)
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Factoring out a z, and then substituting 2.4.14 in for z gives:

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$$ \displaystyle \frac{z}{b^2}[\frac{zlog(z)}{2}-\frac{z}{4}-log(z)+1]=$$
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$$\frac{1+bx}{b^2}[\frac{log(1+bx)}{2}+\frac{bxlog(1+bx)}{2}-\frac{1+bx}{4}-log(1+bx)+1] $$ (2.4.19)
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Multiplying the terms out:

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$$ \displaystyle \frac{log(1+bx)}{2b^2}+\frac{xlog(1+bx)}{2b}-\frac{1}{4b^2}-\frac{x}{4b}-\frac{log(1+bx)}{b^2}+\frac{1}{b^2}+$$
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$$\frac{xlog(1+bx)}{2b}+\frac{x^2log(1+bx)}{2}-\frac{x}{4b}-\frac{x^2}{4}-\frac{xlog(1+bx)}{b}+\frac{x}{b} $$ (2.4.20)
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Plugging into overall equation 2.4.13 and simplifying greatly yields:

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$$ \displaystyle \int \frac{x^2}{1+bx}dx=\frac{log(1+bx)}{b^3}+\frac{x^2}{2b}-\frac{x}{b^2}-\frac{3}{2b^3} $$ (2.4.21)
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Given
Find the following:

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$$ \displaystyle \int \frac{x^2}{a+bx}dx $$ (2.4.21)
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Solution
Using integration by parts, so following Eq 2.4.6:

$$\displaystyle u=x^2, du=2xdx, dv=\frac{1}{a+bx},\text{ and } v=\frac{1}{b}log(a+bx)$$

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$$ \displaystyle\rightarrow \int \frac{x^2}{a+bx}dx=\frac{x^2}{b}log(a+bx)-\int \frac{2x}{b}log(a+bx)dx=$$
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$$\frac{x^2}{b}log(a+bx)-\frac{2}{b}\int xlog(a+bx)dx $$ (2.4.22)
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Looking at the far right integral of Eq. 2.4.13 and using substitution of variables for easier integration:

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$$ \displaystyle z=a+bx $$ (2.4.23)
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Rearrangement yields:

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$$ \displaystyle x=\frac{z-a}{b} $$ (2.4.24)
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Lastly, taking the derivative of 2.4.24:

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$$ \displaystyle dx=\frac{1}{b}dz $$ (2.4.25)
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Subsituting 2.4.23-2.4.25 into the far right integral of 2.4.22 yields (omitting the constant coefficients at this time):

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$$ \displaystyle \int\frac{z-a}{b} log(z)\frac{1}{b}dz=\frac{1}{b^2}[\int zlog(z)dz-a\int log(z)dz] $$ (2.4.26)
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Notice the first integral on the right hand side is the same as part two, Eq. 2.4.10 and the second integral on the right hand side is the same as part one, Eq. 2.4.8. Using these equations on Eq. 2.4.26:

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$$ \displaystyle \int\frac{1}{b^2}[\int zlog(z)dz-a\int log(z)dz]=\frac{1}{b^2}[\frac{z^2log(z)}{2}-\frac{z^2}{4}-azlog(z)+az] $$ (2.4.27)
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Factoring out a z, and then substituting 2.4.14 in for z gives:

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$$ \displaystyle \frac{z}{b^2}[\frac{zlog(z)}{2}-\frac{z}{4}-alog(z)+a]=\frac{a+bx}{b^2}[\frac{alog(a+bx)}{2}+\frac{bxlog(a+bx)}{2}-$$
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$$\frac{a+bx}{4}-alog(a+bx)+a] $$ (2.4.28)
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Multiplying the terms out:

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$$ \displaystyle \frac{a^2log(a+bx)}{2b^2}+\frac{axlog(a+bx)}{2b}-\frac{a^2}{4b^2}-\frac{ax}{4b}-\frac{a^2log(a+bx)}{b^2}+\frac{a^2}{b^2}+$$
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$$\frac{axlog(a+bx)}{2b}+\frac{x^2log(a+bx)}{2}-\frac{ax}{4b}-\frac{x^2}{4}-\frac{axlog(a+bx)}{b}+\frac{ax}{b} $$ (2.4.29)
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Plugging into overall equation 2.4.13 and simplifying greatly yields:

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$$ \displaystyle \int \frac{x^2}{a+bx}dx=\frac{a^2log(a+bx)}{b^3}+\frac{x^2}{2b}-\frac{ax}{b^2}-\frac{3a^2}{2b^3} $$ (2.4.30)
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Given
Find the exact solution of:

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$$ \displaystyle \frac{d}{dx}[(2+3x)\frac{du}{dx}]+5x=0 $$ (2.4.31)
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With the following boundary conditions:

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$$ \displaystyle 1: u(1)=4 $$ (2.4.32)
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$$ \displaystyle 2: -\frac{du}{dx}|_{x=0} =6 $$ (2.4.33)
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Solution
Rearranging 2.4.31 and integrating:

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$$ \displaystyle \int d[(2+3x)\frac{du}{dx}]=\int-5xdx $$ (2.4.34)
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Which gives with rearrangement and integration:

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$$ \displaystyle \int du=\int (\frac{-\frac{5}{2}x^2}{2+3x}+\frac{C1}{2+3x})dx $$ (2.4.35)
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Where C1 is a constant of integration. The first term in the integral on the right hand side is easily seen to be the same as part 4 (Eq. 2.4.30) where a=2 and b=3. This gives the folloing solution to the first term in the integral on the right hand side:

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$$ \displaystyle \int\frac{\frac{-5}{2}x^2}{2+3x}dx=\frac{-5}{2}[\frac{4log(2+3x)}{3^3}+\frac{x^2}{6}-\frac{2x}{9}+C2] $$ (2.4.36) Where C2 is a constant of integration. Solving the second term now:
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$$ \displaystyle C1\int\frac{1}{2+3x}dx=\frac{C1}{3}log(2+3x)+C2=C1ln(2+3x)+C2$$ (2.4.37)
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Notice the 3 is absorbed into the integrating constant C1 after the second equal's sign. Adding 2.4.37 and 2.4.36 and combining constants into C2:

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$$ \displaystyle u=\frac{-5}{2}[\frac{4log(2+3x)}{3^3}+\frac{x^2}{6}-\frac{2x}{9}] +C1log(2+3x)+C2$$ (2.4.38)
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Taking the first derivative of Eq. 2.4.38 gives:

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$$ \displaystyle u'=\frac{-5}{2}[(\frac{4}{27})(\frac{3}{2+3x})+\frac{x}{3}-\frac{2}{9}]+\frac{3C1}{2+3x}$$ (2.4.39)
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Using BC2 (Eq. 2.4.33) gives:

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$$ \displaystyle C1=-4$$ (2.4.40)
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Using BC1 (Eq. 2.4.32) gives:

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$$ \displaystyle C2=4+\frac{5}{2}[\frac{4log(5)}{27}+\frac{1}{6}-\frac{2}{9}]+4log(5)$$ (2.4.41)
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Giving the Final Equation for u:

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$$ \displaystyle u=\frac{-5}{2}[\frac{4log(2+3x)}{27}+\frac{x^2}{6}-\frac{2x}{9}] -4log(2+3x)+4+$$
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$$\frac{5}{2}[\frac{4log(5)}{27}+\frac{1}{6}-\frac{2}{9}]+4log(5) $$ (2.4.42)
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Given
Graph u(x) (Eq. 2.4.42)

title: User:Eml5526.s11.team7.snook/HW2