text:
Talk:Psycholinguistics/Language and Music

Feedback
In an attempt to compare the processing of language and music, the paper does raise some interesting points and questions. However, the overall structure would benefit from more focus and more depth. First of all, in general, there should be a focus on music. It is necessary to summarize the article about the processing of music. This summary is necessary before you actually compare music with language. In fact, there is no strong sense of a comparison between music and language.

The part "Breaking it down" it is a collection of points. These points are interesting and provide a nice summary and background for readers. However, some of the points are vague and the overall structure needs to be reworked. There does not seem to be much connection between the points and some of the points may incorrect. For example, in "Production" it is wrong to say everyone is fluent in language and that some cannot play sing or compose music. Individuals within a culture have a range of abilities in both language and music, from almost none to all. Some people cannot speak their native language fluently; everyone can hum, sing or whistle a then—thought some are not particularly good at it. Some people are experts in music or language, such as professional musicians or linguists one can acknowledge that more people can read language than read in music. In "interpretation", the overall point is not clear. Language includes syntax and semantics. Music includes syntax, but what is the analogue (or equivalent) of semantics for music? The analogue of semantics in language is affect in music. The meaning of music is affective self-expression and communication. As with semantics in language, affect in music may be used to inform others of one’s internal state, or used to change the internal states of others.

The "Language" part in the section "Building on the basics" lacks detailed information. What is stated seems fine, but only three areas of brain structures are mentioned. The associations between these brain structures need to be explained. Also since there is a later reference to the use of music in memory, there should be discussion of the structures of memory in language. The “music” part in section of "Building on the basics" lacks information as well. There is no real discussion of the area of the brain responsible for music processing. Saying that it might be the left or right hemisphere provides absolutely no new information. You may want to check with the studies of Peretz for example. If you are going to say that “'language seems more universal, whereas music is more culturally-based”, you are going to need a reference. This is a strong statement, and in fact, there are good arguments for thinking that music is less culturally defined that language.

In the part "Similar systems", you should avoid to use the words such as "similar", "much", and "many". In scientific writing, one needs to accurately describe amounts: which areas, how many areas, what are the similarities, what are the differences. In this section, the discussion mentions that both the left and right hemispheres are involved with music processing. However, it is not clear which areas are involved. One can also wonder why the motor cortex is mentioned, this should be trivially obvious if one is speaking or performing music. We know it is most important for movement, how there some other particular role with music?

The part "Music and Memory" presents an interesting discussion, but seems have no connection with all other sections. How music helps to understand memory is not clearly defined. For example, children who learn music at an early age seem have better memory than children who did not learn music at an early age. Unfortunately, this may not be an effect of music. Children who learn music at an early age may simply imply that enriched environments help memory. This is, the chance to learn music early also implies an enriched environment. Since musical education is associated with parental affluence, more affluent children can get more chances to learn many things, and all of these may help these children develop better memory. Thus, in this section, there needs to be more references to clarify how music actually helps memory.

Overall, the sections seems to be somewhat disconnected with each other and the actually discussion of musical processing within the brain is weak. Additional material specifying the structures of the brain are needed.


text:
Swahili I/More Grammar

Verb Tenses
The following affices are more tense affices that show when an action is to have happened. They are placed between the subject and object prefix and affix to show the tense or the mood of the verb. This placement applies to all of them except -e, which is actually a marker for the subjunctive, though used only in imperatives. It comes as a suffix to the verb instead of the -a indicative marker. For instance, the verb -imba means "to sing," so to say, "Uimbe," would mean "Sing!" (U- is the subject prefix for the understood "you.")
 * Does not apply to the grammatical rules with regard to placement.

-NGE- & -NGALI-
You can also use -nge- and -ngali- as a sort of tense that is similar to the english subjunctive. You can say, "Ningalimwona, ningalimpenda." This means, "If I had not seen him/her, I would not have loved him/her." This is because -ngali-, as you know, is the conditional perfect, so you put it in both words as a tense marker to express this kind of sentence. (The -m- object affix changes to -mw- in front of -ona.)

Negation of -NGE- & -NGALI-
To negate these tense markers, you simply add the affix -si- in between the subject prefix and the tense marker. So, "Nisingependa" means, "I would not like." Or you can say, "Usingaliyaona machungwa, usingaliyataka." So as you can see, this means, "If you had not seen the oranges," (Oranges = Machungwa, cl. 6) because it was negated by the -si- affix. This means, You would not have wanted the oranges" because it has the -si- affix, negating it.

Noun Class Affices
Here are the noun class pronouns mentioned in lesson 1. In addition, adjective prefices and -a (of) prefices are included for your edification.

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

title: Swahili I/More Grammar


title: File:Fe1.s11.team3.HW4.4.fig1.jpg


text:
Talk:Psycholinguistics/History and Major Theories

Peer Edit

Overall, I think you did well at discussing many of the key contributors to the study of psycholinguistics. I found that the most notable contributors were all listed and had subsections of their own. I thought that perhaps it may be important to discuss or at least list all the major contributors – both the more modern and historic contributors – in your introduction. I think that you should also note the varying schools of thought in psycholinguistics in your introduction as well. By providing this extra information in your introduction, I think your introduction would act as a more thorough overview of the rest of your work and be a more accurate abstract (which could be important for teachers combing through resources).

I think the chapter, as a whole, would benefit from reorganization of topics. It may be more organizationally appropriate to start with the schools of thought, then historic contributors (here you could include which school each contributor fell into), then the more modern contributors, and finally the connection of these contributors to the cognitive revolution. This way the information you present in earlier sections might build upon one another and read more fluidly.

I found some grammatical and spelling issues throughout and some sentences in need of restructuring. I did not want to make specific changes to the document on the internet in case these changes did not reflect your intended meaning. Instead, I highlighted things in need of alteration in yellow on a Word document and beside these highlighted portions I often put suggestions of ways to improve the yellow portions in highlighted purple. I will send this to you in an email (the formatting does not work on Wikiversity) and post a request for your email address on the class discussion board. I know your chapter covers a very broad topic and key information had to be selected and others left out, in order to produce a chapter of decent length. I think the whole work has the content it needs, however, I also think that the chapter would benefit from having you read over the material you have written once more. There are some sentences that are slightly confusing, particularly the part about Broca and Tan. The individual you are referring to needs to be clear and the issue you are addressing must always be well described. I found that at times, such as the section on Wundt, some important aspects were overlooked. This may have been due to trying to keep sections brief and not overdone, however I think at some points clarity was lost due to the brevity. Though if you can go back and reread everything this problem may be fixed on its own.

Finally, I noticed that you cited the textbook by Jay throughout your chapter. Dr. Newman mentioned that we should not reference the text itself, however if we do use a specific source cited by Jay, that we should go directly to that source and cite it instead. Dr. Newman also said to make all citations throughout your work links to the full reference in the reference section at the bottom of the page. I didn’t notice all of your sources in the reference section cited throughout the chapter, so that might be something to check before the final edition of the chapter is due.

Great Job!

Chantel 023:57, 27 February 2011 (UTC)


text:
User:Egm6341.s11.team4.yang/hw4

=HW 4.2 problem=

Refer to lecture slide [[media:nm1.s11.mtg20.djvu|mtg-20]] for the problem statement.

Given
Given that :
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$$ \displaystyle
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A(t)=\int\limits_{-t}^{+t}{F(x)dx=}\int\limits_{-t}^{k}{F(x)dx+}\int\limits_{k}^{+t}{F(x)dx}

$$     (Eq ) Prove:
 *
 * }
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$$ \displaystyle
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{{A}^{(1)}}(t)=F(-t)+F(t)

$$     (Eq )
 *
 * }

=Answer:= Take a derivative on this equation,
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$$ \displaystyle
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A(t)=\int\limits_{-t}^{+t}{F(x)dx=}\int\limits_{-t}^{k}{F(x)dx+}\int\limits_{k}^{+t}{F(x)dx}

$$     (Eq ) We assume that $$\displaystyle \begin{align} & G(t)=\int{F(t)dt},\frac{dG(t)}{dt}=F(t) \\ & \\ \end{align}$$
 *
 * }
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$$ \displaystyle
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\int\limits_{-t}^{k}{F(x)dx}=G(k)-G(-t)

$$     (Eq )
 *
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$$ \displaystyle
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\int\limits_{k}^{+t}{F(x)dx}=G(t)-G(k)

$$     (Eq ) Where k is constant value,
 *
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$$ \displaystyle
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\begin{align} & \frac{dA(t)}{dt}=\frac{d\left( G(k)-G(-t) \right)}{dt}+\frac{d\left( G(t)-G(k) \right)}{dt} \\ & =-\frac{d\left( G(-t) \right)}{dt}\frac{d\left( -t \right)}{dt}+\frac{d\left( G(t) \right)}{dt}=F(-t)+F(t) \\ \end{align}

$$     (Eq ) So,
 *
 * }
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$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{A}^{(1)}}(t)=F(-t)+F(t)

$$     (Eq )
 *
 * }

=Problem 4.9= Refer to lecture slide [[media:nm1.s11.mtg22.djvu|mtg-22]] for the problem statement.

Objectives
A). Redo the proof for two cases: 1).
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$$ \displaystyle
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G(t):=e(t)-{{t}^{4}}e(1)

$$ 2). :{| style="width:100%" border="0" $$ \displaystyle
 *
 * }
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 * style="width:95%" |

G(t):=e(t)-{{t}^{6}}e(1)

$$ Point out where proof breaks down. B). for G(t) as in (1) p.19-1(w/t5), find G(3)(0) and follow same steps in proof to see what happen.
 *
 * }

Solutions
A). Redo the proof for two cases: 1). $$\displaystyle G(t):=e(t)-{{t}^{4}}e(1)$$
 * {| style="width:100%" border="0"

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

e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

$$ Where $$\displaystyle F(t)=f(x(t))$$
 *
 * }
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$$ \displaystyle
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\text{G}\left( 0 \right)=e(0)-{{0}^{4}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

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

$$ Apply Rolle ‘ s theorem, it yields:
 *
 * }
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$$ \displaystyle
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\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
 *
 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-4{{t}^{3}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-4{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-4{{t}^{3}}e(1) \\ \end{align}

$$ So,
 *
 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-4\centerdot {{0}^{3}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *
 * }
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$$ \displaystyle
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\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
 *
 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-12{{t}^{2}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-12{{t}^{2}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-12{{t}^{2}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-12{{t}^{2}}e(1) \\ \end{align}

$$ So,
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-12\centerdot {{0}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *

 * }
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$$ \displaystyle
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\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-24te(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-24te(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-24te(1) \\ \end{align}

$$ So, $$\displaystyle {{\text{G}}^{(3)}}\left( {{\xi }_{3}} \right)=\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]-24\left( {{\xi }_{3}} \right)e(1)=0$$, then using DMVT:
 *

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$$ \displaystyle
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e(1)=\frac{\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]}{24\left( {{\xi }_{3}} \right)}=\frac{\frac{3}[-2{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})]}{24\left( {{\xi }_{3}} \right)}=\frac{-{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})}{36}

$$ Where, $$\displaystyle {{\xi }_{4}}\in \left( -{{\xi }_{3}},{{\xi }_{3}} \right)$$
 *

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

{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})}{36}

$$     (Eq ) The proof will break down here because the magnitude of error is related to $$\displaystyle {{\xi }_{3}}$$in the above equation.
 *

 * }

2)

$$\displaystyle G(t):=e(t)-{{t}^{6}}e(1)$$
 * {| style="width:100%" border="0"

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

e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

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

 * }
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$$ \displaystyle
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\text{G}\left( 0 \right)=e(0)-{{0}^{6}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

$$
 *

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

\text{G}\left( 1 \right)=e(1)-{{1}^{6}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
 *

 * }
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$$ \displaystyle
 * style="width:95%" |
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\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-6{{t}^{5}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-6{{t}^{5}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-6{{t}^{5}}e(1) \\ \end{align}

$$ So,
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-6\centerdot {{0}^{5}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

$$ \displaystyle
 * style="width:95%" |
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\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-30{{t}^{4}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-30{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-30{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-30{{t}^{4}}e(1) \\ \end{align}

$$ So,
 *

 * }
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$$ \displaystyle
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\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-30\centerdot {{0}^{4}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

$$ \displaystyle
 * style="width:95%" |
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\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
 *

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

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

\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-120{{t}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-120{{t}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-120{{t}^{3}}e(1) \\ \end{align}

$$ So, $$\displaystyle {{\text{G}}^{(3)}}\left( {{\xi }_{3}} \right)=\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]-120{{\left( {{\xi }_{3}} \right)}^{3}}e(1)=0$$, then using DMVT:
 *

 * }
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$$ \displaystyle
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e(1)=\frac{\frac{3}[{{F}^{(3)}}(-{{\xi }_{3}})-{{F}^{(3)}}({{\xi }_{3}})]}{120{{\left( {{\xi }_{3}} \right)}^{3}}}=\frac{\frac{3}[-2{{\xi }_{3}}{{F}^{(4)}}({{\xi }_{4}})]}{120{{\left( {{\xi }_{3}} \right)}^{3}}}=\frac{-{{F}^{(4)}}({{\xi }_{4}})}{180{{\xi }_{3}}}

$$ Where, $$\displaystyle {{\xi }_{4}}\in \left( -{{\xi }_{3}},{{\xi }_{3}} \right)$$
 *

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

{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-{{F}^{(4)}}({{\xi }_{4}})}{180{{\xi }_{3}}}

$$
 *

 * }The proof will break down here because the magnitude of error is related to $$\displaystyle {{\xi }_{3}}$$in the above equation.

B)
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$$ \displaystyle
 * style="width:95%" |
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G(t):=e(t)-{{t}^{5}}e(1)

$$
 *

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

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

e(t)=\int\limits_{-t}^{+t}{F(t)dt}-\frac{t}{3}[F(-t)+4F(0)+F(t)]

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

 * }

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

$$ \displaystyle
 * style="width:95%" |
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\text{G}\left( 0 \right)=e(0)-{{0}^{5}}e(1)=\int\limits_{-0}^{+0}{F(t)dt}-\frac{0}{3}[F(-0)+4F(0)+F(0)]=0

$$
 *

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

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

\text{G}\left( 1 \right)=e(1)-{{1}^{5}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

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

\exists {{\xi }_{1}}\in \left( 0,1 \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 1 \right)}}({{\xi }_{1}})=0

$$
 *

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

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

\begin{align} & {{\text{G}}^{(1)}}\left( t \right)={{e}^{(1)}}(t)-5{{t}^{4}}e(1)={{A}^{(1)}}(t)-A{{_{2}^{L}}^{(1)}}(t)-5{{t}^{4}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=F(-t)+F(t)-\{\frac{1}{3}[F(-t)+4F(0)+F(t)]+\frac{t}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-5{{t}^{4}}e(1) \\ \end{align}

$$ So,
 *

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

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

\begin{align} & {{\text{G}}^{(1)}}\left( 0 \right)=F(-0)+F(0)-\{\frac{1}{3}[F(-0)+4F(0)+F(0)]+\frac{0}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]\}-5\centerdot {{0}^{4}}e(1) \\ & \begin{matrix} {} & {} & {} \\ \end{matrix}=0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

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

\exists {{\xi }_{2}}\in \left( 0,{{\xi }_{1}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 2 \right)}}({{\xi }_{2}})=0

$$
 *

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

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

\begin{align} & {{\text{G}}^{(2)}}\left( t \right)={{e}^{(2)}}(t)-20{{t}^{3}}e(1)={{A}^{(2)}}(t)-A{{_{2}^{L}}^{(2)}}(t)-20{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)-\{\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)] \\ & \begin{matrix} {} & {} & {} & {} \\ \end{matrix}+\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]+\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-20{{t}^{3}}e(1) \\ & \begin{matrix} \begin{matrix} {} & {} \\ \end{matrix} & {}  \\ \end{matrix}=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]\}-20{{t}^{3}}e(1) \\ \end{align}

$$ So,
 *

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

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

\begin{align} & {{\text{G}}^{(2)}}\left( 0 \right)=\frac{1}{3}[-{{F}^{(1)}}(-0)+{{F}^{(1)}}(0)]-\frac{0}{3}[{{F}^{(2)}}(-0)+{{F}^{(2)}}(0)]\}-20\centerdot {{0}^{3}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}0 \\ \end{align}

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

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

\exists {{\xi }_{3}}\in \left( 0,{{\xi }_{2}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 3 \right)}}({{\xi }_{3}})=0

$$
 *

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

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

\begin{align} & {{\text{G}}^{(3)}}\left( t \right)=\frac{1}{3}[-{{F}^{(1)}}(-t)+{{F}^{(1)}}(t)]-\frac{t}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-60{{t}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{1}{3}[{{F}^{(2)}}(-t)+{{F}^{(2)}}(t)]-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60{{t}^{2}}e(1) \\ & \begin{matrix} {} & {} & = \\ \end{matrix}-\frac{t}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60{{t}^{2}}e(1) \\ \end{align}

$$ So,
 *

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

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

{{\text{G}}^{(3)}}\left( 0 \right)=-\frac{0}{3}[-{{F}^{(3)}}(-t)+{{F}^{(3)}}(t)]-60\centerdot {{0}^{2}}e(1)=0

$$ Apply Rolle ‘ s theorem, it yields:
 *

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

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

\exists {{\xi }_{4}}\in \left( 0,{{\xi }_{3}} \right)\begin{matrix} {} & st. \\ \end{matrix}{{G}^{\left( 4 \right)}}({{\xi }_{4}})=0

$$
 *

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

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

{{\text{G}}^{(4)}}\left( t \right)=\frac{1}{3}[{{F}^{(3)}}(-t)-{{F}^{(3)}}(t)]-\frac{t}{3}[{{F}^{(4)}}(-t)+{{F}^{(4)}}(t)]-120te(1)

$$ So,
 *

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

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

{{\text{G}}^{(4)}}\left( {{\xi }_{4}} \right)=\frac{1}{3}[{{F}^{(3)}}(-{{\xi }_{4}})-{{F}^{(3)}}({{\xi }_{4}})]-\frac{3}[{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]-120{{\xi }_{4}}e(1)=0

$$ Using DMVT:
 *

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

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

e(1)=\frac{\frac{1}{3}[-2{{\xi }_{4}}{{F}^{(4)}}({{\xi }_{5}})]-\frac{3}[{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{120{{\xi }_{4}}}=\frac{-[2{{F}^{(4)}}({{\xi }_{5}})+{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{360}

$$ Where, $$\displaystyle {{\xi }_{5}}\in \left( -{{\xi }_{4}},{{\xi }_{4}} \right)$$
 *

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

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

{{\text{E}}_{2}}=h\centerdot e(1)=h\frac{-[2{{F}^{(4)}}({{\xi }_{5}})+{{F}^{(4)}}(-{{\xi }_{4}})+{{F}^{(4)}}({{\xi }_{4}})]}{360}

$$ Here, it is observed that the magnitude of error is related to the value of function at these two points$$\displaystyle {{\xi }_{4}},{{\xi }_{5}}$$in the above equation.
 *

 * }

title: User:Egm6341.s11.team4.yang/hw4


title: File:4-9Taylor.jpg


text: File:4.1 all.png

title: File:4.1 all.png


title: File:Fe1.s11.team6.HW4.6.fig1.JPG


text: File:4.1 error.png

title: File:4.1 error.png


text:
User:EML5526.S11.Team1.HS/HW4

3.6
$$ \left. \left( wJG\frac{\text{d}\phi }{\text{d}x} \right) \right|_{x=0}^{x=l} $$

Given
Given the strong form for the heat conduction problem in a circular plate:

$$ k \frac{d}{dr} \left( r \frac{dT}{dr} \right) + rs = 0, 0 < r \leq R. $$

natural boundary condition : $$ \frac{dT}{dr} \left( r = 0 \right) = 0, $$

essential boundary condition : $$ T \left( r = R \right) = 0, $$

where $$ R $$ is the total radius of the plate, $$ s $$ is the heat source per unit length along the plate radius, $$ T $$ is the temperature and $$ k $$ is the conductivity.

Find
Assume $$ k, ~s~ $$ and $$ ~R~ $$ are given:

a. Construct the weak form for the above strong form.

b. Use quadratic trial (candidate) solutions of the form $$ T = \alpha_{0} + \alpha_{1}r + \alpha_{2}r^2 $$ and weight functions of the same form to obtain a solution of the weak form.

c. Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by

$$ T = \frac{s}{4k} \left(R^2 - r^2 \right). $$

a)
First we multiply the equation in the strong form by the weight function and integrate over the domain.

$$ \int_{0}^{R} wk \frac{d}{dr} \left( r \frac{dT}{dr}\right )dr + \int_{0}^{R} wrsdr = 0 ~ \forall w, $$

$$ \int_{0}^{R} k \frac{dw}{dr} \left( r \frac{dT}{dr}\right )dr = \left[wkr \frac{dT}{dr} \right]_0^R + \int_{0}^{R} wrsdr  \forall w(r)  with w(R) = 0 $$

The middle term cancels out when evaluated from $$ 0 to  R $$ which yields

$$ \int_{0}^{R} k \frac{dw}{dr} \left( r \frac{dT}{dr}\right )dr = \int_{0}^{R} wrsdr  \forall w(r)  with w(R) = 0 $$

b)
The trial solution $$ T\left ( r \right )$$ and the weight function $$ w\left ( r \right )$$ can be written as follows:

$$ T\left ( r \right ) = \alpha_{0} + \alpha_{1}r + \alpha_{2}r^{2}$$

$$ w\left ( r \right ) = \beta_{0} + \beta_{1}r + \beta_{2}r^{2} $$

Using the boundary conditions, the coefficients of the trial solutions can be solved. First the you take the derivative of the trial solution and set it to the natural boundary condition.

$$ \frac{dT}{dr} = \alpha_1 + 2 \alpha_{2}r $$

$$ \frac{dT}{dr} \left(r=0 \right) = 0 = \alpha_1 + 2 \alpha_{2}r $$

Therefore $$ \alpha_1 = 0 $$

Now use the essential boundary

$$ T \left(r = R \right) = 0 = \alpha_0 + \alpha_{2}R $$

Therefore $$ \alpha_0 = - \alpha_2 R^2 $$ which solves for our two temperature functions

T \left(r \right) = \alpha_2 \left(r^2 - R^2 \right) $$ \frac{dT}{dr} = 2 \alpha_2 r $$

Now solve for the weight functions using $$ w \left( R \right) = 0 $$

$$ w \left( R \right) = 0 = \beta_0 + \beta_1 R + \beta_2 R^2 $$

Which yields

$$ \beta_0 = -\beta_1 R -\beta_2 R^2 $$

$$ w \left(r \right) = \beta_1 \left( r -R \right) + \beta_2 \left( r^2 - R^2 \right) $$

And the derivative of the weight function is

$$ \frac{dw}{dr} = \beta_1 + 2 \beta_2 r $$

Before substitution, when the middle term is evaluated from $$ \left( 0 to R \right) $$ it goes to zero

$$ w \left( R \right) k r \frac{dT}{dr} \left( R \right) - w \left( 0 \right) k r \frac{dT}{dr} \left( 0 \right) = 0 $$

Therefore we can cancel out the middle term and substitute in the rest

$$ \int_{0}^{R} k \left( \beta_1 + 2 \beta_2 r \right) \left( r \left( 2 \alpha_2 r \right) \right )dr =  \int_{0}^{R} \left( \beta_1 \left( r -R \right) + \beta_2 \left( r^2 - R^2 \right) \right) rsdr \forall w  with w = 0 $$

Integrating we get:

$$ -k \left[ \frac{2 \alpha_2 \beta_1 r^3}{3} + 3 \alpha_2 \beta_2 r^4 \right]_0^R + s \left[ \beta_1 \left( \frac{r^3}{3} - \frac{R r^2}{2} \right) + \beta_2 \left( \frac{r^4}{4} - \frac{R^2 r^2}{2} \right) \right]_0^R = 0 $$

After evaluating the domain, the equation simplifies to:

$$ -k \alpha_2 \left( \beta_1 \left( \frac{2 R^3}{3} \right) + \beta_2 \left( R^4 \right) \right) + s \left( \beta_1 \left( - \frac{R^3}{6} \right) + \beta_2 \left( - \frac{R^4}{4} \right) \right) = 0 $$

Factoring out $$ \beta_1 $$ and $$ \beta_2 $$ we get:

$$ \beta_1\left ( 2\alpha_2 \frac{R^3}{3} + s\frac{R^3}{6} \right ) + \beta_2 \left ( k \alpha_2 R^4 + s \frac{R^4}{4} \right ) = 0 $$

Assuming that $$ \beta_1 $$ and $$ \beta_2 $$ equal zero, $$ \alpha_2 $$ can be solved:

$$ \alpha_2 = \frac{-s}{4k} $$

Now $$ \alpha_2 $$ can be substituted into $$ T \left( r \right) $$ to obtain the solution of the weak form

$$ T \left( r \right) = \frac{-s}{4k} \left( r^2 - R^2 \right) $$

c)
Using the strong form the differential equation can be solved by taking the integral:

$$ k \frac{d}{dr} \left( r \frac{dT}{dr} \right) + rs = 0, 0 < r \leq R. $$

$$ \int \frac{d}{dr} \left( r \frac{dT}{dr} \right) = \int -r \frac{s}{k} $$

$$ r \frac{dT}{dr} = - r^2 \frac{s}{2k} + c_1 $$

Using the natural boundary condition $$ \frac{dT}{dr} \left(r = 0 \right) = 0 \rightarrowc_1 = 0 $$

Integrate once again we get:

$$ T = -r^2 \frac{s}{4k} + c_2 $$

Using the essential boundary condition $$ T \left( r = R \right) = 0 $$ solves for:

$$ c_2 = \frac{s R^2}{4k} $$

substituting in $$ c_2 $$ yields the temperature distribution

$$ T = \frac{s}{4k} \left(R^2 - r^2 \right) $$

=Problem 8=

Given
Based off Problem 3.4

Given Strong Form

$$ \frac{d}{dx}\left ( 2 \frac{du}{dx} \right ) + 2x = 3 \frac{d^{2}u}{dt^{2}}  on  1<x<3 $$

$$ \left. 2 \frac{du}{dx} \right|_{x=1} = 0.1 $$

$$ u \left( x=3 \right) = 0.001 $$

$$ u^{h}\left ( x \right ) = \alpha _{0} + \alpha _{1}\left ( x-3 \right )+ \alpha _{2}\left ( x-3 \right )^2+ \alpha _{3}\left ( x-3 \right )^3 $$

and

$$ w^{h}\left ( x \right ) = \beta{0} + \beta_{1}\left ( x-3 \right )+ \beta_{2}\left ( x-3 \right )^2+ \beta_{3}\left ( x-3 \right )^3 $$

with $$ u^h \left(x =3,t \right) = sin \left(2t \right) $$

Find
1. Assume

$$ A = 1, E = 2, \bar{m} = 3 $$

Find $$ \tilde{M} $$

2. Assume

$$ n = 3 $$

$$ u^{h} \left ( p,t \right ) = g\left ( t \right ) = sin2t $$

Find $$ F \left(t \right) $$

Solution
From the strong form we can easily obtain the weak form. Find $$ u(x) $$ smooth enough and satisfy the essential boundary condition:

$$ u \left( x = 3 \right) = 0.001 $$ such that:

$$ \int_{1}^{3} 3\frac{dw}{dx} 2\frac{du}{dx}dx = -0.1 \left(w \right)_{x=3} + \int_{1}^{3} 2xwdx + \int_{1}^{3} 3 \frac{d^{2}u}{dt^{2}}dx \forall w \left(x \right)  with  w \left(3\right) = 0 $$

since $$ u^h \left(x=3,t \right) = sin \left(2t \right) $$

$$ u^{h}\left ( x = 3 \right ) = \alpha _{0} + \alpha _{1}\left ( 3-3 \right )+ \alpha _{2}\left ( 3-3 \right )^2+ \alpha _{3}\left ( 3-3 \right )^3 \rightarrow \alpha_0 = sin\left(2t \right) $$

Thus $$ \frac{d^2 u}{dt^2} = -4 sin \left(2t \right) $$

$$ w^{h}\left ( x = 3 \right ) = \beta _{0} + \beta_{1}\left ( 3-3 \right )+ \beta_{2}\left ( 3-3 \right )^2+ \beta_{3}\left ( 3-3 \right )^3 \rightarrow \beta_0 = 0 $$

From discrete weak form we get:

$$ \tilde{M} = \int_{1}^{3} \bar{m} w \left ( x \right ) \frac{d^{2}u}{dt^{2}}dx = \int_{1}^{3} 3 \left [ \beta_{1}\left ( x-3 \right )+ \beta_{2}\left ( x-3 \right )^2+ \beta_{3}\left ( x-3 \right )^3 \right ]\left ( -4 sin\left ( 2t \right ) \right )dx $$

$$ \tilde{M} = -12 sin \left(2t \right) \times \left[ -2 \beta_1 - \frac{8}{3} \beta_2 - 4 \beta_3 \right] $$

title: User:EML5526.S11.Team1.HS/HW4


text: User:Darylcook

title: User:Darylcook


text:
User:Doaney/Lexical Access

Suggestions for my classmate
Your opening paragraph is good. I like that you have in-text citations that make it seem less like your opinion and shows more solid evidence towards your information. Although I know that this is an introduction, I do think it should maybe be a little bit longer. It is good that you mentioned the different factors that have an effect on our lexical access, because they are/should be discussed later on in the chapter. For the “What is Lexical Access” section I do however think that you should briefly foreshadow a few other things that you plan to discuss (i.e. lexical ambiguity) because it unconsciously prepares the reader for what they are about to learn about. I also wanted to let you know that your information was clear and to the point, which is good when writing a chapter like this.

The overall organization on this chapter is very well done. Information flows and is not too complicated for the reader to follow. The layout of the page is also great, due to the fact that many wiki pages have similar layouts.

I noticed that there were a few sections that were not entirely filled on in your chapter, and I am not sure if you just forgot to add in the information or something, but I just wanted to remind you, because I do think that the sections missing are going to be very interesting. “Emotional Connotations” is one of the sections that is not filled out, and one that I am looking forward to returning to and reading. Something else that I find adds character and interest to a page is graphics explaining your data. If you are able to find diagrams or graphs online feel free to add them to include excitement to your page.

In the "Lexical Ambiguity" section in the second sentence I do believe there is a word missing, I think you meant to write “One key discovery that has fueled research into this topic is the understanding that words and meanings do not necessarily (have) a 1:1 ratio”. I would have fixed it for you but I was not certain if you would have preferred a more elaborate word. Other than that part I do not see any problems with the Lexical Ambiguity section, in fact, I find it extremely interesting. I also find the Swinney experiment is very interesting. Your section on the “Dominant Meanings and Subordinate Meanings” was to the point, correct and was easy for a person of any reading level to understand and learn from.

At the end of your chapter there are two models that both only consist of one bullet naming the models. I would suggest perhaps doing another sub-bullet explaining what each of the models consists of and more information than just the researchers names.

My personal suggestions would be to add in a conclusion section, because it helps sums up the whole chapter for the reader, and will remind them of what the main points of the chapter are. Sometimes, after reading a lot of information and attempting to process it, people find it easier to have it laid out in front of them.

I think after adding in the necessary information to the sections that are missing some important facts, and elaborating on the sections that could go into more detail, along with adding in additional sources to the thoughts that you have stated. Obviously as Professor Newman stated you do not need to cite the textbook until the end, but if you are able to find other sources than that would really aid in your reliability of information!

And last but not least, do not forget to do your learning activity!

(Doaney 05:51, 2 March 2011 (UTC))

title: User:Doaney/Lexical Access


text:
University of Canberra/RCC2011/RecentChangesCamp Expanded/Sydney

Date and time: 16 March, 6-8pm,

Location: Cancer Council Australia

Ground floor, 120 Chalmers St, Boardroom

Registration
Please register your attendance on our eventbrite page http://rccxsydney.eventbrite.com

Organisers

 * Chris - Wikipedian & co-founder of Appropedia
 * Jutta Cancer Council Australia - Project Manager - Wiki Development
 * Liam Wyatt - Wikimedia Foundation’s Cultural Partnerships Fellow


text:
User:Egm6341.s11.team2.zou/HW4

=Problem 4.1 Verification of Exactness of Simple Simpson ‘ s rule at Cubic Polynomial=

''' This problem was solved without referring to S10 homework. '''

From the lecture slide Mtg18-2

Given
The integrand is

Find
Verify that Simple Simpson ‘ Rule can integrate cubic polynomials exactly by numerically integrating the given function using Simple Simpson ‘ s Rule and compare the result with the same result.

Solution
The exact value of integration $$\displaystyle I$$ is,

The numerical result by simple Simpson ‘ s rule is,

Hence,

In this case we have verified Simple Simpson ‘ s rule can integrate cubic polynomial exactly.

=Problem 4.3 Verification of Value of Goal Function at Particular Point=

From the lecture slide Mtg20-3

Given
The Goal function in proof of error of SSET is,

where,

Solution
In the lecture slide 19 we have already got,

where

Then we have,

Hence,

=Problem 4.8 Comparison of error bound in different estimation method with numerical result=

From the lecture slide Mtg22-2

Background
The maximum error of composite trapezoidal rule is ,

where $$\displaystyle {{M}_{2}}$$ is the maximum of $$\displaystyle \left| {{f}^{(2)}}(.) \right|$$

The maximum error of composite Simpson ‘ s rule is ,

where $$\displaystyle {{M}_{4}}$$ is the maximum of $$\displaystyle \left| {{f}^{(4)}}(.) \right|$$

Given
In the Problem 2.4, we have numerically calculate the order $$\displaystyle n$$ such that the following integration, with Taylor series expansion, composite trapezoidal rule and composite Simpson ‘ s rule, is with error less than $$\displaystyle Q={{10}^{-6}}$$.

where $$\displaystyle [a,b]=[-1,1]$$

And in Problem 2.3 we have the remainder of Taylor expansion of $$\displaystyle f(x)=\frac{{{e}^{x}}-1}{x}$$ is,

where $$\displaystyle \xi \in [0,x]$$

Find
(1)Use different error bound of Taylor series, composite trapezoidal rule and composite Simpson ‘ s rule to calculate the smallest order n needed such that the error of numerical integration of that in Problem 2.4 is less than $$\displaystyle Q={{10}^{-6}}$$. And then compare them with the actual numerical calculation.

Taylor series comparison
Using the same method in Mtg7 we can find the error bound of Taylor series,

Since we have set n to be even,

The error should be within $$\displaystyle Q$$

By WolframAlpha we have,

which coincide with the result in Problem 2.4.

Composite trapezoidal rule comparison
By 4.8.1 we have,

where $$\displaystyle {{M}_{2}}$$ can be calculated by WolframAlpha ,

Then we have,

Calculated by WolframAlpha we have,

While in Problem 2.4 we have $$\displaystyle n=512$$, the order produced by the error bound is larger than the numerical result.

Composite trapezoidal rule comparison
By 4.8.2 we have,

where $$\displaystyle {{M}_{4}}$$ can be calculated by WolframAlpha ,

Then we have,

Calculated by WolframAlpha we have,

In Problem 2.4 we calculated the case $$\displaystyle n=8$$ and $$\displaystyle n=16$$, the former case failed to satisfy the condition that $$\displaystyle \left| {{E}_{n}} \right|\le Q={{10}^{-6}}$$ while the latter succeed. This is a verification of the result above.

Conclusion
From the case of Taylor series and composite trapezoidal rule we can see the order of estimation given by the error bound is larger than or equal to that given by numerical calculation. This can be explained as: the error bound is stricter because it denotes the maximal error hence it is reasonably larger than or equal to the real error and thus requires higher order.

While in case of composite Simpson ‘ s rule we didn ‘ t calculate the case $$\displaystyle n=9$$, but the case $$\displaystyle n=8$$ and $$\displaystyle n=16$$ have demonstrate that if the error bound is satisfied with error limit then the numerical result must be satisfied.

Solution to part (2)
=References0=

title: User:Egm6341.s11.team2.zou/HW4


text:
Talk:Psycholinguistics/Development of Speech Perception

Peer Review

I would thoroughly go through and make sure sentence format doesn’t have any unnecessary errors (i.e. put spaces after every period: “to understand spoken language. Let’s ask ourselves the question how infants learn speech?”) Try not to use apostrophes, use “that is” instead of “that’s”. I would also not separate words like and/or, maybe find an alternate way to say and-or. These are tiny little things that add up to a lot in the end, you wouldn’t want to loose any points on these easy fixes! Make sure to double check for spelling and grammar, as I know myself how easily such errors go unnoticed. Make sure every sentence, and the sequence of sentences collectively make sense in context of the paragraph.

When presenting a question, maybe try using quotation marks, or separate the question with a comma→ (Let’s ask ourselves a question, how do infants learn speech?).

Your introduction of previous theories is a good starting point (genetic sounds vs learned sounds): be sure to include some in-text citations from where these theories and examples come from.

I’m not sure of the exact format of using numbers in writing, but I think numbers 1-10 are to be spelled out (one, two, three) and anything higher it is ok to use the actual number. Best make sure and make the necessary adjustments.

Titles- Some were in capitals, others not. I would make sure they are consistent throughout.

When talking about infant development, I would suggest doing so in chronological order. Make sure you state abilities at three months, followed by abilities at five months. Not vice versa as seen in “A timeline of Speech Perception and Production”. I like that the figure is included, however it is a link. It would be great if you could get the figure to be an actual part of your page.

I think the content is good and is accurate for your topic, now I think you could focus on cleaning up each segment, making sure each is free of grammar, spelling, and typographical errors. I believe the meaning of some paragraphs could be cleared up and more strongly conveyed by simplifying some of the wording. Off to a good start!

Cheers John

Answers to the Questions in Learning Exercises
Part 1

1) (C) High Amplitude Sucking Technique

They use a technique called High Amplitude Sucking where they introduce a speech sound and observe infants sucking responses. This technique can also tell the intensity and frequency the baby is sucking to each sound they hear. It usually takes infants a couple of minutes to get bored and their sucking responses decline. When a new stimulus is introduced, and because infants like novel things, the responses increase again. In order to tell if babies actually remember the info that they are hearing, they introduce a two minute period where no sounds exist and a series of colourful slides are presented and later on the speech sounds are presented again. If the infant remembers the words they heard during the first part of the study, and if there is a change in the sounds, then the frequency and intensity in their sucking responses will remain the same.

2) (B) The Head Turn Technique

The researchers should use the Head Turn Technique to predict an infant’s interest in words presented in the study. Interest in this study is measured by recording the amount of time infant attends to the stimuli presented behind the flashing light. Researchers flash a green light first and observe the infant’s interest, and then they flash a red light, followed by a speech sound. They play this sound until the infant becomes bored and looks away. This is when researchers have an idea of how much the infant is interested in the presented word. This technique also investigates the infant’s habituation by recording how the newborn adapts to new stimuli. Researchers have also discovered in this study an infant will pay more attention when they notice something familiar about the word they are presented to.

3) (A) Visual Habituation Technique

Researchers use the Visual Habituation Technique to determine at what age infants are able to understand the words they hear and refer them to the objects and actions in their environment. More specifically, it investigates what infants comprehension actually consists of. The trial begins with the experimenter presenting visual and auditory stimuli repeatedly to infants. The infant sits on their parents lap and watches two videos that consist of their mother’s and father’s images, and they are also presented with words like "mommy" or "daddy" an unequal number of times. They initially observe the infant’s looking time at their mother’s image when they hear mommy and their father’s when they hear daddy. The results show that infants look more at their mothers’ image when they hear the word “mommy” and more at their fathers’ image when they hear the word “daddy”. Researchers also perceived in this study that infants ultimately get familiarized with both images and the words and thus look away from the screen.

Part 2

QUESTION #1:

Motherese is a slower, more stressed, simplified, and repetitive version of an adult sentence in the native language. When a mother talks to her infant, she uses this language, which has a unique acoustic signature that promotes infant’s processing of speech. When compared to adult-directed speech (AD), child-directed speech (motherese, CD) is slower, has a higher average pitch, and contains exaggerated pitch contours. In CD speech, prosodic cues tend to be exaggerated in the kind of speech that is directed toward learning speech sounds. Likewise, motherese helps infants to analyze the structure of speech by highlighting boundaries between important units, such as words and clauses. When women are speaking to their children, the vowel sounds (the /e/ in 'see' and the /a/ in 'saw' and the /u/ in 'Sue') they are making in CD speech are more clearly articulated, exaggerating all of the acoustic components of vowels, which benefit the infants (Kuhl et al 1994). In contrast, AD speech does not have any emphasis on articulation. Therefore infants generally prefer CD speech to AD speech. There has been an investigation to find preferences for maternal CD speech in 1 and 4-month olds. (Cooper R. et al, 1997) Using a procedure in which infants controlled  access  to voices by fixating  a  visual  display,  infants  listened  to  recordings of  natural  female  CD  and  AD  speech sounds. In Experiment 1, 1-month-olds heard  recordings of  maternal CD  and  AD  speech,  but these  infants showed  no preference  for  maternal CD speech. In Experiment  2,  1-month-olds  heard  the  same CD and AD speech sounds  but were  not familiar  with  the  speakers. In contrast to Experiment 1, these infants preferred CD speech. In the last experiment, 4-month-olds heard  recordings  of  maternal  CD and AD speech and showed  a  significant   reference  for  CD  speech. Collectively, these  results  suggest that infant  attention to CD  speech  depends  on  both  speaker-general  and  speaker-specific  characteristics, with interesting developmental  changes  occurring  during  early infancy. Regarding this study, the role of  maternal  child-directed  speech  is fundamental  in  order  for a more complete understanding of  the  development  of  mother-infant interactions and therefore its importance for infant’s language acquisition.

QUESTION #2:

Categorical perception can either be independent of the listener's language (the auditory system may be more sensitive to some changes than to others) or it can be acquired as part of the process of learning a particular language. For example, animals such as chinchillas whose auditory systems resemble those of humans are found to be language-independent. On the other hand, while the ability to hear a difference between “r” and “l” is easy for English listeners, Japanese have much poorer performance unless they are given extensive training, which demonstrates it to be language specific. These language-specific skills are developed in infant’s speech perception. Newborns are able to distinguish many of the sounds of the world’s languages. They are also able to hear the similarities between sounds such as those that are the same vowels but have different pitches. By 12 months however their ability to respond to such a wide range of distinctions changes dramatically and infants no longer respond to some of those distinctions outside their native language. Both the identification and the discrimination test are tasks involved in demonstrating categorical speech perception and it shows that listeners will have a different sensitivity to the same relative increase in Voice Onset Time (VOT) depending on whether or not the boundary between categories was crossed. VOT is the length of time that passes between when a stop consonant is released and when the voicing, the vibration of the vocal folds, begins. For example, /b/ is perceived if the VOT of the bilabial consonant is less than 25ms, but when the VOT is between 25 and 60 ms then the consonant /p/ is perceived (Jay, 2003). Children, like adults, after varying the VOT tables, showed poor discrimination with a phoneme category but good discrimination across phoneme boundaries. Using the High Amplitude Sucking Technique, it was suggested that children get familiar to sounds within a category. Their pattern of sucking on a nipple signals a perceived change in a repeated sound. They would increase their sucking rate when the sounds shift categories (/b/ to /p/), meaning that there is a new stimuli and therefore a dishabituation. Infants increase their sucking rate after the between category shift and not after the within category shift, exhibiting infants being able to perceive speech in a categorical manner. The High Amplitude Sucking technique therefore can be constructively used to describe discrimination abilities as very young infants. Furthermore, it best illustrates the categorical speech perception in infants.

Part 3

1. FALSE, naturally deaf infants show distinct deficits in their early vocalizations and such individuals fail to develop language if not provided with an alternative form of symbolic expression. If these deaf children are exposed to sign language at an early age, they begin to “babble” with their hands just as a hearing infant babbles audibly. This suggests that regardless of the modality early experience shapes language behavior.

2. FALSE, it is Phase two when the infants' sensitivity to the distributional properties of linguistic input produces phonetic representations. Phase three is when the distortion of perception, which is the perceptual magnet effect, produces facilitation in native and a reduction in foreign language phonetic abilities.

3. TRUE, as we look at the Native Language Magnet Theory, both native and non-native performances at 7 months of age predict future language abilities but in opposite directions. The results support the view that the ability to discriminate non-native phonetic contrasts reflects the degree to which the brain remains in the initial state (Phase 1), open and uncommitted to native language speech patterns.

4. FALSE, Jean-Philippe can detect typical stress pattern in words resulting from statistical learning by 8 months old. When he is 6 months old, he is just introduced to statistical learning and he has only preference to language-specific perception for vowels.

5. TRUE, these infants were presented to words they heard before or nonwords(made up from the actual words) in the study and they found that infants listened the nonwords longer since they prefer novel stimuli. This shows that they knew the difference between them by learning the statistical probabilities of syllables in the words. On the other hand, the nonwords had different transitional probabilities, even though the syllables in the words and nonwords were the same illustrating that infants were counting the statistical frequency of one particular syllable's being followed by another. This study therefore reveals that infants can learn language and they can do it by using the statistical learning strategy.


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Microfluid Mechanics

Objectives
In natural and constructed microsystems, unfamiliar physics of the fluid flow have been discovered and utilized. In this course, we give an theoretical basis of microfluid mechanics and show example studies from the literature and conducted at the institute of fluid mechanics(LSTM-Erlangen) at FAU Erlangen-Nuremberg. In the theoretical part, the fluid models for gases and liquids are discussed, The validity of the Navier Stokes equations is questioned by checking the continuum hypothesis. Different regimes of fluid flow are presented based on continuum considerations. Intermolecular, intramolecular, surface as well as well as particle forces dominating in micro fluid devices are discussed. An overview of modeling strategies on molecular, meso and macro scale is given. Prominent examples and studies from the literature are presented regarding the micro fluid dynamics. Example studies conducted at LSTM-Erlangen are presented.

The students are expected to to understand the multiscaling and multiphysical nature flow phenomena in microsystems. The dominating forces in those systems will be made familiar to students. They should be able to conduct basic dimensional analysis to classify the flow with respect to the different flow regimes regarding the continuum hypothesis. Ultimately, they can decide which forces can be effective under certain gas, liquid and particle laden flows in Microsystems and which numerical modeling strategy is appropriate for a given microsystem. With the given applied examples from the literature and from our institute, it expected that students can establish the bridge between theory and application of the subject matter.

Basic components of the course
This course is dominnatly a theoretical one. However, examples are provided in the form of visual media, recitations and laboratory experiments in the form of experiments and computations. In the last part of the course, relevant examples of research work, which are conducted at LSTM-Erlangen, are provided.

Chapters:

 * Chapter 1. Introduction
 * Chapter 2. Intermolecular and Surface Forces
 * Chapter 3. Models for Gases and Liquids
 * Chapter 4. Fluid Dynamics
 * Chapter 5. Differential Analysis of Fluid Flow
 * Chapter 6. Analytical Solutions of Internal Flows
 * Chapter 7 Dimensional Analysis
 * Chapter 8. Scaling Effects and Governing Equations in Microflows
 * Chapter 9. Flow Phenomena in Microflows
 * Chapter 10. Measurements of Microflows(not offered yet)
 * Chapter 11. Modeling of Microflows
 * Chapter 12. Microflow Devices and Systems
 * Chapter 13. Example Studies Conducted at LSTM-Erlangen

Literature

 * Karniadakis, G., Beşkök, A., Aluru, N. R.: Microflows and nanoflows: fundamentals and simulation, Springer, New York, 2005.
 * Gad-El-Hak, M.: MEMS: Introduction and Fundamentals, CRS Press Taylor & Francis Group, 2006.
 * Israelachvili, J.: Intermolecular and Surface Forces, 2nd Ed., Academic Press, London, 1991.
 * Encyclopedia of Microfluidics and Nanofluidics, Springer, 2008.
 * Gad-el-Hak, M.: The ﬂuid mechanics of microdevices. J. Fluids Eng., 12(1):5–33, 1999.
 * Tabeling, P.: Introduction to Microfluidics, Oxford University Press, ISBN 978-0-19-958816-9,2010.
 * Bruus, H.: Theoretical Microfluidics, Oxford University Press, ISBN 978-0-19-923509-4,2008.
 * Nguyen, N.-T., Wereley, S.: Fundamentals and Applications of Microfluidics, Artech Print on Demand, 2nd ed, ISBN 1580539726, 2006.
 * Abgral, P., Nguyen, N.-T.: Nanofluidics, Artech House Publishers, ISBN 159693350X, 2009.
 * Sinha, S.K. and Dey, T.K.: Molecular Physics, Alpha Science, 2006.
 * Bird, G.A.: Molecular Gas Dynamics and the Direct Simulation of Gas Flows, Oxford Science Publications, 1994.
 * Chapman, S. and Cowling, T.G.: The Mathematical Theory of Non-Uniform Gases, Cambridge University Press, 1970.
 * Durst, F.: Fluid Mechanics: An Introduction to the Theory of Fluid Flows,Springer, 2007.

title: Microfluid Mechanics


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Swahili I/Grammar/Lesson 1 Exercises

Exercise 1, Ngeli
Study the noun classes and do the quiz.

{ What is the plural of mwaka (cl. 3)? { miwaka (i) }
 * type="{}"}

{ What is the singular form of watu (cl. 2)? { mtu (i) }
 * type="{}"}

{ What are the prefices for the gerund noun class (cl. 15)? { ku-, kw- (i) }
 * type="{}"}

{ What is the plural prefix of a noun beginning with n- (cl. 9)? { n- (i) }
 * type="{}"}

{ What do abstractions usually begin with (cl. 14)? { u- (i) }
 * type="{}"}

{ What is the plural of chuo (cl. 7)? { vyuo (i) }
 * type="{}"}

Exercise 2, Pronouns
I love you. -1- na -2- penda. You love me. -3- na -4- penda. We see him. -5- na -6- ona. He sees us. -7- na -8- ona.

Answers

 * 1) Ni-
 * 2) -ku-
 * U-
 * 1) -ni-
 * 2) Tu-
 * 3) -mw-
 * A-
 * 1) -tu-

title: Swahili I/Grammar/Lesson 1 Exercises


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Swahili I/Greetings/Lesson 2 Exercises

Exercise 1, Greeting Responses
Say the correct response in each situation and check your answers below when you are done.
 * 1) Mtu anakusema, "Shikamoo."  (Mtu = person/someone, a-na-ku-sema = HE/SHE/IT-(PRES.)-TO YOU-SAYS (IND.))
 * 2) "Hujambo!"
 * 3) "Hamjambo!"
 * 4) "Habari yako?" (Say you are doing well.)

Exercise 2, Review
Say the correct answer to the question and check your answers below when you are done.

 * 1) What is the plural of kitu?
 * 2) What is the singular form of ndoto?
 * 3) What is the word for school?
 * 4) How do you say, "to be?"
 * 5) How do you say, "childhood?"
 * 6) How do you say, "knife?"

Exercise 1

 * 1) Marahaba.
 * 2) Sijambo!
 * 3) Hatujambo!
 * 4) Nzuri, asante.

Exercise 2

 * 1) Vitu
 * 2) Ndoto
 * 3) Chuo
 * 4) Kuwa
 * 5) Utoto
 * 6) Kisu

Quiz
{ How do you say, "Thank you?" { Asante. (i) }
 * type"{}"}

{ How do you say, "How was your afternoon?" { Habari ya mchana? (i) }
 * type="{}"}

{ How do you say, "church?" { Kanisa (i) }
 * type="{}"}

{ How do you say, "to say?" { Kusema (i) }
 * type="{}"}

{ What is the object affix for you? { -ku- (i) }
 * type="{}"}

{ What is the object affix for he/she/it when in front of -ona? { -mw- (i) }
 * type="{}"}

{ What is the word for childhood? { Utoto (i) }
 * type="{}"}

{ What is the word for child? { Mtoto (i) }
 * type="{}"}

{ What is the groups prefix? { ma- (i) }
 * type="{}"}

{ How do you say, "a person?" { Mtu (i) }
 * type="{}"}

title: Swahili I/Greetings/Lesson 2 Exercises


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User:Eml5526.s11.team04.premchand/HW4

= Problem 4.2 - Weak form for heat conduction in one dimension =

Problem Statement
Obtain the weak form for the equations of heat conduction with the boundary conditions $$T(0)=100$$ and $$\bar{q}=hT$$ .The condition on the right is a convection condition.

Solution

w= weighted function

integrating by part yields

The weak form is

Find $$T(x)\epsilon \upsilon $$ such that

= Problem 4.6 - Strong and weak form of elastic bar problem with variable distributed spring =

Problem Statement
Consider an elastic bar with a variable distributed spring p(x) along its length .The distributed spring imposes an axial force on the bar in proportion to the displacement. Consider a bar of length l,cross-sectional area A(x),young's modulus E(x)with body force b(x) and boundary conditions as follows

a. Construct the strong form

b. Construct the weak form

Solution
First consider an element of this elastic bar.

From the force equilibrium i.e D'Alembert's principle,

taking the limit as $$\Delta x\rightarrow 0$$

for the governing ODE

Boundary conditions are

So the strong form is

The weak form derivation:

integrating by parts for the first term yields

which can be expanded as follows

Let w (0)=0 be the essential boundary condition.

The the weak form is as follows

Find u(x) smooth enough and satisfy u(0)=0,such that

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Talk:Psycholinguistics/Models of Speech Production

Chapter structure/ chapter body: The main section of the chapter is well worded, and for the most part each topic is clear. The body of the chapter provided much information that illustrates the main point of the chapter. The writer did well explaining the example of each model. The writer also did well connecting each subsection to the topic. However, some sections are not separates well and this confuses the reader. The writer might have to work on this part more to make the chapter more clear. I wound give this section an A-.

Organization: Most of the chapter is logical. The topics flow well into each other and each separate subsection is related well with other subsections. Most of the subsections are logical, and most provide the main point. However, there are some points that can be added or changed in the format of the chapter. As a reader, it is sometimes hard to distinguish which example is under the right section. If the writer changed the format of this chapter then this chapter would be better and more clear. I would give this section an A-.

Writing style: I mostly enjoyed reading this chapter, but there were a few grammar and spelling mistakes. However, this chapter was very well done and included much information. I will give this section an A-.

Research literature/ references: the references seemed very relevant to this chapter. The writer did not include many references but all of them are related with each subsection. Each reference presented different main points and some examples of each subsection. However, if the writer included more references that might bring up more viewpoints or drill deeper on each subsection, then it will be better chapter. Finally, the most part of the chapter is great, and I will give this section an A.

== educate yourself on the secrets to How to enhance boost the web advertising and marketing business o ==

The puppy has become a part of the u. s family and really should be treated with a similar think about that you'd treat your young ones or just mother and. normally,the dogs put on hugely popular advantages from the fatty acids in go fishing necessary oil. most may not find that gives like striped bass oil to one dog, even though the reports have shown may possibly complications of your dog end up being a mitigated on among them these healthiness supplements within their natural diet,


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User:Egm6341.s11.team1.arm/HW4

Problem 9
If $$\displaystyle G(t)$$ is given by Eq9.2

$$\displaystyle e(t)$$ is defined as

Therefore,

and

By Rolle's Theorem, this means that there exists a point $$\displaystyle \xi_1$$ such that

Looking at the $$\displaystyle G^{(1)}(t)$$,

As shown in Lecture 20 notes ,

Taking the derivative of $$ \displaystyle e(t) $$,

Since $$\displaystyle G^{(1)}(0) = 0$$ and $$\displaystyle G^{(1)}(\xi_1) = 0$$, by Rolle's Theorm there exists a point $$\displaystyle \xi_2$$ such that

Using the solution from Problem 4.3,

Since $$\displaystyle G^{(2)}(0) = 0$$ and $$\displaystyle G^{(2)}(\xi_2) = 0$$, by Rolle's Theorm there exists a point $$\displaystyle \xi_3$$ such that

Evaluating $$\displaystyle G^{(3)}(t)$$at $$\displaystyle \xi_3$$,

Using the solution from Problem 4.4,

And employing the Derivative Mean Value Theorem

Evaluating for $$\displaystyle e(1)$$, the proof breaks down because $$\displaystyle \xi_3$$ does not cancel out and remains unknown:

If $$\displaystyle G(t)$$ is given by Eq9.3

Analysis is similar to that of Eq9.4 - 9.17.

Eq9.17 then becomes:

The proof again breaks down when solving for $$\displaystyle e(1)$$ because $$\displaystyle \xi_3$$ does not cancel out:

Part B
Inserting Eq9.18 into the third derivative of Eq9.1,

Evaluating at $$\displaystyle t=0$$,

Since $$\displaystyle G^{(3)}(0)=0$$ and $$\displaystyle G^{(3)}(\xi_3)=0$$, by Rolle's Theorem, there exists a point $$\displaystyle \xi_4$$ such that

Evaluating the derivative,

Evaluating the derivative of Eq9.18,

Inserting into Eq9.26 and evaluating at $$\displaystyle t=0$$,

The proof can not be continued at this point since $$\displaystyle G^{(4)}(t)$$ has only one known zero because

Part 1
From Wolfram Alpha ,

Also, the asymptotic error estimate is defined for the trapezoidal rule in Atkinsion eq. 5.1.9 as:

Evaluating 10.2 with the appropriate derivatives for $$\displaystyle f$$,

Using the MATLAB code below, the following table was generated:

 Matlab code:

Part 2
From Wolfram Alpha ,

Also, the asymptotic error estimate is defined for the trapezoidal rule in Atkinsion eq. 5.1.9 as:

Evaluating 10.2 with the appropriate derivatives for $$\displaystyle f$$,

Using the MATLAB code below, the following table was generated:

 Matlab code:

Part 3
From Wolfram Alpha ,

Using the MATLAB code below, the following table was generated:

 Matlab code:

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User:Egm6341.spring-11.5.D/HW4

= Problem 4.6 - = From [[media:nm1.s11.mtg00.djvu|Mtg 00-0]]

Solution
 We solved this our selves

Doubble click on the thumbnail to view.

Author and proof-reader
[Author]

[Proof-reader]

title: User:Egm6341.spring-11.5.D/HW4


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Microfluid Mechanics/Modeling of Microflows

Introduction
The amount of available knowledge on a system and the complexity of the interacting physical effects are decisive role in the selection and development of a model of this system.

The multi-physical nature of the micro flows inflates the number of variables which have to be solved and/or controlled. At some stage, classical data analysis tools or models based on conservation laws can not be feasibly used to understand and/or monitor the process. Analysis tools should be selected depending on our knowledge of a system. If the system is well known, it can be modeled by conservation laws. As our knowledge of the system decreases we should select different approaches and at the low end of our knowledge only classification techniques might help. As regards to a process, there can be many sub-systems which can be modeled and analyzed by a technique appropriate for this system. Therefore, for a complex process having many subsystems, using hybrid models becomes not only feasible but also more accurate. In such a hybrid model, one can train artificial neural networks (ANN) via accurate simulations based on conservation laws and the experiments and integrate knowledge and experience based information to the hybrid by using fuzzy systems. If the hybrid is capable of reproducing the known phenomena and tendencies, this hybrid model can later be used to understand and optimize the process itself. Moreover, it can be further trained during the operation of the process.

As most of the micro flows comprise interacting multiple physical effects, mostly no analytical solution exists for them. Therefore, those flows are modeled mathematically in a form which can be numerically computed. As the knowledge on those flows are relatively high, models are based on mainly atomistic and continuum approaches both for liquid and gas flows.

Some of these methods are briefly explained in the ongoing text.


text:
Microfluid Mechanics/Introduction

>back to Chapters of Microfluid Mechanics

Surface forces vs. body forces
Microdevices tend to behave differently from the objects we are used to handling in our daily life  because of a simple reason: The influences of forces, which are functions of wetted area, decreases slower than those, which are functions of fluid volume.

Let $$ \displaystyle p_1(A) $$ and $$ \displaystyle p_2(V)$$ are any two property which scales with wetted area $$ \displaystyle (A)$$ and fluid volume $$ \displaystyle (V) $$. Their change relative to eachother with decreasing lengh scale of the system $$ \displaystyle(L)$$ can be written as:

$$ \displaystyle \frac{p_1(A)}{p_2(V)}\propto\frac{L^2}{L^3}\propto\frac{1}{L} $$

Typical order of magnitude is $$ \displaystyle 10^6\,\text{m}^2/\text{m}^3 $$. In other words, surface forces clearly dominates the body forces.

The unexpected effects starts to appear because the intermolecular forces either dominate or loose their influence. The former case appears to happen in liquid and multi-phase flows and the decrease of molecular forces happens mainly in rarefied gas flows. Therefore, it is important to understand these flows in terms of their molecular behavior. A key nondimensional parameter for microflows is the Knudsen number:

$$ \displaystyle Kn=\frac{\lambda}{L} $$

where $$ \displaystyle \lambda $$ is the mean free path of the molecules.

Owing to decreasing $$ \displaystyle L $$

 * Reynolds number number decreases $$ \displaystyle (Re<1) $$.

 * Knudsen number increases $$ \displaystyle (Kn>1) $$.

Natural micro devices
Insects are autonomous creatures which have been adopted themselves to the physical effects dominating in the micron world. For example, water strider stays over the water surface owing to the surface tension. The bee flight is totally different than many of the birds.

The locomotion of sperms is not similar to that of big fishes. The alveoli in the lungs ranges between 200 to 300 µm and responsible for the gas exchange. Reinflation of the alveoli following exhalation is made easier by pulmonary surfactant, which is protein mixture that reduces surface tension in the thin fluid coating within all alveoli.

Planktons provide a crucial source of food to larger, more familiar aquatic organisms such as fish and marine mamals.

title: Microfluid Mechanics/Introduction


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Talk:Psycholinguistics/Lexical Access

The plan of your chapter is very well structured. I think that if you stick with this outline, and find research that supports all of your subsections, this could be a strong paper. As it is right now, you are relying too much on the class text. Instead of citing Jay, look up the articles that he references to use as primary sources. You will likely find that these articles give you a fuller understanding of the concepts in the text, which will enable you to convey the material more clearly to your readers.

In your introductory section, I would start off by defining lexical access. I like how you have a link to the Mental Lexicon page, but I think it would be appropriate to premise that with a brief explanation of your own that engages the readers. You could use the “mental dictionary” analogy Dr. Newman talked about in class. This might give readers a basic way of conceptualizing lexical access.

When talking about word frequency, offer explanations for why frequency enhances the ease of lexical access. You could talk about the principle of least effort to support the idea that it takes less effort to access words that we read/use frequently. The following article has some good information on this idea: (http://www.pnas.org/content/100/3/788.full)

I like your bulleted definitions in your section on lexical ambiguity--they are clear and concise. However, you should include an introduction for the section and examples for each type of access and meaning. Also, it might be interesting to delve further into the selective access model by looking at levels of semantic processing. Here, you could describe the Stroop task, and even present the original Stroop task in your chapter for readers to experience how they access one word easier than another. Check out the Stroop task here: http://woldorfflab.ccn.duke.edu/files/pdfs/neuropsychologia2000.38.pdf

Generally, many of your sections simply need more explanation and examples. For instance, explain how concrete versus abstract words influence lexical access (don’t just state that differences exist). Also, a stylistic suggestion is to avoid writing in the first person (I, my). Too many first-person pronouns will make the chapter seem like a personal reflection or project, rather than a general learning resource. Another writing recommendation is to re-word the relationship you describe between words and their meaning in the Lexical Ambiguity section. I understand Jay describes the relationship as a ratio, but I feel like there is a better way to explain that the relationship between words and their meaning(s) is flexible, irregular, undefined, etc.

Like I said, sticking to your outline will give readers a solid understanding of lexical access. However, to make the chapter a little more interesting, you might consider including a section on deficits in lexical access. You could talk a bit about aphasia here, and/or link to the psycholinguistics page on aphasia. The following article explores the topic: Dell, G. S., Schwartz, M. F., Martin, N., Saffran, E. M., & Gagnon, D. A. (January 01, 1997). Lexical Access in Aphasic and Nonaphasic Speakers. Psychological Review, 104, 4, 801.

Finally, while writing, keep in mind that these chapters are intended for people in the general public. To make the content accessible and enjoyable to them, you need to make it relevant to them somehow. Do this by introducing your main ideas in fairly basic terms that anyone can understand. Try to find social references and examples of lexical access to draw readers in (and keep them interested). Then as you proceed, provide more details that help explain and validate your main idea.

I hope this helps as you continue working on your chapter. Good luck with the rest of it!

(129.173.209.105 20:08, 28 February 2011 (UTC))

________________________________________________________________________________________________________________________________

title: Talk:Psycholinguistics/Lexical Access


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Microfluid Mechanics/Models for Gases and Liquids

>back to Chapters of Microfluid Mechanics

>back to Chapters of Fluid Mechanics for Mechanical Engineers

Flowing medium
A fluid is composed of atoms and molecules. Depending on the phase of the fluid (gas,liquid or supercritical), the distance between the molecules shows orders of magnitude difference, being the largest in the gas phase and shortest in the liquid phase. As the distance between the molecules or the mean free path of the flowing medium approaches to the characteristic size of the flow device, the flow can not be treated as continuum.

In a solid, molecules form a regular lattice and oscillate around an equilibrium point. At this state, there is a strong attraction between the molecules and the kinetic energy of the molecules can not overcome this force in this phase of the matter. When enough energy is given to the molecules, e.g. by heating it, the matter melts and consequently becomes a liquid. The molecules gain kinetic energy as a result of added heat and start to move around in an irregular pattern. However, the density of liquids and solids, in other words the mean molecular distances at these two phases do not differ much from each other. When the liquid vaporizes and turns into the gas phase, the density drastically drops as the molecules starts to move freely between the intermolecular collisions.

Gas as a flowing medium
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Macroscopic models for gas flow is better suited when the flow can be accepted as continuum (see continuum mechanics). Gas flows can not be treated as continuum, when their density drops or when they flow through micro or nano channels which have characteristic dimensions at the order of the mean free path of the gas molecules. Therefore, It might be necessary to model gas flows in microdevices in microscopic (molecular) level.

Kinetic theory of gases
Kinetic theory is a molecular model for gas flows. It recognizes the gas as a collection of many discrete molecules and, ideally, provides information on the position, velocity and the state of the molecules. This is done by following each molecule by using Newton's laws of motion and allowing binary collision of molecules with the other molecules and and the bounding walls. Macroscopic properties of gases, like density, pressure and temperature, can be best understood by utilizing this approach.

Let us take a cubic container with a volume of $$\ \displaystyle l^3 $$, which is full of $$\  \displaystyle N $$ simple gas (monatomic) molecules, and derive some of the macroscopic properties. We let the molecules to collide with each other and with the walls of the container. This concept is valid under the condition that collisions can be only of binary type and the molecules do not induce electromagnetic forces on each other. The mathematical expression for this conditions is as follows:

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The number density of the gas molecules in the container is:

$$\ \displaystyle n=\frac{N}{l^3} $$

Hence, the average volume for one molecule in this container is $$\  \displaystyle 1/n $$, so the mean molecular spacing  $$\  \displaystyle \delta $$ reads

$$\ \displaystyle \delta=n^{-1/3} $$

Gas molecules can be modeled by hard spheres, which has the effective molecular diameter $$\ \displaystyle (d) $$. Accordingly, the fraction of volume occupied by a molecule is in the order of $$\ \displaystyle (d/\delta)^3 $$. For very low number densities, mean molecular spacing $$\ \displaystyle (\delta) $$ becomes relatively large compared to the effective diameter of the molecule $$\  \displaystyle (d) $$. Hence, a molecule can freely move outside region of influence of other molecules and almost all collisions of molecules can be accepted to be a binary one or collision with a wall. This situation can be written as:

$$\ \displaystyle \delta>>d $$

and defines the circumstances in a dilute gas.

Density
The mass of each molecule can be calculated from the molar mass of the gas $$\displaystyle M [kg\ mol^{-1}]$$ and the Avogadro constant  $$\displaystyle N_A=6.02214179(30) \times 10^{23} [mol^{-1}]$$

$$\ \displaystyle m=\frac{M}{N_A} $$

Hence, the density of the gas reads:

$$\ \displaystyle \rho=\frac{m N}{l^3} $$

Pressure
Pressure is explained by kinetic theory as arising from the force exerted by fluid molecules impacting on the walls of the container. When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed (an elastic collision), then the change of the momentum of the molecule ($$\Delta p$$) upon wall collision is:

 * $$\ \displaystyle \Delta p = p_{final,x} - p_{initial,x} = -m u -m u=-2 m u$$

where $$\ u $$ is the x-component of the initial velocity of the molecule.

The molecule impacts one specific side wall once every

 * $$\ \Delta t = \frac{2l}{u}$$

Hence the force exerted on the molecule by the wall is the rate of change of momentum per unit time:

 * $$\ F^* = \frac{\Delta p}{\Delta t} = -\frac{m u^2}{l}.$$

The force exerted by the molecule on the wall is in the negative direction:
 * $$\ F = \frac{m u^2}{l}.$$

Not all the molecules moves with the same speed. Say the speed of the molecules varies over the range $$\ u_{max}$$ and $$\ u_{min}$$, so the number of molecules moving over the range $$\ u+\Delta_u$$ is denoted by $$\ N_u$$. Using this distribution, it is possible to calculate the total force exerted by the molecules on the wall in the x-direction is the sum over all molecules:

\begin{array}{lll} F_{total, x} &= &\Sigma N_u \frac{m u^2}{l}\\ &&\\ &=& \frac{m}{l} N \frac{\Sigma N_u u^2}{N}\\ &&\\ &=& \frac{m}{l} N \overline{u^2} \end{array} $$ where $$\ \overline{u^2}$$ is the average of the square of x-component velocity of the molecules. As the pressure is the force per unit area, the x-component of the pressure reads

P_{x} = \frac{m}{l}N \overline{u^2}\frac{1}{l^2}=\frac{N m}{l^3} \overline{u^2} $$ Other components of the pressures can be similarly derived:

P_{y} = \frac{N m}{l^3} \overline{v^2} $$

P_{z} = \frac{N m}{l^3} \overline{w^2} $$

When the cubic container is small enough, the pressure on the walls of the container can be approximated by the average of them

\begin{array}{lll} P &=& \frac{1}{3}\left( P_x+P_y+P_z\right) \\ &&\\ &=& \frac{N m}{3 l^3}\left( \overline{u^2}+\overline{v^2}+\overline{w^2}\right)\\ &&\\ &=& \frac{N m}{3 l^3}\overline{c^2} \end{array} $$ where $$\ \displaystyle \overline{c^2} = \overline{u^2}+\overline{v^2}+\overline{w^2} $$ is the mean square velocity of the molecules of the simple gas.

Rearranging the terms shows that

Since $$\ N m$$ is the mass and $$\ l^3$$ is the volume, the pressure can also be written

$$\ \begin{array}{lll} P &=&\frac{1}{3} \rho \overline{c^2}\\ &&\\ &=& \frac{2}{3} E \end{array} $$

where $$\ E=\frac{1}{2}\rho \overline{c^2}$$ is the kinetic energy of the molecules per unit volume.

Temperature and kinetic energy
After defining the pressure by the kinetic approach, it is possible to define temperature also by utilizing the ideal gas law, which is:
 * $$\ P = \rho R_{\rm specific}T \ \ \text{or} \ \ P=n k_B T$$

where the specific gas constant is defined as
 * $$\ R_{\rm specific}=\frac{R}{M} \ \ \text{and} \ \ R=N_A k_B

\ \ \rightarrow \ \ R_{\rm specific}=\frac{k_B}{m} $$ in which $$\  R=8.314\,472(15)\ [J\ mol^{-1}\ K^{-1}]$$, $$\  M\ [kg\ mol^{-1}] $$, $$\  m\ [kg] $$ and  $$\displaystyle k_B=1.3806504(24) \times 10^{-23} [J K^{-1}]$$  are the gas constant, the molar mass of the gas molecules, mass of one molecule and  Boltzmann constant, respectively. It should be noted that ideal gas law is valid under ideal conditions, viz. at low pressure and at high temperatures, where $$\ \delta >> d $$, so that the molecular interaction is negligible.

Equating the pressure with the pressure derived by using kinetic approach
 * $$\ \frac{1}{3} \rho \overline{c^2}=\rho R_{\rm specific}T,$$

gives the temperature as:

However, this result is not justified as gas ceases to behave as a perfect gas before reaching to the absolute zero of temperature.

Other forms can be obtained by replacing the specific gas constant with $$\ R_{\rm specific}=\frac{k_B}{m}$$ as follows:

 * $$\ T=\frac{1}{3}\frac{1}{k_B} m \overline{c^2}$$

where $$\ m $$ is the mass of one gas molecule.

Mean free path
The distance traveled by a molecule between two successive collision is called the mean free path. Under the assumptions that the molecules are hard spheres and the probability distribution of the molecular velocities follows that of Maxwell distribution law, the mean free path can be derived as:

 * $$\ \lambda=\frac{1}{\sqrt{2}\pi d^2 n}$$

where $$\ d $$ and $$\ n $$ are the effective diameter of the gas molecule and the number density of the gas in the container.

Molecular speeds and the speed of sound
Although the macroscopic velocity under adiabatic conditions in the container is zero, the molecules are still moving around with the mean square velocity of $$\ \displaystyle  \overline{c^2}$$. The root mean square (rms) velocity of the molecules $$\ \displaystyle  c' $$ gives an idea of the speed of the molecules:

$$\ c'=\sqrt{\overline{c^2}}=\sqrt{\frac{3 P}{\rho}}=\sqrt{3 R_{\rm specific} T} $$

Two more statistically interesting speeds can be extracted from the Maxwell distribution of velocity:

\begin{array}{llll} \overline{c}&=&\sqrt{\frac{8}{\pi} R_{\rm specific} T} &\ \ \text{:average molecular speed}\\ c_p&=&\sqrt{2 R_{\rm specific} T} &\ \ \text{:most probable molecular speed} \end{array} $$

The speeds are ordered as follows:

 * $$ 0.886 \overline{c} = c_p < \overline{c} < c' = 1.085 \overline{c} $$

The speed of sound in an ideal gas is
 * $$\ c_s = \sqrt{\gamma \frac{p}{\rho}} $$

where $$\ \gamma=\frac{C_p}{C_v} $$ is the heat capacity ration and it takes the value 5/3 = 1.6667 for monoatomic and 7/5 = 1.4 for diatomic gas molecules. For ideal gas and air $$\ \gamma $$ is 1.4. Thus, in an ideal gas, the ratio between the speed of sound and rms velocity of the molecules is:

$$\ \frac{c_s}{c'}=\sqrt{\frac{\gamma}{3}} \approx 0.68 $$

which reveals that the speed of sound is slower than the all molecular speeds but it is at the same order of magnitude. This fact is not surprising while the waves are carried by molecular motion.

Viscosity
The viscosity of the fluid is responsible for the momentum transport in molecular level. The dynamic viscosity of an ideal gas, which is derived in the frame of kinetic theory, is:

$$\ \mu=\frac{1}{3}\rho \lambda \overline{c} $$

substituting the $$\ \lambda $$ and $$\ \rho $$ in the above equation results in

$$\ \mu=\frac{1}{3}\frac{m}{\sqrt{2}\pi d^2} \overline{c}=\frac{1}{3}\frac{m}{\sqrt{2}\pi d^2} \sqrt{\frac{8}{\pi} R_{\rm specific} T} $$

which states that the dynamic viscosity is only a function of the mass, the diameter and the average speed of the molecules, which is in turn a function of measured temperature. In other words, it is supposedly independent of pressure and density, which is in contradiction to experimental findings showing that viscosity is pressure dependent. In addition to that, this relation does not hold at very low pressures at which the mean free path reaches to the size of the container or flow apparatus and at very high pressures at which intermolecular forces can not be neglected. Furthermore, viscosity was observed to increase much faster than the $$\ \sqrt{T} $$ proportionality suggests. Nevertheless, this relation holds from a few mbars to several bars of pressure. A more accurate equation had been derived by using the intermolecular forces.

Liquid as a flowing medium
The molecules are too close to each other in a liquid, so that strong intermolecular interactions dominates the motion of the molecules. Moreover, the mean free path of the liquid molecules $$\ (\lambda) $$ are in the order of the mean molecular spacing $$\ (\delta) $$.

Definition of pressure
The pressure on a wall in liquids is a result of molecular collisions on a wall and the gravitation. In microfluidic systems, the effect of gravity is insignificant. However, the molecular collisions can not be easily modeled as was done for dilute gases by kinetic theory, because liquid molecules are densely packed and their motion highly influenced by the intermolecular interactions.

Viscosity
The derivation of viscosity using kinetic approach is not straightforward as the electromagnetic molecular interactions play dominant role in the liquids. One rough approximation was derived by Eyring and coworkers, which is:

\mu \approx \frac{N_A h \rho}{M} \exp\left(3.8 T_B / T\right) $$

where $$\ h $$ and $$\ T_B $$ are the planck constant and the boiling temperature of the liquid under one atmosphere, respectively. This equation, though being only an approximation, gives the observed decrease of viscosity with increasing temperature.


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