text:
Katie Harwood

Katherine "Katie" Harwood is a fictional character in the 2002 film Ghost Ship where an innocent young girl goes on a sea voyage of a lifetime, only to be caught up in a living nightmare aboard the ill-fated ocean liner. In the film, Katie is the supporting deuteragonist to the main character (Maureen Epps) and stands in stark contrast to the completely evil and demonic antagonist (Jack Ferriman). In many regards, Katie is just as much of a heroine as Maureen Epps for enduring unfathomable suffering and risking the wrath of Jack Ferriman through her unyielding efforts to save the souls and lives of others on the ship. Katie is portrayed by a young Emily Jane Browning.

Significance
Katie is an iconic representation of childhood from an earlier era, at a time when life was simpler, and childhood more innocent.

Synopsis
In May of 1962, an endearing Katie waves goodbye to her grandparents in Europe and journeys solo aboard the exquisite Antonia Graza on an exclusive cruise to rejoin her family in New York City. Since Katie is the only child onboard, she receives special care and attention. In describing her voyage Katie states, ”The whole ship was my playground. I was the only child onboard, but the ship’s purser and captain took special care of me. I felt so safe and happy with them.”

However, after the Antonia Graza rescues a lone survivor and cargo from a sinking ship, Katie’s exciting journey suddenly takes a horrendous turn for the worse. Unbeknownst to Katie and her shipmates, the single survivor (Jack Ferriman) is literally a demonic henchman for Satan set on destroying lives and collecting souls. Ferriman informs select members of the crew about the millions in gold recovered from his sinking ship, and influences them to launch an elaborate plot to seize control of the Graza and the gold, by killing everyone onboard. Although several crewmembers and passengers try to save her life from the murderous conspirators, Katie is eventually caught and tragically hanged to death, with her body concealed behind the partitioning door of her cabin.

Instead of ascending to her rightful place in heaven, Katie’s spirit is trapped on the ship by demonic forces along with the entire compliment of murdered passengers and crew. However, since Katie is still a young child, her soul is completely innocent and therefore beyond the influence and control of the demonic Ferriman. Over the next 40 years, the evil Ferriman along with the help of the wicked “marked” souls, try to use the ship as a trap to destroy unsuspecting lives, and collect a quota of souls for hell. Despite being the only flicker of good on the ship, Katie bravely opposes Ferriman and the evil spirits, and attempts to warn and save the lives of anyone who has the misfortune of being lured aboard the condemned vessel. Concerning these things, Katie says, “He can’t control me like he does the others and because of this Jack hates me most of all – Always terrorizing me and frightening me away when I try to warn those who come here.”

When Maureen Epps and her ship salving crew are lured aboard the Graza by Ferriman, Katie attempts to warn them about the dangerous ship. Katie has very perceptive eyes, and finds Epps to be more open to the truth than her cohorts. Katie tries to leave Epps messages and clues, and even appears to her on several occasions. When Katie senses that Ferriman has sabotaged the salvage crew’s tugboat, she boldly tries to warn the crew, but is forcibly carried away by Ferriman. Epps witnesses Katie’s warning and sets off in search for the mysterious girl.

Epps locates Katie’s cabin and comes face to face with her remains and her spirit. Katie allows Epps to have her cherished heart-shaped locket, and then proceeds to speak directly about the ship being a demonic trap. However, before Katie can finish explaining, she is overheard by an invisible Ferriman and lets out a frightened scream as he removes her from the cabin. Undeterred, Katie returns and tries to help Epps and her remaining crewmembers escape. Endowed with supernatural power from above, Katie transports Epps back to May 21, 1962 through her memories and reveals to her the horrifying events that led to everyone’s death, including her own. Furthermore, Katie reveals the true identity of Jack Ferriman and his goal to use the ship as a conduit to collect souls for Satan.

After experiencing Katie’s heartbreaking vision, Epps decides to risk almost certain death by destroying the cursed ship with explosives. However, Katie helps Epps escape the rapidly sinking vessel, while emancipated souls rise from the ship. Katie stays faithfully by Epps and gives her a grateful smile as her spirit ascends toward heaven.

Religious & Symbolic Implications
At its core, the story is deeply religious and the plot hinges on the concept of a young girl trying to save the lives (and ultimately souls) of others from demonic forces.

Metaphorical Murals
The religious implications and Christian symbolism surrounding Katie’s story are rich. Although only lightly touched on in the final version of the film, the most obvious religious connections come through the Gustave Dore inspired murals displayed throughout the ship. The murals are depictions based on Dante Alighieri’s Divine Comedy and The Inferno. The metaphorical murals hint at the struggle between Katie and Ferriman over the lives and souls on the ship. Lead actress Julianna Margulies says, “Katie is like Virgil in Dante’s Inferno. Margulies explains that Katie is like a “little guide” trying to lead others through hell and then safely onto the other side. As in the mural, Ferriman is literally a “ferry man” using a boat in the attempt to ferry souls to hell and eternal damnation. Conversely, Katie opposes Ferriman by trying to protect and warn all unsuspecting visitors and sharing the truth about the demonic ship.

Marked Souls
An additional concept presented in the plot are the “marked” souls. Like the murals in the story, the souls of the sinful and the “lost” are marked with a hooked-shaped indention on their hand. Katie says the mark is “a sign of their sins” and that they are bound to the will of Satan and his demons, in addition to eventually sharing in their same flaming judgment. However, not all the murals are somber. The depiction of angels defeating the demons and casting them down foreshadows the eventual defeat of Ferriman and ultimately all evil.

The Dove and Locket
Another subtle, but implicit Christian symbolism is Katie’s heart-shaped locket with the image of a raised dove in flight. In Christian tradition, this type of dove can symbolize innocence, purity, faith, hope, God’s peace, God’s presence through the Holy Spirit and God’s guidance and deliverance. The dove also indicated God’s bestowal of fortitude necessary to bear suffering and death. In this sense, the remains of the dead dove on the bridge of the ship, foreshadows Katie’s death, but at the same time indicates that there could still be hope. When Epps puts on Katie’s locket, it implies that Epps now has a renewed heart, or possibly even salvation as stated in Bible scripture. In this manner, the locket could also represent that Epps’ soul is not "marked".

Parallel of an Innocent and Sacrificial Life
A striking parallel could also be drawn between Katie and Jesus Christ. In the bible, Christ lives a completely innocent life, but is hung on a cross to sacrificially die for the sins of all humanity. Christ tastes the sting of death and hell in order to offer salvation to all who will put their faith in Him. Similarly, innocent Katie is hung by a rope behind a dividing door, but through her death she is able to lead others to "salvation" and safety. Because of her young age and innocent death, Katie is the only voice of hope for all the suppressed souls held captive on the demonic ship. Katie also endures small tastes of hell from Ferriman in her efforts to save others. However, Katie’s love and compassion for the lives of others is greater than her fear and pain, and in similarity to Christ, Katie rises victoriously to heaven in the end.

Forty Years and the Promised Land
40 years is a significant reoccurring number in the Christian faith. The Israelites endured 40 years of great trial and testing in the wilderness before they were able to enter their promised land of rest. In Christian tradition, the earthly life is often compared to a time of trial in the wilderness and the Promised Land a metaphor for heaven. Correspondingly, Katie has endured the “wilderness” and trials of the ship for 40 years, but as the result of Epps’ help she is inevitably freed to enter the “promised land” of heaven.

Concept of Free Will
Free will is both discussed and demonstrated in the story. Ultimately, the crew members that die are led to destruction by their own sins and their selfish desire for greed, lust and power. Epps is spared the fate of her shipmates through listening to Katie’s warnings and humbly responding to the truth.

Elevated Status of Children
In the Christian bible God often uses the most humble and unlikely people to work through – Out of hundreds on the ship, Katie is the meekest, humblest and most unlikely to do anything significant, yet every soul on the ship ends up depending on her.

Katie is also reminiscent of Christ elevating the status of children when he gave them special attention and profoundly declared that unless one humbles them self and has the faith of a little child, they will not enter the kingdom of heaven. This is further illustrated when Christ took a lowly child and placed him in the midst of his followers and stated that only those with childlike faith will be the greatest in the kingdom of heaven. This concept is demonstrated by Epps' willingness to humble herself to Katie's level.

The role of children in Christian scripture – Comparisons could be made between Katie and several unpretentious bible characters such as the lowly shepherd boy David who defeated the warrior Goliath and became king. Outwardly both young Katie and David appear to be nothing more that mere children, however, inwardly they possess vast unforeseen potential and a tremendous strength of character.

Eternal Destinations
In the film, as with Christian soteriology there are two clear eternal destinations which souls go to after death; ether heaven or hell. A heart of child-like faith will put one on the path to heaven, but a sinful life eventually leads to destruction.

The Destructiveness of Sinful Choices
One of the morals of the story is the corrupting dangers of pursuing riches. This theme also fits aptly within the religious concepts of the film. Sinful greed, lust, debauchery and selfish desire can influence ordinarily good people to do horribly evil things (1Timothy 6:9-10). The choices made by the characters in the story illustrate this principle.

“Be sure your sins will find you out” – As the story goes, sinful activities have consequences and will eventually have to be accounted for, and the ultimate payment for sin in the end is both physical and spiritual death (Romans 6:20-23).

Loving Others before Ourselves
“Greater love has no one than this – To lay one’s life down for their friends.” Katie and Epps are both willing put themselves at risk to save lives. Through the process of trying to help others both Epps and Katie experience help for themselves.

“Those who seek to save their life will lose it, but those who are willing to lose their life will save it.” In this sense Epps' life is saved, because she places the needs of others first. Katie also inherits eternal life in heaven by risking herself to help others.

Purgatory
The concept of purgatory (Whalen p. 1034-1039) is also mentioned when Katie states that the ship had become like a “prison” where souls are trapped among the living between heaven and hell.

Life as a Voyage on a Ship
Life itself is comparable to a voyage on ship...illustrating that our lives are like a ship that is bound for one of two possible destinations. In the context of Ghost Ship, the voyage ends in destruction (death and hell) unless we get off. In this sense, Katie pleads with us to get off, but that involves humbly changing our ways and giving up the allure and pursuit of vain things such as wealth and power.

Symbols of Changing Times
The once elegant and decaying ship is a symbol of changing times. In a similar manner, Katie can be viewed as a symbolic representation of childhood from a more innocent era. Just as the ship decays to ruin from its glorious 1950’s splendor, Katie’s deteriorating childhood possessions and even Katie herself, are evocative symbols of changing times and eroding childhood innocence.

Dichotomy and Contrasts
Katie is in some ways a complicated paradox: She has been deeply hurt and traumatized by all she has experienced, but by the same token, she is still very much an innocent child – Even after 40 years of resilient growth as a character. Katie exhibits a wide array of emotions from timid youthful innocence and joy, to extreme terror, sadness, loneliness, and solemn distress.

Characters
Even under tremendously trying circumstances Katie demonstrates an incredible inner strength of character without compromising the sweet kind-hearted "girly" girl she really is. The fact that Katie remains true to herself in spite of all the evil and suffering she experienced transcends their eroding influence and distinguishes her from the other characters.

Ferriman
Ferriman is representative of Satan and the crafty deceptiveness of evil in destroying people’s lives. Katie and Ferriman are at completely opposite ends of the Christian spectrum – The demonic Ferriman is totally depraved and evil while Katie is sweet and innocent. Katie is motivated by love and compassion, Ferriman is motivated by hate and rage.

Epps
Although Katie and Epps are two completely dissimilar characters from different eras, they still grow quite close. Epps is "a woman working in a man's world" and appears somewhat tough and masculine on the exterior like a tomboy. Although not specifically stated, the story indicates that Epps is likely from a broken home. On the other hand, Katie comes from a close, traditional nuclear family. In contrast to Epps, Katie is more of a typical "girly" girl and is slightly timid, but possess a stout inner fortitude despite her diminutive and delicate appearance.

Francesca
The sultry Francesca lives the more representative life of a performing artist. Unlike Epps who feels the need to downplay her femininity, Francesca leans toward the other extreme and attempts to flaunt her femininity to her own advantage, even to the point of her own exploitation. Conversely, Katie is still a virtuous and modest young girl, who has yet to be tainted by any corrupting influences of the world.

Captain Murphy
Captain Murphy appears to be a strong imposing ship captain on the outside, however, when confronted with trials and tribulations, his perceived strength is shown to be superficial. This is evident by his emotional breakdown where he shuts off emotionally and withdraws to himself while attempting to drown his problems in alcohol. In contrast, Katie appears to be very delicate in both physical appearance and personality. However, despite her dainty appearance, Katie proves her true strength and fortitude from within through dealing with her pain and problems directly.

General Themes
Light vs. darkness, good vs. evil, childhood innocence vs. wickedness, tragedy vs. triumph, simpler more wholesome times, vs. the complexity and loss of innocence of today.

Conclusion
On the surface, Katie’s story is an emotional tale of heartache and tragedy – Something that no young person should ever have to endure. However, Katie’s story is also essential to the plot. Through helping each other and being willing to put themselves at risk, Katie and Epps together achieve a triumph out of the tragedy and deliver a major blow to Ferriman and the forces of evil. Although the struggle over the souls of people will continue, this victory of the human spirit gives hope even in the face of terrible suffering, evil and even death.

“Katie is really a very sweet girl. She's completely innocent…“ says Emily Jane Browning, who obviously became completely immersed in her character. “…She's been hoping someone would come onto the ship to be her friend, so when Epps arrives she's very excited – they develop a real friendship."

”Emily did an amazing job,” Director Steve Beck enthuses. “She gave Katie a real complexity…she's not just a little girl caught up in a ghost story…” Both Browning and Beck also added that underneath all her heartfelt emotions, Katie longs for vindication and justice.

The deeply symbolic and metaphorical aspects surrounding Katie’s story along with the close relationships she develops with others lends support to the notion that the film was originally intended to be more than just another Hollywood “slasher” horror. This is further substantiated by Katie’s tearful and heartfelt story – Emotionally poignant and touching drama not usually found in a horror picture.

title: Katie Harwood


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

title: File:C.Repeat.Loop.1.A.20120113.pdf


text: Template:Refbegin

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Template:Refbegin/doc

refbegin and its complementary closing template refend are used to apply any of several optional styles or presentational choices for bibliographies or reference lists—that is, lists of fully-specified references / sources / works cited, alpha-sorted by author/contributor) that appear in Wikiversity resources.

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title: Template:Refbegin/doc


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text:
Ludic Science Club/OTRS

"Equilibrium is soon established by a stream of volunteers."
When I attended the Open-source Ticket Request System organised by Wikimedia UK, I was reminded of a science fiction story I read during the mid-winter break. Red Star was written over a hundred years ago by Alexander Bogdanov. At the time of its first publication in 1908, he was an exile from Imperial Russia who had just broken with Lenin to work with Gorky and others around the Vpered organisation. He was to go on to develop Tectology, which is often consider as a precursor to cybernetics. So perhaps this should come as no surprise.

Red Star gives an account of an earthling, an engineer called Leonid, who visits Mars to discover a socialist society organised around free labour. Leonid asks how production is organised. He was shown an array of production statistics:
 * "The machine-building industry has a surplus of 968,757 man-hours daily of which 11,325 days are skilled labour. The surplus at this factory is 753 hours, of which 29 hours are skilled labour. (...) The clothing industry has a shortage of 392,685 man-hours daily, of which 21,380 hours require experienced repairmen for special machines and 7,852 hours require Organization."''

There were similar figures on a range of other tables covering not merely production but also education, medicine. When Leonid asks how thsystem works his guide ofers an explanation:
 * "The tables are meant to affect the distribution of labor. If they are to do that everyone must beable to see see where there is a labour shortage and just how big it is. Assuming an individual has the same or an approximately equal aptitude for two vocations, he can choose the one with the greater shortage. (...) the figures change every hour. In the course of an hour several thousand workers announce that they want to change jobs. The central statistical apparatus takes constant note of this, transmitting the data to all branches of industry. (...) The Institute of Statistics has agencies everywhere which keep track of the flow of goods into and out of the stockpiles and monitor the productivity of all enterprises and the changes in their work forces. In tah way it can be calculated what and how much must be produced for any given period and the number of man-hours required for the task. the Institute then computes the difference between teh existing and the desired situation for each vocational area and communicates the result to all places of employment. Equilibrium is soon established by a stream of volunteers."

This is, essentially, how the OTRS system works on the Wikimedia projects.

OTRS chez Wikimedia
The role of the OTRS system is to maintain a database of enquiries, whether by e-mail or phone, in such a way that a diffuse network of volunteers from across the world can give suitable replies. The enquiries are organised in different queues which reuire different skill sets to handle. Thus the courtesy list is generally quite straigh forward enquiries which an experienced wikimedia editor should not find to hard to respond to. Other lists such as the quality list are seen as needing a more considered response, although sometimes these can be quite straight-forward, but the ticket got put here because someone wasn't quite sure how to deal with it. There is also a list to deal with Sister projects such as Wikiversity (although there have not been many enquiries concerning Wikiveristy lately). Sometimes it is just a matter of giving a polite response to an enquiry, perhaps pointing the person to the relevant help page, such as Am I allowed to edit articles about myself or my organization?

One thing which comes through OTRS is e-mail confirmation that the producer of an image grants the relevant license for it to be available on the particualr Wikimedia project. On the other hand sometimes people are complaining of the opposite: that an image has been used without their consent and that theyw ould rather like it removed.

Another area concerns Biographies of living persons, often shortened to BLPs. This is a particularly sensitive area:

 * "We must get the article right. Be very firm about the use of high quality sources. All quotations and any material challenged or likely to be challenged must be explicitly attributed to a reliable, published source, which is usually done with an inline citation. Contentious material about living persons that is unsourced or poorly sourced—whether the material is negative, positive, neutral, or just questionable—should be removed immediately and without waiting for discussion. Users who persistently or egregiously violate this policy may be blocked from editing."

Of course, what the person requests is not always the most appropriate response. But what the volunteer aims to do is to work out what the best response is, in accordance with the releavant Wikimedia projects policies, to see what they can do to implement that response and then to explain the response to the person who made the enquiry.

As many of the enquiries require similar responses, a selection of templates have been developed on the OTRS wiki (which is private to accredited OTRS agents). This enables the "agent" to modify the template so as to provide a personalised response.

title: Ludic Science Club/OTRS


text:
Talk:Ludic Science Club/OTRS

 * Metrics Jan 2012
 * Metrics Feb 2012
 * Metrics March 2012
 * Metrics April 2012

title: Talk:Ludic Science Club/OTRS


text:
User:EGM6321.f09.team1.Zhichao Gong/Mtg8

Mtg 8: Thu, 10 Sep 09

page8-1

Application: General non-homogenous L1_ODE_VC

$${\color{red}(1)} \ P(x){y}^{'}+Q(x)y=R(x)$$

$$If \ P(x) \neq 0 \ \forall \ \Rightarrow$$

$${\color{red}(2)}$$

$${\color{blue} \underset{ \overset{ \uparrow}{N(x,y)}}}+{\color{blue} \underset{M(x,y)}{ \underbrace}} \ {\color{blue}p.4-2} \ {\color{red}Eq.(3)}$$

$${\color{blue} \underset{ \overset{||}{1}}}({\color{blue} \underset{ \overset{||}{0}}}-{\color{blue} \underset{ \overset{||}{{a}_{0}(x)}}})=-f(x) \ {\color{red}Case1}$$

$$\Rightarrow$$

$$\Rightarrow$$

page8-2

$${\color{blue}p.8-1} \ {\color{red}Eq(2)} \ {y}^{'}+{a}_{0}y=b \ {\color{red}(1)}$$

$$Mult \ by \ h: \ h({y}^{'}+{a}_{0}y)=hb \ {\color{red}(2)}$$

$$Recall \ \frac{{h}_{x}}{h}= \underset{ \overset{||}{{a}_{0}}}{f} \ {\color{blue}p.6-3} \ {\color{red}(2)}$$

$$\Rightarrow \ h{a}_{0}={h}_{x}={h}^{'} \ {\color{red}(3)}$$

$${\color{red}Eqs \ (3) \ and \ (2)} \ {\color{blue} \underset{{(hy)}^{'}}{ \underbrace}}$$

$$\Rightarrow$$

$$\color{red}(4)$$

Application: Non-homoge L1.ODE.VC

$${y}^{'}+{\color{blue} \underset{{a}_{0}}{ \underbrace}}={\color{blue} \underset{b(x)}{ \underbrace}} \ {\color{red}(5)}$$

$${\color{blue}HW:} \ h(x)=x \ {\color{red}(6)}$$

page8-3

$$Mult \ (5) \ by \ h(x): \ {\color{blue} \underset{\frac{d}{dx}(xy)}{ \underbrace}}+y={x}^{3}$$

$${\color{blue}\underline{HW:}} \ y=\frac{{x}^{3}}{4}+\frac{C}{x}$$

$$C=constant$$

Recall HW on p.4-2 A class ofN1_ODEs

p,6-4 Eq(2): particular case 1.

$$One \ possibility$$

$$\ N(x,y)=N(x)$$

$$\ {N}_{x}=b(x)$$

$$\ {M}_{y}=a(x)$$

$${M}_{y}(x,y)=a(x) \ \Rightarrow \ M(x,y)=a(x)y+k$$

$${N}_{x}=b(x) \ \Rightarrow \ N(x,y)=\int_{}^{x}b(s)ds=c(s)$$

page8-4

$$\ Mdx + Ndy = [a(x)y + k]dx + c(x)dy = 0$$

Exact L1_ODE_VC


text:
User:Egm4313.s12.team17.ying

Reports
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Upload page

Math formulas

title: User:Egm4313.s12.team17.ying


text: The Quellex Method/Assignments

title: The Quellex Method/Assignments


text:
User:EGM6321.f09.team1.Zhichao Gong/Mtg10

Mtg 10: Tue, 15 Sep 09

page10-1

Note: HW p.9-3, p.9-2: "exact"

means "either exact or exactly inte-" grable by int. fact. method" p.9-4: The exact func. phi is called a first intergral of ODE F(x, y, y' ...)=0

$${\color{blue}p.9-4} \ {\color{red}(4):} \ F(x, \ y, \ {y}^{'}, \ {y}^{''})$$

$$=\frac{d}{dx}\phi(x, \ y, \ )={\phi}^{x}+{\phi}^{y}{\color{blue} \underset{p}{ \underbrace}}+{\phi}^{p}{\color{blue} \underset{{p}^{'}}{ \underbrace}}$$

$$F=f(x, \ y, \ p){y}^{''}+g(x, \ y, \ p) \ {\color{red}(1)}$$

$$f(x, \ y, \ p):={\phi}^{p}(x, \ y, \ p) \ {\color{red}(2)}$$

$$g(x, \ y, \ p):={\phi}^{x}+{\phi}^{y}p \ {\color{red}(3)}$$

Eq(1): 1st condition of exactness.

page10-2

Application: N2_ODE not exact

$${\color{blue} \underset{f}{ \underbrace}}{{y}^{''}}^{\frac{3}{2}}+{\color{blue} \underset{g}{ \underbrace}}=0$$

$$2nd. \ condition \ of \ exactness:$$

If both exactness condition hold, then

$$obtained \ from \ {\color{red}(2)} \ {\color{blue}p.10-1}$$

page10-3

Application: Consider

$${\color{blue} \underset{f(x, \ y, \ p)}{ \underbrace}}{y}^{''}+{\color{blue} \underset{g(x, \ y, \ p)}{ \underbrace}}=0$$

1st exactness condition. satisfied.

HW: 2nd exactness condition also satisfied:

$${f}_{x}=y \ \Rightarrow \ {f}_{xx}=0$$

$${g}_{y}=p \ \Rightarrow \ {g}_{yp}=1$$

$$A \ first \ int. \ of \ ODE \ is: \ f=\frac{ \partial \phi}{ \partial p}$$

$$\Rightarrow \ \phi(x, \ y, \ p)=h(x, \ y)+\int_{}^{}(x, \ y)dp$$

$$=h(x, \ y)+xyp$$

Use defn of gp.10-1 Eq(3):

$$g(x, \ y, \ p)={\phi}_{x}+{\phi}_{y}p$$

$$={h}_{x}+{h}_{y}p$$

$$observe \ if \ {\color{red}\underline{{\color{black}h={k}_{{\color{blue}1}}}const}}\Rightarrow \ {h}_{x}={h}_{y}=0$$

page10-4

$$1st \ int. \ \underset{ \overset{||}{{k}_{\color{blue}1}+xyp}}{\phi(x, \ y, \ p)}={\color{blue} \underset{ \overset{||}{{k}_{2}}}}$$

$$\Rightarrow \ (xy){y}^{'}=c-k$$

$$\Rightarrow$$

$$\frac{{y}^{2}}{2}=({\color{red} \underset{ \overset{\uparrow}{Constant}}}-{\color{red} \underset{ \overset{\uparrow}{Constant}}})logx+ {\color{blue} \underset{ \overset{|| \ {\color{red}constant}}{{k}_{3}}}}$$

$$\frac{{y}^{2}}{2}=({k}_{2}-{k}_{1})logx+{k}_{3}$$


text:
User:Egm4313.s11.team11.sheider

I am Jon Sheider and have no idea how to fluidly use wikiversity.

title: User:Egm4313.s11.team11.sheider


text:
User:EGM6321.f09.team1.Zhichao Gong/Mtg12

Mtg 12: Thu, 17 Sep 09

page12-1

p.11-3

$${\color{red}Eq(2):}$$

$${\color{blue}\underline{HW:}} \ use \ {\phi}_{xy}={\phi}_{yx} \ to \ obtain \ {\color{blue}p.10-2} \ {\color{red}Eq(4)}$$

1st rel. in exactness condition 2 for N2. ODEs

Application: ZGenerate exact N2. ODEs (reverse engineering)

$${\color{blue}\underline{Method:}} \ {\color{blue}1)} \ Select \ \phi \ (x, \ y, \ {\color{blue} \underset{ \overset{||}{{y}^{'}}}})$$

$$F(x, \ y, \ {y}^{'}, \ {y}^{''})=\frac{d}{dx}\phi (x, \ y, \ {y}^{'})$$

$$Consider: \ \phi(x, \ y, \ p)=4{x}^{5}{p}^{2}+2{x}^{2}y$$

$${\phi}_{x}=20{x}^{4}{p}^{2}+4xy \ {\phi}_{p}=8{x}^{5}p$$

$${\phi}_{y}=2{x}^{2}$$

$$(8{x}^{5}{y}^{'}){y}^{''}+2{x}^{2}{y}^{'}+20{x}^{4}{{y}^{'}}^{2}+4xy=0$$

page12-2

Show above N2. ODE exact.

$${\color{blue}\underline} \ Condition \ 1:$$

$$f=8{x}^{5}p, \ g=2{x}^{2}p+20{x}^{4}{p}^{2}+4xy$$

$${\color{blue}\underline} \ Condition \ 2:$$

p.10-2 Eqs (4) and (5)HW.

$$A \ first \ int. \ : \ f={\phi}_{p}$$

$$\Rightarrow \phi=h(x,y)+{\color{blue} \underset{b}{ \underbrace{\color{black}\int_{}^{}{\color{blue} \underset{8{x}^{5}p}{ \underbrace}}dp}}}$$

$$g={\phi}_{x}+{\phi}_{y}p \Rightarrow$$

$${\color{red}\underline{\color{black}2{x}^{2}p}}+{\color{blue} \underset{get \ rid \ of}{ \underbrace}}+4xy=({h}_{x}+{\color{blue} \underset{get \ rid \ of}{ \underbrace}})+({h}_{y}+0){\color{red}\underline{\color{black}p}}$$

$$Assume {h}_{y}=2{x}^{2} \Rightarrow \ h=2{x}^{2}y+k(x)$$

$${h}_{x}=4xy=4xy+{k}_{x} \ \Rightarrow \ {k}_{x}=0$$

page12-3

$$\Rightarrow \ k(x)=k \ Constant$$

$$h(x,y)=2{x}^{2}y+k$$

$$\phi(x, \ y, \ p)=2{x}^{2}y+4{x}^{5}{p}^{2}+k$$

$${\color{blue}\underline{\color{black}Non-exact \ L2-ODE-VC:}}$$

$${\color{blue}\underline{\color{black}Application:}} \ \sqrt{x}{y}^{''}+2x{y}^{'}+3y=0 \ {\color{red}(1)}$$

HW: Verify exactness of above ODE.

Recipe: For special power(No sinx, cosx ...) form

$${\color{blue}L1-ODE-VC:} \ {\color{red}(Actually, \ N1-ODE-VC \ for \ b,d \neq 0)}$$

$${x}^{a}{y}^{b}(py+qx{y}^{'})+{x}^{c}{y}^{d}(ry+sx{y}^{'})$$

$$where \ (a, \ b, \ c, \ d), \ (p, \ q, \ r, \ s)\in \ R$$

$${\color{blue}L2-ODE-VC:} $$

$$\ \alpha{x}^{r}{y}^{''}+\beta{x}^{s}{y}^{'}+\gamma{x}^{t}y$$

$$where \ (\alpha, \ \beta, \ \gamma),(r, \ s, \ t) \in \ R$$

$$Consider \ int. \ feet. \ h(x,y)={x}^{'}{y}^{}$$

$$where \ (m, \ n)\in \ R \ to \ be \ determined$$

page12-4

Back to applp,12-3 Eq(1), final (m,n) st:

$$({x}^{m}{y}^{n})[\sqrt{x}{y}^{''}+2x{y}^{'}+3y]=0 \ is \ exact.$$


text:
User talk:Marshallsumter/Keynote lectures (draft)/Astrophysics

Possible Correction in definition
In the Astronomy section of the resource, there is a definition for Luminosity as " let the symbol L_{\odot} indicate the solar mass.". In this definition, I believe that it's meant to be written as " let the symbol L_{\odot} indicate the solar luminosity." since it's a definition of Luminosity. Fotte (discuss • contribs) 10:11, 26 November 2014 (UTC)

Dispute the speedy deletion request
Custodians or Curators considering Speedy deletion of this resource, please see this same section title on Draft talk:Craters by radiation. --Marshallsumter (discuss • contribs) 04:00, 4 August 2018 (UTC)

Moving back to draft space
As per Requests_for_Deletion/Archives/16 I will move this back to Draftspace. If you want, we can revisit that decision.--Guy vandegrift (discuss • contribs) 19:57, 5 August 2018 (UTC)


text:
Talk:Dominant group/Astronomy

Dead link
During several automated bot runs the following external link was found to be unavailable. Please check if the link is in fact down and fix or remove it in that case!

 * http://www.sciencedirect.com/science/article/pii/S0168822708003355
 * In Dominant group/Astronomy on 2012-01-08 08:37:46, HTTP Redirect Loop: ->  ->  ->  ->
 * In Dominant group/Astronomy on 2012-01-15 08:37:46, HTTP Redirect Loop: ->  ->  ->  ->

--Mu301Bot 08:38, 15 January 2012 (UTC)
 * Link working fine. It goes to ScienceDirect and displays the Abstract. I've added pmid from Pubmed to reference. Marshallsumter 17:45, 15 January 2012 (UTC)

title: Talk:Dominant group/Astronomy


text:
User:Egm4313.s12.team17.ying/Test Link

Part 2-1 Solution
Using Eq.3.1.8, the function $$r_n(x)$$ can be developed for each value of n (4,7,11). We will first solve for $$y_n(x)$$ when n = 4.

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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.1) Where n = 4:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$     (Eq.3.2.2) Thus, the particular solution to the Eq. #.#.# in the Part 2 given is then:
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$$  \displaystyle y_{n,p} = \sum_{i=0}^4 c_i x_i = c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.3)
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Deriving Eq.3.2.3 twice will yield the following:

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$$  \displaystyle y'_{n,p} = 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.4)
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$$  \displaystyle y''_{n,p} = 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.5)
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To solve for the particular solution y_{n,p), Eq.3.2.4 and Eq.3.2.5 will be substituted into the following particular equation.

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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = 12c_4x^2 + 6c_3x + 2c_2 - 3(4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$     (Eq.3.2.6)
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Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle (2c_4x^4) + (2c_3x^3 - 12c_4x^3) + (-9c_3x^2 + 2c_2x^2 + 12c_4x^2) + (2c_1x - 6c_2x + 6c_3x) + (2c_2 + c_0 - 3c_1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$     (Eq.3.2.7)
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The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.

The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4$$

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$$  \displaystyle c_0 = -4.0625 $$
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$$  \displaystyle c_1 = -4.1250 $$
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$$  \displaystyle c_2 = -2.1250 $$
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$$  \displaystyle c_3 = -0.5833 $$
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$$  \displaystyle c_4 = -0.1250 $$
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Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:

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$$  \displaystyle y_{n} = y_{n,h}(x) - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.8)
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Where $$y_{n,h}(x)$$ is solved by setting Eq.#.#.# to 0 and finding the roots

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$$  \displaystyle y'' - 3y' + 2y = 0 = \lambda^2 - 3\lambda + 2 = 0 $$     (Eq.3.2.9)
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$$  \displaystyle y'' - 3y' + 2y = 0 = \lambda^2 - 3\lambda + 2 = 0 $$     (Eq.3.2.10)
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$$  \displaystyle (\lambda - 2)(\lambda - 2) = 0 $$     (Eq.3.2.11)
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$$  \displaystyle \lambda = 2,1 $$     (Eq.3.2.12)
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Therefore the homogeneous solution is then:

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$$  \displaystyle y_{n,h}(x) = c_6e^{2x} + c_5e^x $$     (Eq.3.2.13)
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Which then generates the final solution $$y_n(x)$$

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$$  \displaystyle y_{n} = c_6e^{2x} + c_5e^x - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.14)
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Thus to solve for the final coefficients I used a matlab code to generated matrices to solve for the final two unknowns

The values then come out to be:
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$$  \displaystyle c_6 = -5.5868 $$
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$$  \displaystyle c_5 = 8.9001 $$
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$$  \displaystyle y_{n}(x) = -5.5868e^{2x} + 8.9001e^x - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.15)
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Part 2-2 Solution

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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.16) Where n = 7:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} $$     (Eq.3.2.17)
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$$  \displaystyle y_{n,p} = \sum_{i=0}^7 c_i x_i = c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.18)
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Deriving Eq.3.2.18 twice will yield the following:

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$$  \displaystyle y'_{n,p} = 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.19)
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$$  \displaystyle y''_{n,p} = 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.20)
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To solve for the particular solution y_{n,p), Eq.3.2.19 and Eq.3.2.20 will be substituted into the following particular equation.

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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = $$
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$$  \displaystyle 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 - 3(7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$
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$$  \displaystyle - 3(7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$
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$$  \displaystyle + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$ Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle 2c_7x^7 + (2c_6x^6 -141c_7x^6) + (2c_5x^5 - 18c_6x^5 + 42c_7x^5) + (2c_4x^4- 15c_5x^4+30c_6x^4 ) + (2c_3x^3 - 12c_4x^3 + 20c_5x^3) + (2c_2x^2 - 9c_3x^2 + 12c_4x^2) $$
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$$  \displaystyle + (2c_1x - 6c_2x + 6c_3x) + (2c_0 - 3c_1 + 2c_2) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} $$     (Eq.3.2.22) The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.
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The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4,c_5,c_6,c_7$$

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$$  \displaystyle c_0 = 617.6875 $$
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$$  \displaystyle c_1 = 615.3750 $$
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$$  \displaystyle c_2 = 305.3750 $$
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$$  \displaystyle c_3 = 100.4167 $$
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$$  \displaystyle c_4 = 24.3750 $$
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$$  \displaystyle c_5 = 4.6000 $$
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$$  \displaystyle c_6 = 0.6667 $$
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$$  \displaystyle c_7 = 0.0714 $$
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Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:
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$$  \displaystyle y_{n} = y_{n,h}(x) + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.23)
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The homogeneous solution is the same:
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$$  \displaystyle y_{n,h}(x) = c_9e^{2x} + c_8e^x $$     (Eq.3.2.24)
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Which then makes the solution:

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$$  \displaystyle y_{n} = c_9e^{2x} + c_8e^x + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.25)
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Using the same matlab code but adjusting certain formulas, $$c_9,c_8$$ can be found:

The following values for $$c_9,c_8$$ are obtained:
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$$  \displaystyle c_9 = -3.3774 $$
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$$  \displaystyle c_8 = -615.0789 $$
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The final solution is then:
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$$  \displaystyle y_{n}(x) = -3.3774e^{2x} -615.0789e^x + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.26)
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Part 2-3 solution

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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.27) Where n = 7:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} - \frac{x^8}{8} + \frac{x^9}{9} - \frac{x^10}{10} + \frac{x^11}{11} $$     (Eq.3.2.28)
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$$  \displaystyle y_{n,p} = \sum_{i=0}^11 c_i x_i = c_{11}x^{11}+c_{10}x^{10}+c_9x^9+c_8x^8 + c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.29)
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Deriving Eq.3.2.28 twice will yield the following:

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$$  \displaystyle y'_{n,p} = 11c_{11}x^{10} + 10c_{10}x^9 + 9c_9x^8 + 8c_8x^7 + 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.30)
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$$  \displaystyle y''_{n,p} = 110c_{11}x^9 + 90c_{10}x^8 + 72c_9x^7 + 56c_8x^6 + 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.31)
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To solve for the particular solution $$y_{n,p}$$, Eq.3.2.29 and Eq.3.2.30 will be substituted into the following particular equation.

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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = $$
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$$  \displaystyle 110c_{11}x^9 + 90c_{10}x^8 + 72c_9x^7 + 56c_8x^6+42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$
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$$  \displaystyle - 3(11c_{11}x^{10} + 10c_{10}x^9 + 9c_9x^8 + 8c_8x^7 + 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) $$
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$$  \displaystyle + 2(c_{11}x^{11}+c_{10}x^{10}+c_9x^9+c_8x^8 + c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$     (Eq.3.2.32)
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Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle 2c_{11}x^{11} + (2c_{10}x^10-33c_{11}x^{10}) + (2c_9x^9 - 30c_10x^9 + 110c_11x^9) + (2c_8x^8 - 27c_9x^8 + 90c_10x^8) $$
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$$  \displaystyle + (2c_7x^7 - 24c_8x^7 + 72c_9x^7) + (2c_6x^6 -21c_7x^6 + 56c_8x^6) + (2c_5x^5 - 18c_6x^5 + 42c_7x^5) + (2c_4x^4- 15c_5x^4+30c_6x^4) $$
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 * style="width:95%" |
 *

 * }

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

$$  \displaystyle + (2c_3x^3 - 12c_4x^3 + 20c_5x^3) + (2c_2x^2 - 9c_3x^2 + 12c_4x^2) + (2c_1x - 6c_2x + 6c_3x) + (2c_0 - 3c_1 + 2c_2) $$     (Eq.3.2.33) The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4,c_5,c_6,c_7,c_8,c_9,c_{10},c_{11}$$

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

$$  \displaystyle c_0 = -1.4204 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_1 = -1.4170 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_2 = -0.7052 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_3 = -0.2328 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_4 = -0.0571 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_5 = -0.0110 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_6 = -0.0017 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_7 = -0.0002 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_8 = -0.0000 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_9 = 0.0000 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_{10} = 0.0000 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_{11} = 0.0000 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{n} = y_{n,h}(x) -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The homogeneous solution is the same:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{n,h}(x) = c_9e^{2x} + c_8e^x $$     (Eq.3.2.35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which then makes the solution simplifies Eq. 3.2.34

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

$$  \displaystyle c_{13}e^{2x} + c_{12}e^x -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the same matlab code but adjusting certain formulas, $$c_{13},c_{12}$$ can be found:

The following values for $$c_{13},c_{12}$$ are obtained:
 * {| style="width:100%" border="0"

$$  \displaystyle c_{13} = -0.0034 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_{12} = 1.4237 \times 10^7 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution is then:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{n}(x) = -0.0034 \times 10^7e^{2x} + 1.4237 \times 10^7e^x -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 $$ $$   \displaystyle -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.37)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Below are the three corresponding graphs to the n=4,7,11

title: User:Egm4313.s12.team17.ying/Test Link


text:
User:Egm4313.s12.team2.sleshko

About Me
Hey team. My name is Stephen.

--Egm4313.s12.team2.sleshko 21:39, 22 January 2012 (UTC)

=Report 1=

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in figure 53. on page 85 of Kreyszig's Advanced Engineering Mathematics (2011).

Kinematics
$$ \bar{r} = y \bar{E}_y $$ $$ \bar{v}=y' \bar{E_y}$$ $$ \bar{a} = \bar{v}' \bar{E_y}=y''\bar{E_y}$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 53 (also pictured above) are derived, first, from the position function of the mass with respect to time $$\bar{r}(t) = y\bar{E}_y $$. The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$\bar{r}(t) = y\bar{E}_y $$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

From Newton's 2nd Law of Motion
$$\bar{F} = m\bar{a} $$ $$\bar{F}_s = -ky\bar{E}_y$$ $$\bar{F}_d = cy'\bar{E}_y $$ $$ mg = m\bar{a} = my''\bar{E}_y$$ Where $$\bar{F}_s$$ is equal to the force in the spring and $$\bar{F}_d$$ is equal to the force in the dashpot
 * Sum the Forces

$$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = r(t) $$

 * From Kreyszig pg. 86

$$ \bar{r}(t) = \bar{F}_0{cos}(wt) $$ $$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = \bar{F}_0{cos}(wt) $$ --Egm4313.s12.team2.sleshko 20:43, 27 January 2012 (UTC)

Problem (5)
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

=Report 2=

Problem (5)
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

Problem 6

 * $$ 10y'' -32y' +25.6y = 0 \ $$

Which can be written in standard form as:
 * $$ y'' -3.2y' +2.56y = 0 \ $$

The characteristic equation is then:
 * $$ \lambda ^2 - 3.2\lambda +2.56 = 0 \ $$
 * $$ \lambda_1,\lambda_2 = \frac{3.2 \pm \sqrt{(-3.2)^2 - 4(2.56)}}{2} \ $$

Therefore the roots $$ \lambda_1, \lambda_2 = 1.6 \ $$ are actually a double root at $$ 1.6 $$ The form for the solution of the differential equation with double-real roots is of the following:
 * $$ y_h(t) = C_1e^{1.6t} + C_2te^{1.6t} \ $$

And then in order to check the solution, find the first and second derivatives to plug back into the original equation and check their equality to zero.
 * $$ y'_h(t) = 1.6C_1 e^{1.6t} + C_2[e^{1.6t} +1.6te^{1.6t}] \ $$
 * $$ y''_h(t) = 2.56C_1e^{1.6t} + C_2[1.6e^{1.6t} + 1.6e^{1.6t} + 2.56te^{1.6t}] \ $$

 * $$ 10[1.6C_1e^{1.6t} +C_2[3.2e^{1.6t} +2.56te^{1.6t}]] - 32[1.6C_1e^{1.6t} +C_2[e^{1.6t} + 1.6te^{1.6t}]] + 25.6[C_1e^{1.6t} +C_2te^{1.6t}] = 0 \ $$

 * $$ 25.6C_1e^{1.6t} + 32C_2e^{1.6t} + 25.6C_2te^{1.6t} - 51.2C_1e^{1.6t} -32C_2e^{1.6t} - 51.2C_2te^{1.6t} + 25.6C_1e^{1.6t} + 25.6C_2te^{1.6t} = 0 \ $$

The solution then checks out

Problem Statement
Find an ODE $$ y'' +ay' + by = 0 \ $$ for the given basis.

Problem (16)
Basis: $$ e^{2.6x}, e^{-4.3x} \ $$

 * $$ \lambda_1,\lambda_2 =2.6, -4.3 = \frac{-a \pm \sqrt{a^2 - 4b}}{2} \ $$

Therefore multiplying each $$ \lambda \ $$ by 2 yields:
 * $$ 5.2 = -a + \sqrt{a^2 - 4b} \ $$ ............. (1)
 * $$ -8.6 = -a - \sqrt{a^2 -4b} \ $$ ............. (2)

Adding the equations together gives:
 * $$ -3.4 = -2a \ $$ so $$ a = 1.7 \ $$

Plug $$ a \ $$ back into equation (1) to get:
 * $$ 5.2 = -1.7 + \sqrt{1.7^2 - 4b} \ $$

Solving for $$ b \ $$:
 * $$ b = \frac {(5.2 +1.7)^2 - 1.7^2}{-4} = -11.18 \ $$

So then the differential equation would be:
 * $$ y'' +1.7y' - 11.18 y = 0 \ $$

Problem (17)
Basis: $$ e^{-\sqrt{5}x} ,xe^{-\sqrt{5}x} \ $$

From the basis it can be seen that $$ -\sqrt{5} \ $$ is a double root of the characteristic equation
 * $$ \lambda_1, \lambda_2 = - \sqrt{5} \ $$

So then the characteristic equation must have root such that:
 * $$ (\lambda + \sqrt{5})^2 = 0 \ $$

Multiplying out gives:
 * $$ \lambda ^2 + 2 \sqrt{5} \lambda + 5 = 0 \ $$

Which corresponds to a differential the equation:
 * $$ y'' + \sqrt{20} y' + 5y = 0 \ $$

title: User:Egm4313.s12.team2.sleshko


text:
User:Egm4313.s12.team2.ladin

Hello my name is Jonathon.

= Problem R1.3 =

Position Function
$$ y(t)=y_k + y_c\ $$

$$ y'(t)=y'_k + y'_c\ $$

$$ y(t)=y_k + y''_c\ $$

$$ f(w)=f(k)=f(c)=:\ f(I) $$

$$ \overline{f}(t)-\overline{f}(I)=my'' $$

$$ \overline{f}(t)=\overline{f}(I)+my'' $$

= Problem R2.6 =

Problem Statement
Realize spring-dashpot-mass systems in series, as shown in Figure 2.6, with a double root $$ \lambda = -3 $$. Find the values for the parameters k, c, m.

Figure 2.6

Solution
The equation of motion for spring-dashpot-mass system in series is: $$ m(y''_k+\frac{k}{c}y'_k)+ky_k=f(t)\ $$ Distribute m: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The standard L2-ODE-CC with excitation is: $$ Ay''+By'+Cy=r(x)\ $$ The homogeneous equation is: $$ Ay''_h+By'_h+Cy_h=0\ $$ Given the double root: $$ \lambda=-3\ $$

The characteristic equation is: $$ [\lambda-(-3)]^2=(\lambda+3)^2=\lambda^2+6\lambda+9=0\ $$ The homogeneous L2-ODE-CC for this characteristic equation is: $$ y''+6y'+9y=0\ $$ Compare coefficients with: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The result is: $$ m=1\ $$

$$ k=9\ $$

$$ 6=\frac{mk}{c}\ $$

$$ 6=\frac{mk}{c}\rightarrow c=\frac{mk}{6}=\frac{(1)(9)}{6}\ $$

$$ c=\frac{3}{2}\ $$

title: User:Egm4313.s12.team2.ladin


text:
User talk:Egm4313.s12.team2.williams.ccw

I am a brand new Wikiversity user that has created this account for my Intermediate Engineering Analysis class.--Egm4313.s12.team2.williams.ccw 22:58, 15 January 2012 (UTC)This account will be used to share information between myself and members of my team, team number 2.--Egm4313.s12.team2.williams.ccw 22:59, 15 January 2012 (UTC)I am looking forward to learning more advanced mathematical techniques.Egm4313.s12.team2.williams.ccw

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text:
User:Egm4313.s12.team2.williams.ccw

I am a brand new Wikiversity user that has created this account for my Intermediate Engineering Analysis class. This account will be used to share information between myself and members of my team, team number 2. I am looking forward to learning more advanced mathematical techniques. The reports contained in my Wikiversity page will hopefully benefit other engineering students.Egm4313.s12.team2.williams.ccw

About Me
My name is Corey Williams and I am a mechanical engineering student at the University of Florida.University of Florida home pageEgm4313.s12.team2.williams.ccw My hobbies include, being with my friends, playing guitar and singing, learning new things, anything dealing with soccer, and do-it-yourself projects. I am about to start my senior year in college. I earned my associates degree from Daytona State College in Daytona Beach, Florida, in 2009.--Egm4313.s12.team2.williams.ccw 23:20, 15 January 2012 (UTC) The link for the main page for team 2 can be found here:Team 2 --Egm4313.s12.team2.williams.ccw 02:41, 22 January 2012 (UTC)

Side Note

 * The wiki format is something that I am not used to using, especially not in a college classroom. Is this the new future of education? Is the vast, faceless internet destined to replace the face to face, personal transfer of knowledge that we have all been instinctually drawn to since the beginning of our species? Will education become a case of more quantity and less quality? It is up to educators and students alike to ensure that we use the internet responsibly in our lives, and in the context of our education.--Egm4313.s12.team2.williams.ccw 03:17, 22 January 2012 (UTC)

= R 1.1 =

Problem Statement

 * Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$f(t)$$.

Free Body Diagram
Fig. 1.
 * In Fig. 1. above, the spring (in green) is represented by K, the dashpot (in blue) is represented by C, the mass (in red) is represented by M, and the applied force (in black) is

represented by $$\bar{r}(t)$$.

Kinematics
$$\bar{r}=y\bar{E}_y$$ $$\bar{v}={y}'\bar{E}_y$$ $$\bar{a}=y''\bar{E}_y$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 1. are derived, first, from the position function of the mass with respect to time ($$r(t)$$). The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$r(t)$$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics- Free Body Diagram
Fig. 2.

Using Newton's Second Law
$$F=m\bar{a}$$ Summing Forces $$\bar{F}_k+\bar{F}_c+m\bar{g}=r(t)$$ $$ky\bar{E}_y+cy'\bar{E}_y+my''\bar{E}_y=r(t)$$ --Corey Williams 20:42, 27 January 2012 (UTC) =R2.2=

Problem Statement
Given initial conditions $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$ and excitation $$ r(x) = 0 \ $$ , find the solution to $$ y'' - 10y' + 25y = r(x) \ $$

Solution

 * $$ y'' - 10y' + 25y = r(x) \ $$
 * $$ \lambda^2 - 10\lambda + 25 = 0 \ $$
 * $$ (\lambda - 5)(\lambda - 5) = 0 \ $$
 * $$ \lambda = 5 \ $$

Double real root
 * $$ \lambda = -\frac 1 2 a $$
 * $$ y = (C_1 + C_2 x)e^{- \frac {ax} 2} \ $$
 * $$ y = C_1 e^{5x} + C_2 xe^{5x} \ $$

Apply initial conditions
 * $$ 1 = C_1 e^{5(0)} + C_2 (0)e^{5(0)} \ $$
 * $$ 1 = C_1 e \ $$
 * $$ C_1 = 1 \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 (1*e^{5x}) + x*5xe^{5x} \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 e^{5x} + 5x^2 e^{5x} \ $$
 * $$ y' = 5(1) e^{5(0)x} + C_2 e^{5(0)} + 5(0)^2 e^{5(0)} \ $$
 * $$ 0 = 5+ C_2 + 0 \ $$
 * $$ C_2 = -5 \ $$
 * $$ y = (1) e^{5x} + (-5) xe^{5x} \ $$

$$ y = e^{5x}-5xe^{5x} \ $$ --Corey Williams 17:55, 6 February 2012 (UTC)

Matlab Code
x = 0:0.01:1; >> ft = exp(5*x) - 5*x.*exp(5*x); >> plot(x,ft)

Effort Distribution
--Corey Williams 15:58, 31 January 2012 (UTC)

Archived Link to Reports
|Archived Report 1 | Report 2

title: User:Egm4313.s12.team2.williams.ccw


text:
User:Egm4313.s12.team14.djp

= Problem 5.3 =

Given

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

b_{1} = 2e_{1} + 7e_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

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

b_{2} =1.5e_{1} + 3e_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Find
Show that Eq. 1 and Eq. 2 are linearly independent using the Gramian.

Solution
The Grammian is defined as

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

\Gamma(b_{1},b_{2}):=\det\begin{bmatrix} \langle b_{1},b_{1}\rangle & \langle b_{1},b_{2}\rangle \\ \langle b_{2},b_{1}\rangle & \langle b_{2},b_{2}\rangle \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Where the angle brackets imply the dot product of the two equations located within them. If the two equations are linearly independent, then:

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

\Gamma(b_{1},b_{2})\ne 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Evaluating the dot products:

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

\langle b_{1},b_{1} \rangle = \left \| b_{1} \right \|^2 \cos(0)=53 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\langle b_{2},b_{2} \rangle = \left \| b_{2} \right \|^2 \cos(0)=11.25 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

To find the angle between the two vectors given, we must first find the angle formed by each individual vectors with respect to the $$e_{1}$$ and $$e_{2}$$ directions.

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

\theta_{1} = \arctan(7/2) = 74.05 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\theta_{2} = \arctan(3/1.5) = 63.43 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\phi = \theta_{1} - \theta_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

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

\langle b_{1},b_{2} \rangle = \langle b_{2},b_{1} \rangle = \left \| b_{1} \right \| \left \| b_{2} \right \|\cos(\phi)=24 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

Plugging these values into Equation 3, we obtain:

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

\Gamma(b_{1},b_{2}) = 53(11.25) - 24^2 = 596.25 - 575 = 21.25 \ne 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

= Problem 5.8 =

Given

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

\int x^n \log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Solve the integral for n=0 and n=1, using integration by parts.

Solution
For n=0:

As per the suggestion in the problem, we'll solve this problem using integration by parts:

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

\int udv = uv - \int vdu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Selecting the u and dv:

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

u = \log(1+x), du=\frac{1}{1+x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

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

dv = x^n, v=\frac{x^{n+1}}{n+1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Plugging into Eq. 2, we obtain:

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

\frac{x^{n+1}}{n+1}\log(1+x) - \int\frac{x^{n+1}}{n+1} (\frac{1}{1+x}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

But n=0, so this can be simplified to:

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

x\log(1+x) - \int x (\frac{1}{1+x}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

The remaining integral can be simplified using polynomial division. The result is 1 with a remainder of -1/(1+x). Thus:

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

x\log(1+x) - \int 1 -\frac{1}{1+x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Evaluating this final integral leaves us with a final answer of:

$$\int x^n \log(1+x)dx = x\log(1+x) - x + \log(1+x)$$

title: User:Egm4313.s12.team14.djp


text:
User:Egm4313.s12.team1.durrance

Important Links

______________________

Lecture Notes

Almost WYSIWYG Manual

Google Course Site

Report Grading Criteria

Report Table

Writing Tools

Links to Other Reports Click on other reports and then click on edit and the format can be referenced for that particular report.

Team Page

$$\pi=\frac{3}{4} \sqrt{3}+24 \int_0^{1/4}{\sqrt{x-x^2}dx}$$

title: User:Egm4313.s12.team1.durrance


text:
User:Egm4313.s12.team1.durrance/r1

R1.1: Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force f(t).

R1.2: Derive the equation of motion of the spring-mass-dashpot ing Fig. 53, in K 2011 p.85, with an applied force r(t) on the ball.

R1.3: For the spring-dashpot system on p.1-4 (of sec.1 notes), draw the FBD's and derive the equation of motion (2) p.1-4.

title: User:Egm4313.s12.team1.durrance/r1


text:
User:EGM6321.f09.team1.Zhichao Gong/Mtg14

Mtg 14: Thu, 24 Sep 09

page14-1

$${\color{blue}p.13-3} \ {\color{red}(Eq.(1))} \ {\phi}_{y}(k)=\frac{\partial \phi}{\partial y}(k) \ {\color{red}(1)}$$

$$\ 0, \ ..., \ n$$

$${\color{blue}\underline{HW:}} \ {\color{blue}p.13-3} \ Case \ n=1.$$

$$1) \ Find \ {f}_{0} \ in \ terms \ of \ \phi$$

$$2) \ Find \ {f}_{1} \ in \ terms \ of \ \phi \ {\color{blue}{f}_{1}={\phi}_{y}}$$

$$3) \ show \ {f}_{0}-\frac{{f}_{1}}{dx}=0 \Leftrightarrow {\phi}_{xy}={\phi}_{yx} \ {\color{red}(2)}$$

$${\color{blue}\underline{HW:}} \ Case \ n=2.$$

$$1) \ show \ {f}_{1}=\frac{{f}_{2}}{dx}+{\phi}_{y} \ {\color{red}(3)}$$

$$2) \ show \ \frac{d}{dx}({\phi}_{y})={f}_{0} \ {\color{red}(4)}$$

$$3) \ {f}_{0}-\frac{d}{dx}{f}_{1}+\frac{{d}^{2}}{d{x}^{2}}{f}_{2}=0 \ {\color{red}(5)}$$

$$Relate \ {\color{red}(5)} \ to \ {\color{red}Eqs \ (4) \ and \ (5)} \ <\color{blue}p.10-2>$$

page14-2

$${\color{blue}\underline{HW:}} \ {\color{blue}Legendre \ L2-ODE-VC}$$

$$F=(1-{x}^{2}){y}^{''}-2x{y}^{'}+n(n+1)y=0 \ {\color{red}(1)}$$

$${\color{blue}Ref:} \ {\color{red} \underset{K}{ \underbrace}} \ (2003) \ p.31$$

$$1) \ Verify \ exactness \ of \ {\color{red}(1)} \ using \ 2 \ method$$

$$1a) \ {\color{blue}p.10-3} \ {\color{red}Eqs. \ (4) \ and \ (5)}$$

$$2a) \ {\color{blue}p.14-1} \ {\color{red}Eqs. \ (5)}$$

$$2) \ If \ not \ exact, \ see \ whether \ it \ can$$

$$be \ made \ exact \ using \ int. \ fact.$$

$$meth. \ w/ \ h(x,y)={x}^{m}{y}^{n}$$

$$Superposition: \ {\color{blue}p.3-3} \ {\color{red}Eq(3)} \ Linearity$$

$${\color{blue}\underline{Read:} \ K.(2003)}$$

$${\color{blue} \ App.3: \ Power \ series}$$

$${\color{blue} \ App.5: \ ODEs \ const.}$$

$${\color{blue}coeff. \ \begin{cases} & int. \ fact. \\ & trial \ soln \  \equiv \ undetermined \ coeff. \end{cases}}$$ page14-3

$${\color{blue}\underline{\color{black}Alternative}} \ defn \ of \ {\color{blue}\underline{\color{black}Linearity:}}$$

L(.) is linear iff ('if and only if' or "equivalent f0")

$${\color{blue}\underline{HW:}} \ show \ {\color{blue}[{\color{red}(1) \ and \ (2)}]} \ \Leftrightarrow \ {\color{blue}[{\color{red}(3)} \ p.3-3]}$$

$${\color{blue}\underline{K. \ p.4}} \ {L}_: \ {C}^{2}(I) \rightarrow \ {C}^{0}(I)$$

 * PEA1.F09.Mtg14.pg3.fig1.svg

$$\ {C}^{0}(I)=space(set)$$

of continous funcs defined on I (superscript "0" means zeroth deriv.)

page14-4

C2(I)= space of funcs on I that are cont. , having cont. first and 2nd(superscript"2") deriv.

$$ \begin{cases} & \ sinx, \ cosx, \ exp(x)   \\ & \ {x}^{2}, \ {x}^{3}, \ ... \end{cases}$$

$$\in \ {C}^{\infty}$$

inf. diff.('smooth') duncs

 * PEA1.F09.Mtg14.pg4.fig1.svg


text:
University of Canberra/RCC2012/Thursday dinner

Please join us for dinner in Dickson the night before RCC starts, if you can. Please add your name to the list below so we can ensure we have enough places booked.

Where: Fekerte's Ethiopian Cuisine. Address: 74/2 Cape St, Dickson, ACT. A short bus or car trip from Civic (city centre), or a nice long walk. When: We've booked for 7 pm, Thursday 19 Jan 2012. Vegetarian/vegan options: Yes (traditional Ethiopian Orthodox Church "fasting food") Cost: Main dishes are $21 for veg, or $25-28 for meat/fish; entrees and salads are $7-14. Contact: call Chris if lost or confused - 0439 471 632.

Are you coming?
Please add yourself. (Note, Laura has some other people's names, so it will be more than just those here):
 * Laura Hale
 * Chris Watkins
 * Nigel Mitchell
 * Tom Worthington
 * Marghanita DaCruz
 * Gillian White
 * Jack Greenmaven
 * Bidgee
 * Hawkeye7
 * Danny_B
 * Liam
 * Chris from Indonesia
 * Anne
 * JJ Harrison
 * Anna-Maria Sviatko

''Current reservation is for 13 people. If more people show up, they will probably be at a separate table. If you are planning to show up, please contact LauraHale on the phone (number available on WM-AU members list or by e-mailing her at lhale at wikimedia dot org dot au. She will need to modify the reservation.''


text:
User:Egm6322.s12.TEAM1.WILKS

=SPRING 2012 Team 1 Homework Archived= Wiki Editing Help Link: Help:Editing Wiki Formula Editing Help Link: Help:Formula How to link the archived version of your HW report to the Report table. https://docs.google.com/Doc?docid=0AY5m1lfNbLroZGM4Mm5mZ2JfMTQzZDVkNnFkZGc&hl=en link to "ideal" reports http://en.wikiversity.org/wiki/User:Egm6341.s10/Report_table/Good_model_reports Link to class HW Report Table for Spring 2012 http://en.wikiversity.org/wiki/User:Egm6322.s12/Report_table Link to Term Paper Term Paper

title: User:Egm6322.s12.TEAM1.WILKS


text:
User:Egm6322.s12.TEAM1.HW

=Egm6322 Spring 2012 Homework = Wiki Editing Help Link: Help:Editing Wiki Formula Editing Help Link: Help:Formula How to link the archived version of your HW report to the Report table. https://docs.google.com/Doc?docid=0AY5m1lfNbLroZGM4Mm5mZ2JfMTQzZDVkNnFkZGc&hl=en link to "ideal" reports http://en.wikiversity.org/wiki/User:Egm6341.s10/Report_table/Good_model_reports Link to class HW Report Table for Spring 2012 http://en.wikiversity.org/wiki/User:Egm6322.s12/Report_table

title: User:Egm6322.s12.TEAM1.HW


text:
Technical writing/Rukmini Gorur

Rukmini Gorur's Home Page in Wikiversity for Technical Writing

title: Technical writing/Rukmini Gorur


title: File:Sampling.2.B.BandPass.pdf


title: File:Iea.s12.sec2.djvu


text:
User:Egm6322.s12.TEAM1.yang

=Homework=

Homework-1

 * HW1

=Links for this course=
 * EGM6322-Team1
 * EGM6322-Lecture Plan
 * EGM6322-Course

title: User:Egm6322.s12.TEAM1.yang


text:
User:Egm6322.s12.team1.zheng.zx

Welcome To My Personal Page!

EGM6322 Course Homepage

Report Table Report 8 Report 9 Report 10 Report 11 Report 12 Report 13 Report 14

Term Paper

Team Page's Here TEAM1.HW

Team Category

Previous Workspace Egm6321.f11.team1.zheng.zx EGM6321 Course Homepage

Useful Links Help:Wikitext examples

Online LaTeX Equation

Manual:New account, upload files, team work

References: PEA1 F10

SandBox

title: User:Egm6322.s12.team1.zheng.zx


text:
User talk:Egm4313.s12.team1.silvestri

First edit. Here's to seeing what will happen.

playing with equations

First, we identify the property of periodic functions, as shown on p9-2 of the notes, equation (1)

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

$$\displaystyle{f(x+np)=f(x)}$$ (1.0)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

So for $$ \cos n \omega x $$, the following determines the smallest period.

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

$$\displaystyle{\cos (n\omega x)}$$ (1.1) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{\cos (n\omega x + 2 \pi)}$$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\cos[n\omega ( x + \frac{2 \pi)}{n \omega}]}$$ (1.3) This means the smallest period is $$ frac{2 \pi}{n \omega} $$.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For $$ \sin n \omega x $$, the following determines the smallest period.

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

$$\displaystyle{\sin (n\omega x)}$$ (1.4) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{\sin (n\omega x + 2 \pi)}$$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\sin[n\omega ( x + \frac{2 \pi)}{n \omega}]}$$ (1.6) This means the smallest period is $$ frac{2 \pi}{n \omega} $$, just as for its cosine equivalent.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Egm4313.s12.team1.silvestri 20:09, 25 January 2012 (UTC)

What is the copyright of File:Report1.1.jpg?
Alpha = .5

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MaintenanceBot (discuss • contribs) 03:41, 11 November 2013 (UTC)

title: User talk:Egm4313.s12.team1.silvestri