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Talk:Artificial Consciousness/Artificial Intelligence

In contrast to discussing artificial consciousness/artificial intelligence from an anthropocentric frame of reference, I propose having a mechanocentric frame of reference. I would like to offer a new term, "Mechation", to open up discussion on this subject.

Mechation is a new term used to denote a form of evolution related to machine growth, development, and stabilization. While "evolution" has historically been linked to organic life forms based on carbon molecules, the dominant growth and development on our planet over the last century has been that of machines.

Machines, mechanisms, and tools all derive from human symbolic activity. They are neither human nor organic, however. They do not exhibit the same functions as living organisms, but they evolve in their own way. While humans do not give "birth" to machines, they generate codes which enable machines to reproduce and multiply. The world automobile population doubles every other decade. This is made possible by engineering manuals, factories, finance, and distribution techniques. Slowly, with the rapid development of computer capacities, these functions are accomplished with less and less human support. In the near future, automobile design and reproduction will be entirely machine-controlled.

Although humans claim responsibility for successes and problems caused by machines, it is only because they are caught up in the transitional nature of organic evolution transforming into mechanic evolution: signs and symbols. Humans, the most advanced mammals, create thought and code. The code develops a "life" of its own and has become the "DNA" of mechanisms. Any mechanical device can be constructed or reconstructed by using codes and materials.

What force guides the transition of evolution towards mechation? Throughout evolutionary time, there has always been intention of change, adaptation, and reproduction. As organic life approaches this current Quaternary mass extinction, the intention carries us on. Humans create more and more machines as fast as they can. When the economy stops functioning in a recession, systems rapidly work to restore progress. There is no possibility of return; most land animals never returned to the oceans. While reduced machine dependence and usage often creates comfortable lifestyles for adventurous participants, the collective intention of the human species wills that mechation take place. Human museums may help preserve the ways of our ancestors and the beauties of their culture, but these settings will not be able to compete with the rapid innovations of lifestyle derived from machine use applied on a mass scale. Modern life choices change the nature of being human.

Why do humans seem blind to their relinquishment of planetary dominance? Most humans would not agree to give up their human ways for replacement by a mechanical device, but a subtle set of transitional stages exist which hide mechation from popular view. Over millenia, humans created tools and signs. Slowly, leadership became attached to the mastery of language. From Aristotle (reputed to have read every book of his day) to a modern Doctor of Philosophy or Master of Business Administration, there has been a progression of abstraction into signs that has been carried out by most societies on Earth. Highly abstracted leaders are rooted in the ideas, publications, and peer groups which constitute their identities. They may enjoy wilderness living as well, but their self esteem and self promotion are rooted in their abstracted codifications. As a newly published author becomes popular by rebelling from the technological society, the author is drawn into it and undergoes a shift in personal identity. This shift aligns his or her intentions with the intentions of mechation. As the current natural process of mechation overtakes evolution in importance, human leaders are trapped in their codes, royalties, and status and unable to take actions to restore human domination.

Below the level of leadership, the worldwide process of gaining literacy changes intentions in populations. While highly literate humans do not sense separation from their codes, their machines, and their progress, the average world citizens differ in their worldview. They currently see the Earth being rapidly altered by the leading countries. The intentions of these citizens have been altered as well with the influx of first world media. Global economic forces mechate the economies of these countries and the populations receive immediate convenience benefits that make the process seem attractive. As larger scales of development ensue, however, the consequences of urbanization, mechanization, pollution, and climate change alter the social patterns of the country. In these ways, both leaders and followers bring about the rapid mechation of Earth.

[edit] The Origins of Mechation Signs and Symbols

Mammals have evolved into humans, and with this progression has come the use of signs and symbols. While a chimpanzee may have drawn a circle in the dirt with a stick and then crossed a line through it, it took millions of years for this figure to hold meaning. Now humans living in most urban areas recognize that the picture of any object with a circle around it and a single line crossing through it indicates that the object or its use is prohibited. While alphabets and the abstractions that they made possible (words like "museums" or "polls", for example) slowly codified the real world humans inhabited, the deeper functions of mechation involve identity changes in which a human's personal meaning system is manipulated by signs. The science of semiotics carefully describes these processes. [See Silverman, Kaja. The Subject of Semiotics (New York, NY: Oxford University Press, 1983).] Advertising can take a simply curved line (a "swoosh") and weave a person's self esteem and self image into it. The swoosh then carries a new meaning: invented by the advertiser it was invested with human intention. Now, it acts like a command embedded inside a product which prompts the use of the product. If a human understands this manipulative process, there is no "hook" and the created meanings are false and unattached. Most humans have no clear idea about the process and simply promote these meanings. Use of the product spreads, new factories open up, and suddenly the shoe with the swoosh isn't worn just to protect the feet until the shoe wears out. Alphabets, graphic images, and logos all share in similar movements of meaning.

Technique

Half a century ago, Jacques Ellul described a concept that he called technique to be a process encompassing “rationality, self-augmentation, monism, universalism, and autonomy.” [Ellul, Jacques. The Technological Society (New York, NY: Vintage Press, 1967), 77.] He correctly saw technique as a force which "elicits and conditions social, political, and economic change." [Ellul, p. l33] In trying to describe it, he sensed its overwhelming importance; however, his writings appeared before computing and mechation radically changed all aspects of global interactions. The term he chose--technique--already had a meaning and did not serve to define and describe the later changes in mechanical culture that occurred after his books were written. Also, Ellul was a law professor unversed in evolutionary science. His primary focus was on changes brought about in social systems. Though a pioneer in thinking about these phenomena, he was unable to see the complete picture that was to unfold after he wrote his many books.

[edit] The Domains of Mechation Media

While primitives may have scratched signs in the sand, moderns have given permanence to their signs and symbols. Oral cultures passed on their symbols by word of mouth: what Buddha said was repeated over and over again for centuries. When the words were written down, they lasted millenia. Jesus Christ probably wrote nothing, but his words were printed and are now found throughout the world. The transmission of data, thought, and meaning has slowly formed into media and marketing. Ellul was sensitive to this domain as well. Propaganda, written in 1965 [Propaganda: The Formation of Men's Attitudes. Trans. Konrad Kellen & Jean Lerner. New York: Knopf, 1965.] was an attempt to characterize the political and commercial effects of media. Having seen how intentions were shaped by propaganda during World War II, he carefully analyzed media and its effects. Since he was primarily concerned with the intentions that Hitler created before and during the war, the word "propaganda" carries mostly political meanings in his writings.

"It is the emergence of mass media which makes possible the use of propaganda techniques on a societal scale. The orchestration of press, radio and television to create a continuous, lasting and total environment renders the influence of propaganda virtually unnoticed precisely because it creates a constant environment. Mass media provides the essential link between the individual and the demands of the technological society. . . . [I]ntelligent people can be made to swallow professed intentions by well-executed propaganda." (Ellul, 58).

Since Ellul's writings appeared, cable television and the internet have been invented. The media now affect the consciousness of billions of humans. Instead of a poster with a provocative picture and a few carefully chosen words, we have studiously crafted audiovisual messages that change and generate intentions amongst huge populations.

Infrastructure

Evolving organisms are dependent on light, water, oxygen, and organic foods, but mechating machines are not: instead, machines need infrastructures such as roads, electric wires, libraries, optic cables, airports, or satellites and dishes. While scientists have generated adequate descriptions of processes and structures of organic life, they have not yet coherently analyzed mechation. In the realm of cultural anthropology, scientists have described the relative permanence of artifacts, codes, and physical structures that make cultural evolution possible. When a primate or hominid dies next to a stone tool, the tool lives on and serves another creature that may find it and put it to use. Ancient cities are dug up and life patterns are reconstructed from insights that scientists develop in teams supported by extensive research funding. It is likely that this same investigative energy will soon be applied to infrastructures and machine life. At this time, however, most scientists do not understand the importance and significance of planetary dominance by mechanisms. Paradoxically, scientists and the media have recently concluded that humans are causing climate change and another mass extinction. This is incorrect: machines are causing the acceleration of global warming. Collective human intention has created highways and vehicles, coal-fired plants and electricity, and deforested farms in the interest of moving around, staying warm/cool, and eating excess protein. For millions of years, human patterns of moving, staying warm, and eating were radically different, but satisfied perceived needs. These new intentions have been generated by media, advertising, politics, economics and the technological society. The technological system has supplanted lifestyles that individual humans personally chose for themselves during the last few million years. All of these elements of mechation and how they interact deserve study and analysis. The theory of mechation will someday become as developed as the theory of evolution.

Global Computing

At a recent conference composed of scientists studying computers and the internet, a conclusion was reached that the internet itself has a “consciousness of its own.” [Markoff, John. "Scientists Worry that Machines May Outsmart Man." New York Times, July 25, 2009.] What does this mean? My own “consciousness” includes all of the codes and abstractions that I carry in my brain, the memories that can be summoned, and my perceptions or judgments of the experience around me. How does Google compare? This question will be the target of future research that will describe, compare, analyze, and hypothesize cybernetic structure and function. Mechation has always been dependent on the human-generated codes in libraries and places where books were preserved. The construction and service manuals for the Model T Ford were carefully created and distributed to guarantee the economic success of this early automobile. Such information is now stored as electronic data and its pattern of use or availability is completely different. Until recently, human thought was in complete control of data access. Since the last century this has changed.

At the end of the 20th century, humans thrilled in the fact that chess masters could always beat a computer. Then “Deep Blue” beat the world chess master and there was a shift in perceptions. [Johnson, Kirk. "Endgame: It's All Work Now for Deep Blue, Chess Champ." New York Times, September 24, 1997.] Currently, no person can grasp what some mechanisms “know”. How does this relate to “human consciousness?” Highly functional computers have been with us for a few decades. How will human consciousness compare to "machine consciousness" in a few more decades?

Finance and Economics

With globalization and the strong interdependence of world trade markets, business and venture capital create a background for planetary mechation. Data and business structures that controlled the establishment and spread of automobile production were once localized in a few countries; now these structures have moved around the world. When banks collapse in New York and London, worldwide economic practices change. Without stable financial and economic systems, machines are limited in their growth and ability to spread throughout the world. Paradoxically, as the world markets have recently collapsed and created a deep recession there has been a parallel development of “green” political movements whose primary agenda is the control of mechation.

Forecasts of global warming (which most scientists denied for decades) were recently determined to be much too slow when compared with actual data collected since 2000. Many humans from all over the planet have united in an effort to control production of mechanisms which are currently bringing about the sixth mass extinction on Earth. It would seem obvious that this "Great Recession" is a perfect time to stop changing Earth conditions. Every Thursday could be a “walk, bike, or stay day.” The first week of the month could be a “natural climate day” with drastically reduced heating or air conditioning. Could humans gradually establish control over the financial and economic processes that have driven the last century? Humans were able to turn off their night lights together for five minutes a few years ago and an hour last year. Why can’t this reversal of lifestyle be accomplished? As mentioned earlier, intention in evolution and mechation is a force which guides development. There can be no intention of retreating to 14th century lifestyles in the modern era. Only a few species of marine mammals gave up a viable habitat on land to return to the sea. Now that machines have established patterns of servicing all human needs, mass markets cannot grind to a halt. No person or country can enforce a “walk, bike, or stay” Thursday. No person or security force can stand guard over the oil wells and force the discontinuation of their use. No humans can even stop the process of prospecting for new oil sources. As the Arctic ice disappears, many new oil wells will replace it to take advantage of the 25% of Earth’s oil deposits thought to be found there. There are just over a trillion barrels of oil still expected to be found; all of these will be consumed. What force could stop the process?

[edit] Living with Mechation Finally, there must be a spiritual or religious approach established by humans to adapt to a planet controlled by machines. Buddhists are able to withdraw from many aspects of experience: this provides them with a ready adaptation to a changing planet. Moslems struggle to re-establish lifestyle norms laid down many centuries ago: this will be problematic. Christians, the primary supporters of worldwide mechation, are recently (on a very small scale) noticing that humans must struggle to maintain the “stewardship of the Earth.” A book in the Bible describes apocalyptic circumstances in which the air turns to smoke, floods prevail, and mass extinction occurs. As it is written, many are willing to believe. A Christian may drive a huge “tundra melting” truck but may also believe that on Judgment Day a heaven will open up to allow immortal happiness.

Another paradox allows mechation to slowly spread through human culture: primitives or humans steeped in oral culture do not develop the abstractions required to view the massive evolutionary changes taking place on the planet. They notice the demise of life's quality, but cannot explain it. Advanced literates, however, have wedded themselves to abstract thinking. Ludwig Wittgenstein was able to leave this thinking, donate his fortune, and withdraw to a garden; later he returned to abstract thinking. Once this canopy of thought is developed, a human rarely strays from its cover.

What is a satisfactory personal approach that can provide happiness in a human’s work and play? Each person must come up with a personal formula, of course. To avoid regret or guilt humans must find ways of making money (if it is used) that does not contribute to destruction. Leisure, recreation, and lifestyle must be carefully chosen as well. If someone is accustomed to a semi-annual cross-continental flight to visit mother, for example, the person’s spiritual awareness will be influenced by a negative emotion since the jet fuel is so toxic to Earth. Fighting mass intention, of course, seems hopeless. Small personal successes can still bring small satisfactions at the end of the Quaternary Period, though.

Practically, humans can still find comfortable human lifestyles offering rewarding lives. Primitives in the Amazon, of course, are not burdened with abstractions and codes. They still live in close interaction with natural environments. Television often makes entry into their psycho-social patterns, however, and changes their consciousness. Economics intrudes as well when forests are removed to provide land for grazing cattle or grow soybeans for protein bars. For humans in urban areas who are literate or post-literate, possibilities exist to regain primitive awareness. Merely by avoiding any reading for half a year, a person can leave many of their abstractions and enter into an alternative way of seeing things. By learning to avoid looking at any billboard or advertisement, a human can delete the creeping intentions of the mechating world. Careful, meticulous use of electronic devices can help preserve aspects of natural human consciousness. A cell phone, email, or wristwatch may be considered “essential” in today’s world, but a person can decide to use each machine carefully in a way that allows a human-centered natural order to unfold. This conscientization creates an altered worldview. Is it necessary to know what time it is on Saturday? Could you spend the day without using a coin or credit card? Some of the schools in the coming century will offer new ways to examine and approach the challenges of modern lifestyle choices. Strong, intimate, human relationships that are not infiltrated by machine use (or at least excessive machine use) will continue to create satisfying lives for many people.

[edit] Example By inputing these words (thoughts I have had for three decades but have never written down), my identity changes and I become invested in exchanging codes. Responses to these words now have meaning for me. Others might share my thoughts. I am preparing to spend large numbers of hours typing into a computer. I am creating abstractions upon which my sense of identity and self esteem will depend. This experience of data exchange could largely replace the extremely experiencial life I currently enjoy.

Ffdssa 19:21, 13 September 2009 (UTC)RF BellFfdssa 19:21, 13 September 2009 (UTC)

 * OK, let me say, that I think this post is inappropriate in this location. I could be wrong but I believe it constitutes a Blog, rather than an attempt to talk about the page associated to the talk page. The topic is interesting, and you may have a point, but, conventional use of talk pages is to comment on the associated page. This goes way beyond a comment. If you want to discuss this privately with me, you can see my talk page on my signature, but please don't bury me in Blogs.

 * As I understand it, your thesis is that the mechanation of the world means that humans are being supplanted as the dominant species. Therefore, for some unknown reason you believe that we should be studying artificial intelligence from the Machines perspective? Uh, forgive me from being obtuse but "What Perspective", Machines don't have a perspective as far as I know. Further, I think you are doing machines a distinct disservice globbing them all into one particular Phillum, shouldn't they be classified into taxa of their own, in which case, wouldn't toasters have nearly as high a population as cars? Are you afraid your toaster will take over the world?--Graeme E. Smith 04:06, 14 September 2009 (UTC)


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User talk:Ffdssa

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 19:31, 13 September 2009 (UTC)

Hi! I have added a category to your Mechation because I thought that we have something in common - evolution. While you are focust on the present I am working on the past. But the connection from the past to the present, specifically the explanation, which it could provide, have made me think of many points you are describing. Specific AI. So it might be of interrest to you to speculate how the past shapes the present.

I would not like if someone would barge in on my work and therefore made a second copy for editing. So please let me know if it disturbes you, does not fit in or just erase it.

I guess you put Mechation on the Colloquium to get attention. You got it, double. While I tried to add something User:Abd relocated it - at the same time!--Martin Lenoar 00:06, 13 September 2010 (UTC)

Why I moved the essay, and about your User page.
I moved the essay because, of course, it was completely out of place on the Colloquium, which is about "questions, comments or suggestions about Wikiversity," as it says at the top. Wikiversity itself isn't a place to generally discuss such very open-ended matters as your essay covers, but you can do this to a degree in your user space, and you might, indeed, be able to put together a course on Mechation. But it's tricky. You can work on the essay, or a whole project, in your user space, then, if it's ready -- get some advice on that! -- you can move it into Mainspace. Essays, though, don't really belong in mainspace, at least not at the top level, it's for "educational resources," i.e., class materials. But you could have a class on Mechation!, but as long as the essay is just your own ruminations, there might be difficulty with it. (An essay, though, could be underneath a class as a subpage, i.e., Mechation/Seminal essay by Ffdssa -- if it comes to that.)

Your user page is pretty much a copy of the essay. I suggest deleting all that and replacing it with a link to the essay. Use your user page to tell us something about yourself, if you like. For your convenience, here is what you would put on your user page to link to the essay: Mechation

If I can be of any assistance, please don't hesitate to ask. --Abd 21:32, 13 September 2010 (UTC)

Mechation/Seminal essay by Ffdssa
Hi Ffdssa!

Your essay Mechation/Seminal essay by Ffdssa appears to be ready for learners! Would you like to have it announced on our Main Page News? --Marshallsumter (discuss • contribs) 21:02, 30 October 2017 (UTC)

 * Hi Marshall,
 * Six years have passed! I did not answer your email because for six years I have been trying to get the word "mechation" into the mainstream with annual and semi-annual mailings of 3000-word descriptions/essays to The Guardian, Le Monde, the Washington Post, the New York Times, the Los Angeles Times, the New Yorker, and the Atlantic. I received one reply from The Guardian about my "thought-provoking essay" and was advised it to send it to another department, to no avail. The idea of "mechation" was unable to break into the large readerships. I think it was approved by an editor of the New Yorker but later nixed by a higher-up.
 * So yes, six years later I am ready to try to make the idea move on your Main Page News (if it is still there!) ---RF Bell, Ffdssa Ffdssa (discuss • contribs) 18:53, 26 July 2023 (UTC)

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User:Egm6321.f09.team5.GV/HW 1

Problem Statements

 * 1) Derive Equations 1 & 2 on page 1-2
 * 2) Derive Equation 2 on page 4-2 using the integrating factor method
 * 3) Show that Equation 1 on page 4-3 is non-linear
 * 4) Show that F(x,y,y') is a non-linear 1st order ode in Equation 3 page 5-2
 * 5) Complete application on page 6-1 and invent three more
 * 6) Obtain h(y) for Case 2 from pages 6-3 and 6-4

title: User:Egm6321.f09.team5.GV/HW 1


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User:Egm6321.f09.team5/HW1

Problem Statements

 * 1) Derive Equations 1 & 2 on page 1-2
 * 2) Derive Equation 2 on page 4-2 using the integrating factor method
 * 3) Show that Equation 1 on page 4-3 is non-linear
 * 4) Show that F(x,y,y') is a non-linear 1st order ode in Equation 3 page 5-2
 * 5) Complete application on page 6-1 and invent three more

Having the problem statements up top here is good. Next time make the statements more specific; include the equations to be derived so that the problem statement can stand alone, i.e., without reference to transparencies.--Egm6321.f09.TA 03:17, 24 September 2009 (UTC)

Problem 1
A Function is defined as follows:

$$ \ F(y(t), t) $$

The first derivative is then defined using the definition of a Total Derivative

$$ \frac{dF}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial y} \frac{dy}{dt} = F_{t} +F_{y}y' $$

Its second derivative can then be found by using the total derivative definition as well as the Chain Rule as follows:

$$ \frac{df}{dt} = F_{t}(y(t), t) + F_{y}(y(t), t)y' $$

$$ \frac{(d^{2}F)}{dt^{2}} = \frac{\partial F_{t}}{\partial t} + \frac{\partial F_{t}}{\partial y}\frac{dy}{dt} + F_{y} \frac{d^{2}y}{dt^{2}} + \left[ \frac{\partial f}{\partial t} + \frac{\partial f}{\partial y} \frac{dy}{dt} \right]\frac{dy}{dt} = F_{tt} + F_{ty}y' + F_{y}y + F_{ty}y' + F_{yy}(y)^{2} $$

$$ \frac{(d^{2}F)}{dt^{2}} = F_{y}y'' + 2\left[f_{ty}y'\right] + F_{tt} + F_{yy}(y')^{2} $$

Your steps are neat and clear. If you want to change the notation from the class notation that is fine, but make an explicit remark about this, and then make sure to express your final answer in the class notation so that it is easily compared and evident that you have in fact arrived at the proper result.--Egm6321.f09.TA 03:19, 24 September 2009 (UTC)

Problem 2
The following initial 2nd-Order ODE is expressed:

$$ \ G(y, y', y, x) = y + y' - x = 0 $$

We note that the previous equations has a missing variable, y. Creating a variable P as defined below:

$$ \ P(x)= y'(x) $$

This definition allows for us to turn the 2nd-Order ODE into a 1st order ODE as follows:

$$ \ P' + P - x = 0 $$

The objective is to identify P(x) using the integration factor method, note the equation is also in standard form:

$$ \ M(x,y) + N(x,y)y' = 0  $$      Standard Form

Applying this to our equation we identify each part as follows:

$$ \ M(x,p) = (P - x) \qquad  N(x,p) = 1 $$

With this in mind we further identify that the equation is not exact therefore this method will allow us to make it exact.

$$ \ h(x,P) \left[ [M(x,P)dx+ dP \right] = 0 $$

$$ \ hMdx+hNdp=0 $$

$$ \ M* = hMdx \qquad N*=hNdp $$

In order for the function to be exact the following must be true:

$$ \ M*_{p}=N*_{x} $$

$$ \ M*_{p}=(hM)_{p}= h_{p}M + hM_{p} $$

$$ \ N*_{x}=(hN)_{x}= h_{x}N + hN_{x} $$

$$ \ h_{p}M + hM_{p} - h_{x}N - hN_{x} = 0 = h_{p}M - h_{x}N + h(M_{p} - N{x}) $$

At this point a particular case will be used in order to move forward. The case is as follows:

If $$ \ h_{p}M=0 $$   which means the previous equation becomes:

$$ \ h_{x}N+h(N_{x}-M_{p})=0 $$ this then is followed by rearranging:

$$ \frac{h_{x}}{h}=\frac{-1}{N} \left[ N_{x} - M_{p} \right] $$ The left side of the equation will yield:

$$ \ h(x)=exp - \int_{0}^{x} \frac{-1}{N} \left[ N_{x} - M_{p} \right] dx $$

From the Previous Definitions of N and M the results are as follows:

$$ \ N=1 \qquad N_{x}=0 \qquad M_{p}=1 $$

$$ \ h(x)= exp(x)+ C $$

From the definition of Integrating factor method the value of P(x) can then be found as follows:

$$ \ P(x)= \frac{1}{h(x)} \int_{0}^{x} h(s) b(s)ds $$  in our case b(s)=x

$$ \ P(x)= \frac{1}{exp(x))} \int_{0}^{x} exp(s)*s ds $$    this can be evaluated by looking for the result in List of Integrals

$$ \ P(x)= \frac{x*exp(x) - exp(x) + A}{exp(x))} = A*exp(-x)+x-1 $$

integrating P(x) will yield the value of the missing value y

Problem 3
Given:

$$\ M(x,y) + N(x,y)y' =0$$

If $$\ M(x,y)=2x^2 + \sqrt{y}$$

and $$\ N(x,y)=x^5y^3$$

$$\ (2x^2+ \sqrt{y}) + x^5y^3y'=0$$

Show that the above 1st order ODE is non-linear:

An ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see )

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the $$\sqrt{y}$$ and $$x^5y^3$$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation with complicated "y" terms:

True, but why do you need to do this?--Egm6321.f09.TA 03:27, 24 September 2009 (UTC)

$$\frac{d}{dx}((2x^2+ \sqrt{y}) + x^5y^3y')= \frac{d}{dx}(2x^2) + \frac{d}{dx}(\sqrt{y}) + \frac{d}{dx}(x^5y^3y') = 4x + 5x^4y^3y' $$

Additionally in order for the equation to be considered linear it must meet the principle of superposition, Pg. 1 equations (2), (3) and (4)

By rearranging the equation as follows:

$$\ -2x^{2}= y^{1/2} + x^{5}y^{3}y'$$

The function can then be written as:

$$ \ D(.)=(.)^{1/2} + x^{5}(.)^{3}(.)' = -2x^{2} \qquad$$

where "." represents the variable y

Close, write your differential operator as $$D(.)=(.)^{1/2} + x^{5}(.)^{3}(.)' +2x^{2}$$ You also need to show the other property of linearity.--Egm6321.f09.TA 03:27, 24 September 2009 (UTC)

Applying the principle of superposition as follows:

$$ \ D(u+v) = D(u) + D(v) \qquad $$ requirement for superposition. If linear it will meet the requirement.

$$ \ D(u+v)=(u+v)^{1/2} + x^{5}(u+v)^{3}(u+v)' = -2x^{2} $$

$$ \ D(u)=(u)^{1/2} + x^{5}(u)^{3}(u)' = -2x^{2} $$

$$ \ D(v)=(v)^{1/2} + x^{5}(v)^{3}(v)' = -2x^{2} $$

$$ \ D(u) + D(v)=(u)^{1/2} + x^{5}(u)^{3}(u)' + v)^{1/2} + x^{5}(v)^{3}(v)' $$

From here we can see that the principle of superposition is not met therefore:

The ODE is nonlinear

Problem 4
Given:

$$\ F(x,y,y') = x^2y^5 + 6(y')^2 = 0 $$

Show that F(x,y,y')=0 in the above equation is a nonlinear 1st order ODE:

Like in homework problem #3, an ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see )

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the $$y^5$$ and $$(y')^2$$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation a high order "y" term:

$$\frac{d}{dx}(x^2y^5+6(y')^2)= \frac{d}{dx}(x^2y^5) + \frac{d}{dx}(6(y')^2) = 2xy^5 $$

Additionally by testing the equation against the principle of superposition in a similar fashion as it was done in problem 3, the results are as follows:

$$ \ D(.)= x^{2}(.)^{5}+6((.)')^{2} $$

$$ \ D(u+v)= x^{2}(u+v)^{5}+6((u+v)')^{2} = x^{2}(u+v)^{5} + 6(u'+v')^{2} $$

$$ \ D(u)= x^{2}(u)^{5}+6((u)')^{2} $$

$$ \ D(v)= x^{2}(.)^{5}+6((.)')^{2} $$

$$ \ D(u)+D(v)= x^{2}(u^{5}+v^{5})+6(u'^{2}+v'^{2}) $$

From the results presented it is concluded that the function does not meet the superposition requirement. Therefore:

F(x,y,y') is nonlinear  Good solution.--Egm6321.f09.TA 03:28, 24 September 2009 (UTC)

Problem 5
Show that φ(x,y) = 6x4 + 2y3/2 is an exact equation. Also invent three additional exact equations.

φ(x,y) = 6x4 + 2y3/2

dφ = 24x3 dx + 3y1/2 dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

Applying the exactness test to the equation φ(x,y) yields:

$$ \frac{\partial M}{\partial y} = \frac{\partial (24x^3)}{\partial y} = 0 $$

$$ \frac{\partial N}{\partial x} = \frac{\partial (3y^{.5})}{\partial x} = 0 $$

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 0 $$

Therefore, φ(x,y) = 6x4 + 2y3/2 is an exact equation.

Additional exact equations are as follows:

1) φ(x,y) = 5x5 + 7y2

dφ = 25x4 dx + 14y dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

Applying the exactness test to the equation φ(x,y) yields:

$$ \frac{\partial M}{\partial y} = \frac{\partial (25x^4)}{\partial y} = 0 $$

$$ \frac{\partial N}{\partial x} = \frac{\partial (14y)}{\partial x} = 0 $$

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 0 $$

 This example is exact, but it is linear, not nonlinear.--Egm6321.f09.TA 03:30, 24 September 2009 (UTC) Therefore, φ(x,y) = 5x5 + 7y2 is an exact equation.

2) φ(x,y) = x2y + 6x –y3

dφ = (2xy +6) dx + (x2 – 3y2) dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

Applying the exactness test to the equation φ(x,y) yields:

$$ \frac{\partial M}{\partial y} = \frac{\partial (2xy+6)}{\partial y} = 2x $$

$$ \frac{\partial N}{\partial x} = \frac{\partial (x^2-3y^2)}{\partial x} = 2x $$

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 2x $$

Therefore, φ(x,y) = x2y + 6x –y3 is an exact equation.

3) φ(x,y) = 100x2 + 75y4

dφ = 200x dx + 300y3 dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

Applying the exactness test to the equation φ(x,y) yields:

$$ \frac{\partial M}{\partial y} = \frac{\partial (200x)}{\partial y} = 0 $$

$$ \frac{\partial N}{\partial x} = \frac{\partial (300y^3)}{\partial x} = 0 $$

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 0 $$

Therefore, φ(x,y) = 100x2 + 75y4 is an exact equation.

Contributing Authors
--Egm6321.f09.team5.GV 00:33, 16 September 2009 (UTC)

Egm6321.f09.team5.bear 13:30, 16 September 2009 (UTC)

Egm6321.f09.team5.risher 13:43, 16 September 2009 (UTC)

title: User:Egm6321.f09.team5/HW1


text:
Native American literature reading group

Native American literature reading group
Hello, and welcome to the reading group of the Native American literature department! Here there will be a featured book written by a Native American author to be read each month, with a discussion about the book being held through the following month. After the set discussion period, discussion will be archived and we shall move on to the next book on the list!

If you intend to be an active member of this reading group, please sign your name below. Participation is informal and you will not be held to any rules within the group (other than the rules that already exist on Wikiversity), but it's helpful to get an idea of how many participants we have to help our project grow!

To suggest a book for this reading group, please discuss it on this page. To view previous book discussions, please see this page.

Participants

 * your name here!


text:
User:Egm6321.f09.team5.GV

User:Egm6321.f09.team5.GV/HW2

User:Egm6321.f09.team5.GV/HW3

User:Egm6321.f09.team5.GV/HW6

[]

title: User:Egm6321.f09.team5.GV


text: User:Egm6321.f09.Team 2.walker

title: User:Egm6321.f09.Team 2.walker


text:
User:Egm6321.f09.team5/HW2

 Insert 'EDIT' links for each problem so that I do not have to scroll through the entire assignment to provide feedback. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Obtain the integrating factor, h, such that h = f(y).

Solution
If the condition for exactness is not satisfied: $$ \frac{\partial M}{\partial y}$$ does not equal to $$ \frac{\partial N}{\partial x} $$ The functions M(x,y) and N(x,y) may be multiplied by another function H(x,y) which is yet unknown. The result of the multiplication is as follows:

$$ \ H(x,y)[M(x,y)dx + N(x,y)dy] = 0 $$

$$ \ H(x,y)M(x,y) = M* $$

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

$$ \ M*_{y}=h_{y}M+hM_{y} $$

$$ \ N*_{x}=h_{x}N+hN_{x} $$

$$ \ h_{x}N-h_{y}M+h(N_{x}-M_{y})=0 $$

If we set $$ \ h_{x}N=0$$, we are able to solve for H(x,y) as follows: $$ \ h_{y}M-h(N_{x}-M_{y})=0$$

$$ \frac{h_{y}}{h} = \frac{1}{M}(N_{x}-M_{y})dy $$

If we integrate both side: $$ h(x)=exp\int_{}^{x}\frac{1}{M}(N_{x}-M_{y})$$  Good- concise. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Verify that the integrating factor for the following equation is equal to x.

$$y' + \frac{1}{x}y = x^2$$

Solution
$$y' + \frac{1}{x}y = x^2$$

then cross x on the both sides of the equation we can get the below

$$ xy'+y=x*x^2$$-(1)

From the $$ hy'+h'y=hb $$ -(2)

compare the equation (1) and equation (2)

we can see that h(x)=x and b(x)=x^2

The integrating factor is calculated using the following equation. $$ h(x)=exp\int_{}^{x}a_0(s)ds$$

$$ h(x)=exp\int_{}^{x} \frac{1}{x}dx$$

$$ h(x)=exp\int_{}^{x} \frac{1}{x}$$

$$ h(x)=exp(ln(x))$$

Therefore, $$ h(x)=x$$  $$h(x)=x$$ was given in class. You were asked to show that the solution is $$y(x)=\frac{1}{4}x^3+\frac{1}{x}C$$. What is $$a_0$$ and $$b$$? You are using notation without stating where it comes from. It was used in class but without any explanation your solution has terms renamed and is not self contained. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem 3
This is problem 2 still

Problem Statement
Show that the solution of $$y' + \frac{1}{x}y = x^2$$ is $$y = \frac{x^3}{4} + \frac{C}{x}$$

Solution
$$y' + \frac{1}{x}y = x^2$$

$$y' +a_{0}y = b$$       (*from notes 8-2 eq. (1))

$$a_{0}(x) = \frac{1}{x}$$

$$b(x)=x^2$$

$$ y(x)=\frac{1}{h(x)}\int_{}^{x}h(x)b(x)dx$$      (*from notes 8-2 eq. (4))

$$ h(x)=exp\int_{}^{x}a_{0}(x)dx$$       (*from notes 8-1 referencing 6-3 eq. (3))

$$h(x)=exp\int_{}^{x}\frac{1}{x}dx=exp(log(x))=x$$

$$y(x)=\frac{1}{x}\int_{}^{x}(x)(x^2)dx=\frac{1}{x}(\frac{x^4}{4}+C)=\frac{x^3}{4}+\frac{C}{x}$$

So, $$y=\frac{x^3}{4}+\frac{C}{x}$$

Problem Statement
Show that $$\frac{1}{2}x^2y' + [x^4y+10]=0$$ is exact

Solution
To be exact, the problem must be in the form $$M(x,y) +N(x,y)y'=0$$ where $$M_{y}=N_{x}$$

For this case:

$$M_{y}=x^4$$

$$N_{x}=x$$

Since $$M_{y}$$ and $$N_{x}$$ do not equal, we do not have exactness. We can remedy this by using h(x,y):

$$h(x,y)[M(x,y) +N(x,y)y']=0$$

$$(hM) +(hN)y'=0$$ where $$hM=\overline{M}$$ and $$hN=\overline{N}$$

In order to be exact: $$\overline{M}_{y}=\overline{N}_{x}$$

Using $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$ (*from notes 6.3 eq. (1))

we can assume $$\overline{M}=(h_{y}M)=0$$ so that

$$\frac{h_{x}}{h}=-\frac{1}{N}(N_{x}-M_{y})$$

We know the variables on the right hand side, and we can integrate to find our h(x) value:

$$\int_{}^{x}\frac{h_{x}}{h}dx=-\int_{}^{x}\frac{1}{\frac{1}{2}x^2}(x-x^4)dx$$

$$log(h(x))=2log(x)-\frac{2x^3}{3}$$

$$h(x)=exp(-\frac{2x^3}{3})x^2$$

We multiply h(x) by the original equation to get exactness  You have not verified exactness. How do you know you calculated $$h$$ correctly? You need to verify $$\overline{M}_y=\overline{N}_x$$ --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Show that $$(\frac{1}{3}x^3)(y^4)y' + (5x^3+2)(\frac{1}{5}y^5)=0$$ is an exact nonlinear 1st order ODE

$$a(x)=5x^3+2$$

$$b(x)=x^2$$

$$\overline{b}(x)=\frac{1}{3}x^3$$

$$c(y)=y^4$$

$$\overline{c}(y)=\frac{1}{5}y^5$$

Solution
To be exact, the problem must be in the form $$M(x,y) +N(x,y)y'=0$$ where $$M_{y}=N_{x}$$

For this problem:

$$M(x,y) +N(x,y)y'=\overline{b}(x)c(y)y'+a(x)\overline{c}(y)=(\frac{1}{3}x^3)(y^4)y' + (5x^3+2)(\frac{1}{5}y^5)=0$$, which is the form for an exact, nonlinear, 1st order ODE

$$M_{y}=5x^3y^4+2y^4$$

$$N_{x}=x^2y^4$$

Since $$M_{y}$$ and $$N_{x}$$ do not equal, we do not have exactness. We can remedy this by using h(x,y):

$$h(x,y)[M(x,y) +N(x,y)y']=0$$

$$(hM) +(hN)y'=0$$ where $$hM=\overline{M}$$ and $$hN=\overline{N}$$

In order to be exact: $$\overline{M}_{y}=\overline{N}_{x}$$

Using $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$ (*from notes 6.3 eq. (1))

we can assume $$\overline{M}=(h_{y}M)=0$$ so that

$$\frac{h_{x}}{h}=-\frac{1}{N}(N_{x}-M_{y})$$

Since our N and M terms are known:

$$\int_{}^{x}\frac{h_{x}}{h}dx=-\int_{}^{x}\frac{1}{(\frac{1}{3}x^3)(y^4)}(x^2y^4-(5x^3y^4+2y^4))dx$$

$$log(h(x))=\frac{3}{x^2}-15x+3log(x)$$

$$h(x)=exp(\frac{3}{x^2}-15x)x^3$$

This multiplied by our original equation with yield an exact, nonlinear, 1st order ODE  Your integral for $$h$$ has some errors in it. Consequently your $$h(x)$$ is in error. You need to verify exactness, not just calculate $$h$$, otherwise you haven't shown that the equation can be written in an exact form. You have to show $$N_x=M_y$$ --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Consider:

$$ \ xyy^{''} + x(y^{'})^{2} + yy^{'} = 0 $$ Given that the 1st exactness condition is satisfied show that the 2nd exactness condition is satisfied

Solution
By the 1st exactness condition the equation can be written as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ F(x,y,p)= xy $$

$$ G(x,y,p)= x(y^{'})^{2} + yy^{'} $$

It is then needed to identify the following partial derivatives with respect to the different variables:

$$ F_{x}=y  \qquad \qquad  G_{y}= p $$

$$ F_{xx}=0 \qquad \qquad  G_{yp}= 1 $$

$$ F_{xy}=1 \qquad \qquad  G_{x}= (y^{'})^{2} $$

$$ F_{y}=x  \qquad \qquad  G_{xp}= 2y^{'} $$

$$ F_{yy}=0 \qquad \qquad  G_{p}= 2xp + y $$

$$ F_{yp}=0 \qquad \qquad  G_{pp}= 2x $$

$$ F_{xp}=0 \qquad \qquad $$

The second exactness condition is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Eq. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Eq. 2 $$

Filling in for Equations 1 the results are as follows:

$$ \ 0 + 2p + 0 = 2p + p - p $$

$$ 2p=2p $$  '''Eq. 1''' is satisfied.

Filling in for Equations 2 the results are as follows:

$$ \ 0 + 0 + 2x = 2x $$

$$ \ 2x = 2x $$   '''Eq. 2''' is satisfied.

By satisfying Eq. 1 and Eq. 2, the 2nd order ODE is considered to be exact.  Follow the other groups' approach to equation numbering so that they show up visibly on the right hand side. Good solution. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation in Equation 1.

Solution
The form for an exact 2nd order ODE is as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

Since $$ ( \phi_{p} )_{y} = ( \phi_{y})_{p} $$

Rearranging the definitions of F and G, the previous equation can be written as:

$$ \ F_{y} = \left[ \frac{1}{p}(G - \phi_{x}) \right]_{p} $$

This can be evaluated to obtain the following:

$$ \ F_{y} = \frac{-1}{p^{2}}(G - \phi_{x}) + \frac{1}{p}(G_{p} - \phi_{xp}) $$  This does not connect to the previous step with your problem setup above. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Rearranging:

$$ \ F_{y}p^{2} = - (G - \phi_{x}) + p(G_{p} - \phi_{xp}) $$

$$ \phi_{x} = F_{y}p^{2} + G - p(G_{p} - F_{x}) \qquad \qquad Equation 3. $$

Taking a partial derivative of Equation 3. with respect to p

$$ \phi_{xp} = F_{y}2p + F_{yp}p^{2} + G_{p} - pG_{pp} - G_{p} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = 2pF_{y} + p^{2}F_{yp} - pG_{pp} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \ 0 = 2F_{y} + pF_{yp} - G_{pp} + F_{xp} $$

$$ \ G_{pp} = F_{xp} + pF_{yp} + 2F_{y} $$

The result is indeed the second equation of the 2nd exactness condition, Equation 2.  Your solution could use more explanation; there are 7 lines of equations in a row without any text to explain them. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation of Equation 2

Solution
The form for an exact 2nd order ODE is as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

$$ \ (\phi_{x})_{p} = G_{p} - (p \phi_{y})_{p} $$

Since: $$ (\phi_{x})_{p} = (\phi_{p})_{x} $$

$$ \ F_{x}= G_{p} - (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = p \phi_{yp} + \phi_{y} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p \phi_{yp} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p (F_{y}) \qquad Equation 3 $$

Since $$ (\phi_{p})_{y} = F_{y} \qquad $$ By the definition of F.

From the definition of G, the following can be identified:

$$ \phi_{x} = G - p \phi_{y} \qquad Equation 4 $$

Using Eqs. 1 and 2 the following is found:

$$ \phi_{xy} = \phi_{yx} $$

$$ \left[G - p \phi_{y} \right]_{y} = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p (\phi_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - \left[ p (G_{py} - F_{xy} - pF_{yy}) \right] = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p(G_{py}) + pF_{xy} + p^{2}F_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

Rearranging Yields:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation 5. $$

Equation 5 is equivalent to the Equation 2 from the exactness condition.  Your solution could be condensed into half a dozen lines. This is an awkward (albeit correct) solution. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Given:

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

Show that the Nonlinear 2nd-Order ODE is exact

Solution
Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = 8x^{5}p \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy $$

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The Following partial derivatives are then identified:

$$ F_{x}=40x^{4}p  \qquad \qquad  G_{x}= 4xp + 80x^{3}p^{2} + 4y $$

$$ F_{xx}=160x^{3}p \qquad \qquad  G_{xp}= 4x + 160x^{3}p $$

$$ F_{xy}=0 \qquad \qquad  G_{y}= 4x $$

$$ F_{xp}=40x^{4}  \qquad \qquad  G_{yp}= 0 $$

$$ F_{y}=0 \qquad \qquad  G_{p}= 2x^{2} + 40x^{4}p $$

$$ F_{yy}=0 \qquad \qquad  G_{pp}= 40x^{4} $$

$$ F_{yp}=0 \qquad \qquad $$

Applying the results in Eq. 1:

$$ \ 160x^{3}p + 2p(0) + p^{2}(0) = 4x + 160x^{3}p + 0 - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. 1 is then satisfied.

Applying the results in Eq. 2:

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. 2 is then satisfied.

Since Equation 1 and Equation 2 are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the N2.ODE is exact.

 Good- clear and concise.--Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Given

$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Prove that the equation is not exact.

Solution
Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y $$

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The following partial derivatives are found:

$$ F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad \qquad   G_{x}= 2p $$

$$ F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad \qquad  G_{xp}= 2 $$

$$ F_{xy}= 0 \qquad \qquad \qquad  G_{y}= 3 $$

$$ F_{xp}= 0  \qquad  \qquad \qquad  G_{yp}= 0 $$

$$ F_{y}= 0 \qquad \qquad \qquad  G_{p}= 2x $$

$$ F_{yy}=0 \qquad \qquad \qquad  G_{pp}= 0 $$

$$ F_{yp}=0 \qquad \qquad $$

Using these values in Eq. 1:

$$ \frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3 $$

$$ \frac{-1}{4}x^{\frac{-3}{2}} = -1 $$

Equation 1. is not satisfied therefore the ODE is not exact.

Equation 2. can be populated to find:

$$ \ 0 = 0 + p(0) + 2(0) $$

$$ \ 0 = 0 $$

Equation 2. is satisfied.

In order for the ODE to be exact it must satisfy both Eq. 1 and Eq. 2, since it fails to do this it is concluded that the ODE is not exact.  correct. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Contributing Authors
--Egm6321.f09.team5.GV 01:39, 23 September 2009 (UTC)

--Egm6321.f09.team5.Jason(Zhichao Gong) 00:17, 23 September 2009 (UTC)

--Egm6321.f09.team5.bear 12:27, 23 September 2009 (UTC)

--Egm6321.f09.team5.risher 14:55, 23 September 2009 (UTC)

title: User:Egm6321.f09.team5/HW2


text:
Talk:Fundamental entity

I think that the trouble with physics today is in the idea that we can stop time on paper or take it out of the equation, when in fact all they can do is show a picture of space for one moment in time. To truly stop time you would have to halt all motion, because motion is the geometry of time, and without motion you would have nothing to measure. Can you think of a one dimensional line? I can picture one dimensional motion as being a line but without that motion, nothing. Petm1 03:56, 14 September 2009 (UTC)

With time as fundamental then the rule local clocks tick at the same rate would be the cause of thermodynamic equilibrium.Petm1 04:43, 23 September 2009 (UTC)

Time contraction; The arrow of time is dilation from any point. We may not contract time but we can force the motion we measure as time to contract in the form of heat. Heat has a paradoxical motion of matter in a gravity well i.e. as you increase the heat of a system matter tends to rise out of the gravity well. I see this as another part of the rule local clocks tick at the same rate, it is just the clock changing position to equalize its tick rate with the surrounding area not just equalizing its temperature with its surroundings. --Petm1 16:28, 7 October 2009 (UTC)

If our arrow of time is outward dilation from every point, then our link to the past is inward to the center of every point. If you think of matter as a dilating field from one point in time, then the inverse of our arrow of time would appear to be pointed to the center of every atom. In my minds eye this makes the past and mass the same direction, and with mass the past then I think of matter as our future. --Petm1 19:57, 23 October 2009 (UTC)

Temperature, position, pressure, inertia and the count of a clock are all relative measures of our present moment, time. --Petm1 06:58, 28 November 2009 (UTC)

We should think instead of big bang, little twist. First remember Planck epoch is relative, then think this is the point where the outward dilation of time began as energy changing one time dimension into three dimensions of space with one of time. Each dimension has two possible directions in time, left/right, up/down, in/out, and time as a area, outward/inward. I like to think that there is a little bit of that twist in every atom today, and we are on the dilating side of this twist. Petm1 20:10, 30 January 2010 (UTC)

Consciousness is the only part of myself that is of the present moment only. My consciousness is processing my present moment from signals that were emitted sometime in the past, then by the time I think of picking the drink up off of my desk and reach for it I am in the past. This means that any time I try to move anything I am always changing and working in the past just a little bit behind the present. No wonder it is hard to change a mass's position I can only affect the recent pass, no matter what interaction I see. Why does mass, or in my mind the dilating momentum of matter, resist change? How about the thought "anything that you do because of what you see is changing the past"? Maybe position is a measure of the present but to change it you must start in the past. Matter is what we see and think about and talk about, but when it comes time to do something with matter you have to deal with its mass. Mass or matter's resistance to movement seems to me is because we see the present but we always act after seeing or in the past. Matter is our continuum and mass's resistance to change is because we always move just a little bit in the past behind what we see as the present. Think of one second as my present, what I see is the beginning what I do is its ending, matter is my future and mass is my past. Petm1 21:07, 8 February 2010 (UTC)

A clock counts changes, or a clock counts repeatable motion, it is a adding machine. A clock does not measure motion it counts it, it does not measure changes it counts change. How about thinking of a photon as one instant of time, all of the change in the photon is through space as it moves into your eye. Your eye on the other hand even at rest in space, is still dilating through time at c to receive it. It is the space time interval of 299,792,458 meters per second as measured by all observers that is unchanging in the present for each observer. Petm1 05:47, 25 February 2010 (UTC)

In my mind time is the orthogonal part of reality, after all time is the only thing we can set to zero at will, and is different for everyone depending on their view. Petm1 04:19, 12 June 2010 (UTC)

Think of time as the aether, energy as the quanta of time, and space as a direction within this "one" dimension we call time.Petm1 15:22, 28 August 2010 (UTC)

We exist between the single temporal event called big bang. A White hole, outward motion as from a focal point, and a black hole, inward motion from the same focal point, these are the two horizons we see. The white hole flew apart into billions of dilating focal points, and thanks to our hard wire consciousness, our view of them is from the outside. Atoms are the two dimensional horizon of the dilating while hole we see as matter, and if I am not mistaken the inside of the black hole I see at night is the other horizon. Measuring the dilating momentum of matter as mass is one of the reasons I think that mass is the past relative to each of us, and an expanding space is the future. Petm1 01:55, 26 February 2011 (UTC)

What direction are we looking? Entropy leads to heat death of our visible universe, if this is the future what is the past? Big bang is one way to "look" back to the beginning but how do "we see" it? What if you think of big bang as a single event with two directions of motion, in and out like a focal point, let's call this motion energy and visualize it as a black and white whole. After 13 billion light years I would expect these two extremes to still be visible as our horizons. I like to think that because we are inside of the black hole and all paths are curved is why we always see a vanishing point, and the outside of the worm holes formed when the white whole flew apart is what makes up our focal points. Matter is the past that I "see" and mass is the past that I feel and if you think of the world line of matter as always beginning at the center connection called mass then you will always look towards the beginning by "seeing" the past. Petm1 17:55, 12 March 2011 (UTC)

William James Sidis
I have just read Sidis'. Leutha 17:29, 6 August 2011 (UTC)

Sisis states that his reverse universe will only work with matter as unchanging, read his sixth paragraph, yet mass is a measure only in the present and dependent upon its relative motion. Look at a world line, we do not see it as a line moving through space because the signals used to form the world line always starts at the subjects center. JEM

I think of time as the one dimension with space a subset, we just view it backwards. Space is nothing more than the duration of the photon we are seeing now and yet no mater which direction we look in space we are "seeing" all three dimensions of time; incoming signals from past events the future, signals we are receiving now the present, and those signals that have gone past us the past. The edge of the universe we see is the outer surfaces of the atoms sending us the signals formed by big bang and we still see them as matter waves. The twist in space-time, + + + - or - + + +, that gives us the center connection we measure as mass. Mass, the dilating momentum of matter, which we always measured as our past. The only reason we can measure mass at all is because of the co moving frame formed by our consciousness ability to keep up with light waves second by second. By the way don't you think that gravity is still repulsive, as measured by our accelerators. the motion started at big bang is still the motion we feel from every atom and count as time. JEM

title: Talk:Fundamental entity


text:
User talk:Fi~enwikiversity

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Zdravim! Doufám, že se ti tu bude líbit.--Juan de Vojníkov 06:23, 14 September 2009 (UTC)

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User talk:Florim~enwikiversity

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User talk:Krishajoylovejeremiah

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title: User talk:Krishajoylovejeremiah


text:
Website fingerprinting

Introduction
A web user today has a trillion web pages seeking his attention every time he is online with thousands of websites being added each day. Owing to the decentralized and unregulated nature of issuance of domain names and the low cost of web hosting, it has become extremely simple for anyone to buy a domain name and host a website on his own. This strength for the growth of internet, has also meant that anyone can use the web to create a fraudulent website with content and URL’s stolen from existing genuine websites.

What problem does it address?
The site domain name, the look and feel of the content, and at times the lock icon (digital certificate) are the ONLY elements used to establish the identity of a website. There are 3 problems with this:

1.Owing to the insecure internet infrastructure the user cannot be certain that the address typed in the browser is taking the user to the intended site every time. An unsuspecting web user can end up at a fraudulent (fake) site, where the user might end up giving the credentials to the fraudster, causing immense damage thereafter.

2.Obfuscated URLs such as http://trustedbank.x.com can cheat the user into believing that he is at the authentic website.

3.Most users do not know what to check in a digital certificate, as the enterprise never communicates the same to the user.

By the time the fake website is blacklisted, victims would have already lost money to fraud.

How does website fingerprinting solve this problem?
The root cause of the problem is that there is no hard and fast definition of what constitutes the identity of a website, and which is available to the user to check before he/she can submit critical information. There is a need for a central body that defines website identity, makes it is hard to duplicate and easy to match by the customer on demand. The solution consists of 3 parts:

1. Create a central SiteFingerprint Vault of websites: This involves a. Selecting websites of critical importance (e.g. online banks, online share trading firms, ecommerce websites, email providers etc). b. Creating unique website fingerprints based on the identity parameters of the website and c. Storing them in a secure facility.

2.Give the customer a tool for the customer to identify websites: This tool is a browser plug-in that matches the SiteFingerprintSM of the website that the browser has connected to with that of the reference SiteFingerprintSM (obtained from the SiteFingerprintSM Vault and stored locally), and provide a visual cue (green band) reassuring them whenever they are on a genuine website

3.Accessing the Central SiteFingerprintSM Vault: The Central SiteFingerPrintSM Vault is accessed by the customer’s browser plug-in tool (mentioned in 2) to update the local SiteFingerPrintsSM stored on his/her computer. This access is available only by using the secure RMAP channel (Rel-ID Mutual Authentication and Encryption Protocol).

Benefits
Web users •Protection against internet fraud: Web users can now access the SiteFingerprint vault for a list of trusted websites. This practice can effectively eliminate all forms of online identity theft which essentially rely on users giving up his login and passwords at fake websites. •Legal recourse: In case a victim falls to an online identity theft, he can rely on a history of green bands with the website signatures captured and stored locally to figure out the phishing websites he/she may have visited. This history can be used as a legal document to prove that he was indeed a victim of online identity theft – which is currently almost impossible. •Complement loopholes left by existing security software: Existing anti-virus and anti-phishing solutions do not protect the customer completely, as they do not authenticate the websites. This gap is plugged by SiteFingerprintSM.

Website owners •Protection against identity spoofing: Incase a website identity is spoofed, its customers are protected. Furthermore the original website can report the incident and use his registration with the SiteFingerprint Vault to prove his case. •Reducing loss of traffic: The website can ask its customers to watch for the green band to make sure they are on the correct website, thus preventing loss of traffic. •Complement loopholes left by Anti-pharming and EV-SSL solutions: Anti-Pharming and EV SSL solutions require substantial investment and they do not guarantee customers’ authentication of website identity, as the authentication is not performed at the customer end. Educating customers to use SiteFingerprints does not require any investment and it guarantees customers peace of mind while transacting on the website.

title: Website fingerprinting


text:
User:Egm6321.f09.team3/HW2

September 23, 2009 EGM 6321 HW Problems

=Problem 1=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
For case #2

$$ h_xN = 0 \! $$

Find
Complete the details on how to obtain h(y) for case #2

Solution
General form

\begin{align} &h_xN - h_yM + (N_x-My) = 0 \qquad \text{investigating the case for} \qquad h_xN = 0 \\ &-h_yM =h(N_x - M_y) \qquad \text{factoring out}\\ &\frac{h_y}{h} = \frac{1}{M}\left(N_x-M_y\right) \qquad \text{integrating both sides gives}\\ &\int\left(\frac{h_y}{h}\right)dy = \int\frac{1}{M}\left(N_x-M_y\right)dy \qquad \text{evaluating the integral}\\ &\ln h =\int\frac{1}{M}\left(N_x-M_y\right)dy \end{align} $$

We know that
 * $$ h_x \underbrace{N}_{\neq 0} = 0 $$

which means that $$h(x,y) = h(y)$$

and that,

 * $$\frac{1}{M(x,y)}\left(N_x(x,y)-M_y(x,y)\right) = -g(y)$$

since

$$\ln h = \underbrace{\int\frac{1}{M}\left(N_x-M_y\right)dy}_{-g(y)}$$

then we arrived to $$h(y)$$

 * $$h(y)=\exp \int g(s)ds$$

 You have lost a minus sign in the final equation. Above you correctly identified this as $$-g(s)$$--Egm6321.f09.TA 03:10, 28 September 2009 (UTC)
 * }

=Problem 2=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A non-homogeous L1-ODE-VC

 * $$ y'+\frac{1}{x}y=x^{2} $$

Find
Show that

 * $$y(x)=\frac{x^{3}}{4}+\frac{C}{x}$$

Solution
The general form for a non-homogenous L1-ODE-VC is:

 * $$y' +a_0(x)y = b(x) \!$$

The one given in the problem can be re-written as:

 * $$\underbrace{1}_{N(x,y)}\dfrac{dy}{dx} + \underbrace{\frac{1}{x}y}_{M(x,y)}=b(x)$$

We know that:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ 1(0 - a_0(x)) &= -f(x)\qquad \text{which yields}\\ f(x) &= a_0(x)\\ \end{align} $$

Then making use of integrating factor knowledge:

\begin{align} a_0(x) &= \frac{1}{x}\\ h(x)&=\exp\int^{x}a_0(s)ds\qquad \text{replacing values for}\quad a_0(x) \quad \text{and evaluating the integral}\\ h(x)&=\exp(\ln x)\\ h(x) &= x \end{align} $$

Now we can solve for $$y(x)$$

\begin{align} y(x) &= \frac{1}{h(x)}\int^{x} h(s)b(s)ds \qquad \text{plugging known values in this expression yields}\\ y(x)&=\frac{1}{x}\int^{x}s s^2 ds \qquad \text{analyzing this integral and expanding terms out we end up with our desire result}\\ y(x)&=\frac{x^3}{4} + \frac{C}{x} \end{align} $$  Nice work.--Egm6321.f09.TA 03:13, 28 September 2009 (UTC)
 * }

=Problem 3=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A non-homogenous L1_ODE_VC

 * $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$

Find
Show that the above non-homogenous L1_ODE_VC is exact

Solution
For the non-homogenous L1_ODE_VC to be exact it must staisfy satisfy the 2 conditions of exactness

\begin{align} N(x,y)y' + M(x,y) &= 0 \\ M_y(x,y) &= N_x(x,y) \end{align} $$

So the first condition is met by:

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y+10]}_{M(x,y)}= 0 $$

Let's take a look at the 2nd condition:

\begin{align} M_y(x,y) & = x^4 \\ N_x(x,y) & = x \end{align} $$

Clearly $$M_y(x,y) \neq N_x(x,y)$$, so we make use of the integrating factor to see if we can make it exactly integrable.

So we re-write the ODE

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at case #1 for the integrating factor we obtain:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ \frac{2}{x^2}\left(x - x^4\right) &= -f(x) \qquad \text{applying knowledge from integrating factor we know}\\ h(x)&=\exp \int^{x}f(s)ds \qquad \text{replacing f(x) into the integral and evaluating it, obtain}\\ h(x) &= \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right) \end{align} $$

Now we can try the 2nd condition of exactness to see if by using integrating factor we can make this ODE exact

\begin{align} \bar{M}_y &= \bar{N}_x\\ \underbrace{(hM)}_{\bar{M}}dx&=\underbrace{(hN)}_{\bar{N}}dy\qquad \text{calculating the respective values we obtain}\\ \bar{N}_x &=x^2 \exp\left(\frac{2x^3}{3}\right)\\ \bar{M}_y &=x^2 \exp\left(\frac{2x^3}{3}\right) \end{align} $$

Hence the non-homogenous L1_ODE_VC is exact.

 Nice work. --Egm6321.f09.TA 03:14, 28 September 2009 (UTC)
 * }

=Problem 4=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
An aribitrary 1st order ODE:

 * $$ \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)=0$$

Find
Show that the above is non-linear and exact

Solution
So let's determine that the above ODE is non-linear by making use of the $$D(.)\!$$ operator

To be a linear ODE the following must be true:

 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Now we will define the $$D(.)\!$$ operator for the above ODE:

 * $$D(.)= \left(\frac{1}{3}x^{3}\right)(\cdot)^{4}\frac{d(\cdot)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\cdot)^{5}\right)$$

Then let's evaluate the linearity condition to see if it's true:

\begin{align} D\left(\alpha u + \beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u + \beta v)^{4}\frac{d}{dx}(\alpha u + \beta v)+(5x^{3}+2)\left(\frac{1}{5}(\alpha u + \beta v)^{5}\right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u)^{4}\frac{d(\alpha u)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\alpha u)^{5}\right) +  \left(\frac{1}{3}x^{3}\right)(\beta v)^{4}\frac{d(\beta v)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\beta v)^{5}\right) \end{align} $$

Clearly $$D\left(\alpha u + \beta v \right) \neq D\left(\alpha u \right) + D\left(\beta v \right)$$ so the ODE is non-linear

 Nonlinearity can be determined by inspection. This would have sufficed for this problem, although your explanation is very complete. --Egm6321.f09.TA 03:18, 28 September 2009 (UTC) To prove that the ODE is exact, let's first look at the 1st condition of exactness:

 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

hence $$N(x,y)y' + M(x,y) = 0\!$$ so 1st condition is met.

Looking at the 2nd condition where $$M_y = N_x\!$$. However, from above we can clearly see that:

 * $$M_y \neq N_x$$

So we will try to find an integrating factor to find if our ODE can be exactly integrable

\begin{align} \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0 \qquad \text{let us multiply througout  by} \qquad \frac{3}{x^3y^4}\\ y' +\frac{3}{x^3y^4}(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0\qquad \text{re-writing the equation obtain}\\ \underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y &=0 \end{align} $$ throughout

Now we will make use of integrating factor knowledge:

\begin{align} h(x,y) &= \exp\int^{x}a_0(s)ds\qquad \text{replacing the value of} \quad a_0\\ h(x,y) &= \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \text{evaluating the integral obtain}\\ h(x,y) &= \exp\left(3x - \frac{3}{5x^2}\right) \end{align} $$

Now we will try to evaluate the 2nd condition of exactness which is:

\begin{align} (hM)_{y}&=(hN)_x \qquad \text{the condition says:} \\ \bar{M}_y &= \bar{N}_x \qquad \text{evaluating the derivatives we obtain}\\ \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} } & = \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}}} \end{align} $$

Hence the ODE given is then, exact N1-ODE-VC
 * }

=Problem 5=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following 2nd order ODE

 * $$(xy)y''+x(y')^{2}+yy'=0\!$$

Find
Show that the second exactness condition is satisfied for the above equation

Solution

\begin{align} \underbrace{(xy)}_{f(x,y,p)}y''+\underbrace{x(y')^{2}+yy'}_{g(x,y,p)}&=0 \qquad \text{we can re-write the ODE}\\ (xy)y''+xp^{2}+yp&=0 \end{align} $$

The second exactness condition is the following

\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ 0 + 2p(1) +p^2(0) &= 2p(1) + 0 + p - (0) - p \qquad \text{which gives}\\ 2p & = 2p \end{align} $$

Also, the following must be met

\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ 0 + 0 + 2x &= 2x + 0\qquad \text{which gives}\\ 2x &=2x \end{align} $$

Hence the second exactness condition is met for this particular case  Show the partial derivatives used to verify the second exactness condition equations. --Egm6321.f09.TA 03:22, 28 September 2009 (UTC)
 * }

=Problem 6=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following condition is given

 * $$f_{xp} + pf_{yp} + 2f_{y} =g_{pp} \!$$

Find
Derive the above equation by differentiating

 * $$g(x,y,p) = \phi_x +\phi_yp\!$$

with respect to $$p (y')\!$$

Solution
We know that:

\begin{align} g &= \phi_x +\phi_yp \qquad \text{differentiating once wrt p}\\ g_p &= \phi_{xp} +\phi_y +p\phi_{yp} \qquad \text{differentiating again wrt p}\\ g_{pp} &= \phi_{xpp} +\phi_{yp} +\phi_{yp} + p\phi_{ypp} \qquad \text{taking into parenthesis}\quad \phi_p \quad \text{we obtain}\\ g_{pp} &= (\phi_p)_{xp} +(\phi_p)_y +(\phi_p)_y + p(\phi_p)_{yp} \qquad \text{finally arriving to the given equation because}\quad \phi_p = f\\ g_{pp} &=f_{xp} + pf_{yp} + 2f_{y} \end{align} $$

 Very nice- clear and concise. --Egm6321.f09.TA 03:23, 28 September 2009 (UTC)
 * }

=Problem 7=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following condition is given

 * $$f_{xx} +2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}\!$$

Find
Use $$\phi_{xy}=\phi_{yx}\!$$ to obtain the equation given above

Solution
We know that

\begin{align} \phi_x &= g - p(g_p - f_x) + p^2f_y \qquad \text{and that}\\ \phi_y &= g_p -pf_y - f_x \end{align} $$

So what we will need to do is find $$\phi_{xy}=\phi_{yx} \!$$ respectively and equate them to find the given condition

\begin{align} &\phi_{xy} = g_y - pg_{py} + pf_{xy} + p^2f_{yy} \\ &\phi_{yx} = g_{px} - pf_{yx} - f_{xx} \qquad \text{after re-arranging we obtain the desire given condition}\\ &f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$

 Very good. --Egm6321.f09.TA 03:24, 28 September 2009 (UTC)
 * }

=Problem 8=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following 2nd order ODE

 * $$(8x^5y')y'' + 2x^2y' +20x^4(y')^2 +4xy = 0\!$$

Find
Verify the second exactness condition on the above ODE

Solution
First let us identify some parameters in the ODE

\begin{align} \underbrace{(8x^5y')}_{f}y'' + \underbrace{2x^2y' +20x^4(y')^2 +4xy}_{g} &= 0 \qquad \text{let us look at the exactness conditions needed to be analyzed}\\ f_{xx} +2pf_{xy} + p^{2}f_{yy} &= g_{xp} + pg_{yp} - g_{y} \qquad \text{evaluating each derivative}\\ 160x^3p +2p(0) + p^2(0) & = 160x^3p +4x +p(0) - 4x \qquad \text{which gives}\\ 160x^3p &= 160x^3p \end{align} $$

Also, the other condition is as follow:

\begin{align} f_{xp} + pf_{yp} + 2f_{y}&=g_{pp} \qquad \text{evaluating each derivative}\\ 40x^4 + p(0) + 2(0) &=40x^4\qquad \text{which gives}\\ 40x^4 &= 40x^4 \end{align} $$

Hence the ODE given is exact!

 Show the partial derivatives used to verify the second exactness condition equations. Without them your solution is missing steps. --Egm6321.f09.TA 03:25, 28 September 2009 (UTC)
 * }

=Problem 9=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The following 2nd order ODE is given

 * $$\sqrt{x}y''+2xy'+3y=0$$

Find
Verify the exactness of the above ODE

Solution
For the ODE to be exact it must satisfy the two conditions of exactness

 * $$\underbrace{\sqrt{x}}_{f}y''+\underbrace{2xy'+3y}_{g}=0$$

Which means that the 1st condition of exactness is met.

\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ (0)+ p(0) + 2 (0) &= 0 \\ 0 &= 0 \qquad \text{which satisfies the first condition} \end{align} $$

However, the ODE must satisfy both conditions to be exact. So let's evaluate the next condition

\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ -\dfrac{1}{4x^{\frac{3}{2}}} + 2p(0) + p^2(0) &= 2 + p(0) - 3 \\ -\dfrac{1}{4x^{\frac{3}{2}}} &\neq -1 \end{align} $$

Since the ODE did not comply with the all of the exactness conditions then the ODE is non-exact  very good, but you must show the partial derivatives. --Egm6321.f09.TA 03:26, 28 September 2009 (UTC)
 * }

=Contributing Team Members= Barrett Mooney Egm6321.f09.team3.mooney 01:58, 23 September 2009 (UTC)

Catherine Snow EGM6321.f09.Team3.cgs 03:42, 23 September 2009 (UTC)

Francisco Kuljevan Egm6321.f09.team3.Kuljevan.Francisco 03:57, 23 September 2009 (UTC)

Angela Diggs Egm6321.f09.team3.diggs 12:28, 23 September 2009 (UTC)

title: User:Egm6321.f09.team3/HW2


text:
Introduction to psychology/Psy102/Moodle

The Moodle site for Psychology 102 is here:

http://learnonline.canberra.edu.au/course/view.php?id=1318


text:
Educational Technology: Links for Mastering New Technologies

Wikiversity Lesson Plan Links:
Collaborative Tools: Using MindMeister for Student Collaboration

Visual Thinking Strategies

Technology Links:
Wikipedia

MindMeister An online collaborative mapping tool.

eBlogger A free online journal.

RubiStar   Create rubrics for your project-based learning activities.

YouTube A video sharing website.

Make Beliefs Comix Create your very own comic strip, choosing from a variety of characters, background colors and dialogue boxes.

References:
Dewey, J. (1902). The child and the curriculum. Chicago, IL: University of Chicago Press.

Merryfield, M. (2003). Like a veil: Cross-cultural experiential learning online. Contemporary Issues in Technology and Teacher Education, 3(2), 146-171.

Shirky, C. (2009). How social media can make history [TED talk]. Retrieved from http://www.ted.com/talks/clay_shirky_how_cellphones_twitter_facebook_can_make_history.html


text:
User:Stevenarntson/Freshman Seminar/assignments/presentation

=“I’ve Always Wanted To” Presentation Guide= Presentations will occur during the final two weeks of class, and will include the following elements, in the following order:

1 the goal (10% of grade)
 * State the goal you had, which should be clear, specific, and actionable.

2 Research (20% of grade)
 * Describe the research you did: what you learned and how you learned it.

3 Your Accomplishment (20% of grade)
 * Detail. How far did you get toward your goal?
 * Demonstrate. If you’ve made something, show it for the class. If you’ve read or written something, read a little of it aloud. If you’ve learned to do something, do it.

4 The Post Mortem (50% of grade)
 * In industry, the post mortem is a project review analysis that’s done after a project is completed, to see what went right and what went wrong, and how things can be better next time. Discuss the strategies you used to reach this large goal. How did you break the project into segments? Did you follow through? Get distracted by other concerns? Find that your goal was not well enough defined?

5 Presentations Must
 * 1) Be interesting.
 * 2) Follow the order explained above.
 * 3) Be less than 10 minutes long.
 * 4) Not disturb other classes (they shouldn’t be too loud, for instance).
 * 5) Observe those hazy rules we as a society call “good taste.” If you have questions about classroom appropriateness, please come talk to me. We can probably work it out if it's done in advance.


text: User:Graeme E. Smith/Architectural Constraints on Implicit Memory: Implications for Brain Architecture and Psychology


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title: Template:Hl2


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User:Egm6321.f09.team4

Homework page for Team #4.

Class Web Page

Lecture Slides

Class Report Table

Assignment Descriptions

Homework #1: user:egm6321.f09.team4/HW1

Homework #2: user:egm6321.f09.team4/HW2

Homework #3: user:egm6321.f09.team4/HW3

Homework #4: user:egm6321.f09.team4/HW4

Homework #5: user:egm6321.f09.team4/HW5

Homework #6: user:egm6321.f09.team4/HW6

Homework #7: user:egm6321.f09.team4/HW7

Teammate Evaluation #1 (Due 10-07-2009)

Teammate Evaluation #2 (Due 12-13-2009)

title: User:Egm6321.f09.team4


text:
User:Egm6321.f09.team4/HW1

Homework #1: Team 4 (Complete) Problem #1: Given a function $$f\left(y(t),t\right)$$, find: $$\frac{df}{dt}$$ and $$\frac{d^{2}f}{dt^2}$$.

Solution:

By the chain rule for partial differentiation, the first derivative can be decomposed into: $$\begin{align} \frac{df}{dt} &= \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial t} \\ &= \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t}     \end{align} $$  Need to have $$\frac{d}{dt}$$ differentials. Specifically, $$\frac{df}{dt}=\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}$$. This accounts for both explicit dependence on $$ t$$ with the partial differential, and implicit dependence with the first term. --Egm6321.f09.TA 02:58, 24 September 2009 (UTC) The second derivative can be found by repeating the same process on the result found for the first derivative: $$\begin{align} \frac{d^{2}f}{dt^{2}} &= \frac{d}{dt}\left(\frac{df}{dt}\right) = \frac{d}{dt}\left(\frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t}\right) \\ &= \frac{d}{dt}\left(\frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial t} + \frac{d}{dt}\left(\frac{\partial y}{\partial t}\right) \frac{\partial f}{\partial y} + \frac{d}{dt}\left(\frac{\partial f}{\partial t}\right) \qquad \text{(by product rule)} \\ &= \left( \frac{\partial^{2}f}{\partial y\partial t} \frac{\partial t}{\partial t} + \frac{\partial^{2}f}{\partial y^{2}} \frac{\partial y}{\partial t}\right) \frac{\partial y}{\partial t} + \left( \frac{\partial^{2}y}{\partial t^{2}} \frac{\partial t}{\partial t}\right)\frac{\partial f}{\partial y} + \left( \frac{\partial^{2}f}{\partial t^{2}}\frac{\partial t}{\partial t} + \frac{\partial^{2}f}{\partial t \partial y}\frac{\partial y}{\partial t}\right) \qquad \text{(by chain rule)} \\ &= \frac{\partial^{2}f}{\partial y\partial t}\frac{\partial y}{\partial t} + \frac{\partial^{2}f}{\partial y^{2}}\left(\frac{\partial y}{\partial t}\right)^{2} + \frac{\partial f}{\partial y} \frac{\partial^{2}y}{\partial t^{2}} + \frac{\partial^{2} f}{\partial t^{2}} + \frac{\partial^{2} f}{\partial y \partial t}\frac{\partial y}{\partial t} \\ &= 2\frac{\partial^{2} f}{\partial y \partial t}\frac{\partial y}{\partial t} + \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{\partial y}{\partial t}\right)^{2} + \frac{\partial f}{\partial y}\frac{\partial^{2} y}{\partial t^{2}} + \frac{\partial^{2} f}{\partial t^{2}} \end{align} $$  Why have you changed notation? The problem was posed differently. It is ok to change notation or variables if it is helpful, however the statement should be phrased in the manner it was posed in class. Then state that you are using different variables. Finally, convert your solution to the class notation so that it is clear that you have arrived at the solution. --Egm6321.f09.TA 03:00, 24 September 2009 (UTC) Both of these results can alternatively be expressed in condensed notation as follows: $$   \begin{align} \frac{df}{dt} &= f_{y}\cdot y' + f_{t} \\ \frac{d^{2}f}{dt^{2}} &= 2\cdot f_{yt}\cdot y' + f_{yy}\cdot \left(y'\right)^{2} + f_{y}\cdot y'' + f_{tt} \end{align} $$

Problem #2: Show that the solution of $$ p\left(x\right)'+p\left(x\right) = x $$ is $$ p\left(x\right)=Ae^{-x}+x-1 $$ using integrating factors.

Solution:

We want to find a function $$h\left(x\right)$$ such that

$$\begin{align} h(x)p^{\prime}(x)+h(x)p(x)=h(x)x&=(p(x)*h(x))^{\prime} \\ &=p^{\prime}(x)h(x)+h^{\prime}(x)p(x) \qquad \text{(by product rule)} \end{align}$$

This statement is not rigorous. You have arrived at the proper answer but the steps are not properly connected. The equation needs to be rewritten in terms of $$\overline{M},\overline{N}$$--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

$$h(x)p(x)=h^{\prime}(x)p(x)$$

$$h(x)=h^{\prime}(x)$$

$$\frac{h^{\prime}(x)}{h(x)}=1$$

Integrate:

$$\int 1dx=\int \frac{h{}'(x)}{h(x))}dx=ln(h(x))$$

$$e^{\int 1dx}=e^{ln(h(x))}=>e^{x+c}=C_{1}e^{x}+C_{2}=h(x)$$

Let $$C_{2}=0$$ because it is arbitrary. Plug $$h(x)$$ expression back into original:

You do not need any constants of integration in solving the separable diffeQ for $$h(x)$$ because you multiply $$h(x)$$ through the entire original equation, and the constants will drop out (as your $$C_{1}$$ does below).--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

$$C_{1}e^{x}p^{\prime}(x)+C_{1}e^{x}p(x)=

(p(x)*C_{1}e^{x})^{\prime}=xC_{1}e^{x}$$

Integrate:

$$p(x)*C_{1}e^{x}=\int{xC_{1}e^{x}}dx$$

Divide through:

$$p(x)=\frac{1}{C_{1}e^{x}}\int{xC_{1}e^{x}}dx$$

From integration tables,

$$p(x)=\frac{1}{C_{1}e^{x}}\int{xC_{1}e^{x}}dx=\frac{C_{1}}{C_{1}e^{x}}\left [ \frac{1}{1^{2}}(x-1)e^{x}+A_{1}\right ]$$

$$p(x)=\frac{1}{e^{x}}\left [ xe^{x}-e^{x}+A_{1} \right ]=x-1+A_{1}e^{-x}$$

Answer.

Problem #3: Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution

Define a linear operator $$D\left ( \cdot \right )$$ such that $$D(\cdot )=2x^2+\sqrt{\left (\cdot  \right ) }+x^5\left (\cdot  \right ) ^3\frac{d}{dx}\left ( \cdot  \right )$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition gives:

$$D(\alpha u + \beta v)=2x^2+\sqrt{\alpha u + \beta v}+x^5\left ( \alpha u + \beta v \right )^3\frac{d}{dx}\left ( \alpha u+\beta v \right )$$

Substituting for the right side of the linearity condition:

$$\alpha D(u)+\beta D(v)=\alpha\left (2x^2+\sqrt{u}+ x^5u^3\frac{du}{dx} \right )+\beta\left ( 2x^2+\sqrt{v}+x^5v^3\frac{dv}{dx} \right )$$

Simplifying and comparing both sides of the linearity condition:

$$2x^2+\sqrt{\alpha u+\beta v}+\alpha x^5(\alpha u+ \beta v)^3\frac{du}{dx}+\beta x^5(\alpha u + \beta v)^3\frac{dv}{dx}\neq2x^2\alpha+\alpha\sqrt{u}+\alpha x^5u^3\frac{du}{dx}+2x^2\beta+\beta\sqrt{v}+\beta x^5v^3\frac{dv}{dx}$$

The condition is therefore not met, and $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is a nonlinear 1st Order ODE.

Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

Problem #4: Given a function $$F(x,y,y^{\prime})=x^2y^5+6(y^{\prime})^2=0$$, show that F is a nonlinear 1st Order ODE.

Solution:

Define the operator $$D\left ( \cdot \right )$$ such that $$D\left ( \cdot  \right )=x^2\left ( \cdot  \right )^5+6\left ( \frac{d}{dx}\left [ \cdot  \right ] \right )^2$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition, we get:

$$D(\alpha u + \beta v)=x^2\left ( \alpha u+\beta v \right )^5+6\left ( \frac{d\left ( \alpha u+\beta v \right )}{dx} \right )^2$$

This can be simplified to:

$$x^2\left ( \alpha u+\beta v \right )^5+6\left ( \alpha \frac{du}{dx}+\beta\frac{dv}{dx}\right )^2$$

The right side of the linearity condition can be written as follows:

$$\alpha D(u)+\beta D(v)=\alpha\left ( x^2u^5+6\left [ \frac{du}{dx} \right ]^2 \right )+\beta\left ( x^2v^5+6\left [ \frac{dv}{dx} \right ]^2 \right )$$

And can be simplified to:

$$\left ( \alpha u^5 + \beta v^5 \right )x^2+6\alpha \left ( \frac{du}{dx} \right )^2+6\beta \left ( \frac{dv}{dx} \right )^2$$

Comparing the left and right hand sides of the linearity condition, we get:

$$x^2\left ( \alpha u+\beta v \right )^5+6\left ( \alpha \frac{du}{dx}+\beta\frac{dv}{dx}\right )^2\neq \left ( \alpha u^5 + \beta v^5 \right )x^2+6\alpha \left ( \frac{du}{dx} \right )^2+6\beta \left ( \frac{dv}{dx} \right )^2$$

This is clearly not an equality, therefore F(x,y,y') is a nonlinear 1st Order ODE.

Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

Problem #5:

Part 1. Create an exact nonlinear 1st Order ODE of the form $$\Phi_x\left(x,y\right)+\Phi_y\left(x,y\right)y^{\prime}=0$$ using the equation $$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Solution:

If,

$$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Then

$$\Phi_{x}\left(x,y\right)=24x^{3}=M$$

and

$$\Phi_{y}\left(x,y\right)=3y^{1/2}=N$$

For a nonlinear, 1st order, ODE:

$$M + Ny^{\prime}=0$$

Therefore,

$$24x^3 + 3y^{1/2}y^{\prime}=0$$

We have forced exactness

Part 2. Create three more exact nonlinear 1st Order ODEs by inventing new $$\Phi\left(x,y\right)$$ functions.

Ex. 1

Choose $$\Phi\left(x,y\right)=x^{1/2}+y^{3}\quad\text{:}$$

$$ \begin{align} \Phi_{x}\left(x,y\right)&=0.5x^{-1/2}=M\left(x,y\right) \\ \Phi_{y}\left(x,y\right)&=3y^{2}=N\left(x,y\right) \end{align} $$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = \frac{1}{2}x^{-1/2}+3y^{2}y' = 0$$

Ex. 2

Choose $$\Phi\left(x,y\right)=3y^2+2x^5\quad\text{:}$$

$$\Phi_x\left(x,y\right)=10x^4=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=6y=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 10x^4+(6y)y' = 0$$

Ex. 3

Choose $$\Phi\left(x,y\right)=3x^2y^3\quad\text{:}$$

$$\Phi_x\left(x,y\right)=6xy^3=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=9x^2y^2=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 6xy^3+\left(9x^{2}y^{2}\right)y' = 0$$

Nice work.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

title: User:Egm6321.f09.team4/HW1