text:
User talk:FrequentIntenseMatureSuggestiveThemes

Post me messages below.

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

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! Jtneill - Talk - c 08:51, 23 April 2012 (UTC)

 Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. Even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

Custodian nomination
Thanks for your edits. FYI, I've speedy deleted your self-nomination for full custodianship, assuming it was in jest. Happy to discuss. Sincerely, James -- Jtneill - Talk - c 08:51, 23 April 2012 (UTC)
 * Why did you delete my request for custodianship? I am 8 years old, about as mature enough to become a custodian. I also changed my signature. --FIMST (talk 2 me or contributions) 16:35, 23 April 2012 (UTC)
 * I deleted it because I interpreted it as a jestful/facetious nomination not a serious nomination. Feel free to self-nominate it again and let the community discuss and decide. I imagine the community would like to see more evidence of experience with contributing to Wikiversity. -- Jtneill - Talk - c 19:13, 23 April 2012 (UTC)
 * Uh, given his disruption across Wikimedia, that's a no-no.--Jasper Deng (talk) 23:37, 23 April 2012 (UTC)

User rights request for confirmed
Since I'm 1 day old and I've made over 17 edits, I want to bypass the wait for 3 days and become confirmed. --FIMST (talk 2 me or contributions) 17:27, 23 April 2012 (UTC)

Globally locked
This account is globally locked. It shows only contributions at Wikiversity, but it is possible that other contributions have been deleted or revision deleted. plus there are many socks. See CentralAuth. This user is certainly Brandon Sky, per the email address above and [].

Brandon, if you want to contribute to Wikiversity, register an account for that purpose, and use it properly. You may verify your identity with me by email. (This FIMST account should still have email access here.) —Abd (discuss • contribs) 18:38, 5 January 2014 (UTC)
 * Locked accounts cannot use Special:EmailUser. I can say that he has emailed me before and confirmed that he is abusing multiple accounts as in the page you linked.--Jasper Deng (talk) 03:01, 6 January 2014 (UTC)
 * I've heard differently, from locked users, but this stuff changes sometimes. The user can email me, one way or another. I'll be able to identify him by the mail address, if he uses that gmail account. The mail doesn't have to come from this FIMST account. Thanks for the note. --Abd (discuss • contribs) 03:11, 6 January 2014 (UTC)
 * Those users probably already had your email address, locked users can't access their accounts at all.--Jasper Deng (talk) 04:37, 6 January 2014 (UTC)
 * Well, I was told the exact behavior they saw, but I'm not going to look back. It's quite possible that the behavior has been changed. It is trivial for a locked user to bypass the restriction to gain access to email, and no locked user who has any clue at all would be unable to do it. In any case, Jasper, how you know that the behavior is actually as you say? Is there a MediaWiki documentation page on this? I did search and found nothing, so far. --Abd (discuss • contribs) 17:46, 6 January 2014 (UTC)
 * The only way a locked user can do that is to make a sockpuppet. After his account was locked he posted at the local custodians' noticeboard, even giving out his password. This is something of mw:Extension:CentralAuth and is also mentioned at Global locks - login attempts across all Wikimedia projects fail with the error "Incorrect password entered" (even if the password is correct). They can email you if they know your address, but you have to login to use Special:EmailUser, so a lock prevents that.--Jasper Deng (talk) 21:06, 6 January 2014 (UTC)
 * I don't know, but he might be starting a new start, I believe in new starts. --Goldenburg111 21:42, 6 January 2014 (UTC)
 * No, he first must appeal the existing sanctions, evading them is sockpuppetry and not allowed.--Jasper Deng (talk) 22:28, 6 January 2014 (UTC)
 * Ok. Understood. --Goldenburg111 22:31, 6 January 2014 (UTC)

 * Actually, Jasper, this is the problem, and some sophisticated users know it. We cannot stop him from doing that. We had a user here who was globally locked, and he had an unlinked account here. He was not disruptive *here* and was active here with positive editing. This user, however, had serious enemies elsewhere. (He'd done some disruptive stuff years earlier.) As long as he had an active account here, any steward could pull up his IP and user agent data and could see if he was editing elsewhere. I argued for allowing him to continue editing here, partly on this basis.

 * I was actually desysopped here because he was locally blocked without local disruption, violating long-standing policy here, and I unblocked him, and there was not an adequate consensus here to maintain this.

 * The result of his block, when his unblock template went unanswered for more than a year, may have been that he began socking again. (He hadn't actually socked for a long time, he did have a different account at another wiki, with permission from 'crats there, he'd done it right.) In any case, we had a probationary custodian here, candidate for 'crat, and it appears to have been a WMF office action that identified him as this globally banned user. They would not say what the evidence was. He claims, privately, that it wasn't him, but that it was indeed someone he knew, and that they may have shared a computer at one point. Bad Idea. This guy was known for shooting himself in the foot. How they identified him, I'm sure they don't want to say, because *legitimate checkuser* might not have been able to accomplish this. Checkusers are not supposed to keep old data, for example. This is what I've learned: what are stated as rules and policy are not necessarily what is actually followed, if nobody is watching. Democratic societies have learned the necessity of whistle-blowers and watchdogs, because power corrupts. And those in power always think that, surely, with their good intentions, they will be exceptions.

 * Global locks, by policy, are quite restricted in usage. I am *not* commenting about this particular user. However, I just saw a global lock that passed by with little notice. . I don't see, in the evidence presented, anything clear that establishes the kind of behavior normally required for a lock. It might be there, but ... it's not clear. Much is made of a Wikia ban, which should be totally irrelevant. And this is what is so remarkable: Central Auth. Take a look at the es.wikipedia edit count. 330,094. This is not your average vandal or ordinary disruptive sock master! I don't read Spanish, but it's looking to me like the user lost a political battle, was upset, and may have made a set of abusive edits at meta. So they are getting back. The *substance* of the report was entirely, AFAIK, with regard to one project, es.wikipedia. Meta is being abused, it appears to me. It used to be difficult to get a lock, for anything other than obvious global vandalizing and spamming accounts. It's become easy. Supposedly the accounts vandalized meta. However, the evidence appears to be, with the exception of one edit, abusive usernames. In spite of being hidden, they can be easily seen.

 * I wonder if you would be willing to look at the human story here, Japser. This es.wikipedia user did an *enormous* amount of work, and has been tossed on the trash heap, whether or not he "deserved it." Wouldn't you be angry? It used to be understood on Wikipedia that users, when blocked, would be angry, and so if they swore at the blocking administrator, it was to be disregarded. Of late, it seems that punishment is more in order. The wiki ideals are being lost. --Abd (discuss • contribs) 00:52, 7 January 2014 (UTC)
 * To be honest, most of the Wikimedia Projects focus on harm than helping the wiki (adding pages, according to the project's scope). Well, an example is me, rights are not power, which most users, who I talk about this, usually say. Rights are for the benefit of a wiki, not disruption. Here's an example:

 * After being stalked for at least up at July, I hade to "not request user rights". So you can literally see the message here, they are actually pretending these rights are powers, not benefits for a wiki.

 * Here on Frequent, we are just making him feel bad, don't you think? Talking about his global lock. I recommend you to close this discussion, thus, it may be causing anger for this user. Believe me, I have been angered by many people across. --Goldenburg111 01:24, 7 January 2014 (UTC)
 * Goldenburg, I appreciate your concern for the feelings of the user. However, this user is apparently quite sophisticated, and the purpose of the discussion here was to invite him to participate here, if he chooses to do that. Your experience was not unusual, my guess is that many like you encounter similar experiences.
 * I don't think you have yet understood it, not fully. The other wikis are focused on the content, and are not about helping the community. The concept of "educating" someone who looks like a vandal or POV-pusher is not on their mind, and they reject it when it's mentioned. "Not Therapy" is commonly asserted. Wikipedia is about building an encyclopedia, and "learning community" is not what they are.
 * However, this is Wikiversity, and our goal is education, but not *only* educational materials, but also learning-buy-doing, and there was once some substantial interest in Wikiversity in resources and projects for children. We *are* a learning community.

 * The wiki community, in general and for the most part, isn't trained or experienced in the education of children, and what it takes, and *especially* what it takes in a wiki environment. For a beginning, it takes patience.
 * I know that some prominent users at Simple English Wikipedia had an attitude similar to mine. I don't see that this attitude is protected there, though.
 * I don't think FIMST is a child. I have not researched his history, how he came to be what he is known for and was apparently locked for. I don't really care, because I've seen many "disruptive users" from elsewhere become cooperative participants in Wikiversity. There is little reason to fight here, if a person is actually interested in developing educational materials and in learning. "Educational materials" can represent all kinds of points of view, they can be imbalanced and even misleading, but, here, we balance them by inclusion, with criticism, not with deleting them. At least that's the way it usually goes. People may debate issues here. That is theoretically prohibited on Wikipedia. Wikversity is a kind of haven, it's famous for that. --Abd (discuss • contribs) 02:44, 7 January 2014 (UTC)
 * I, for one, know his real age (his mother allegedly did email me back in 2012). There is only a finite amount of w:en:WP:ROPE we can give to disruptive users. Our global locking policy isn't perfect and stewards aren't always down to do locks like this one, but it's the best we have in lieu of a proper global blocking policy.--Jasper Deng (talk) 20:25, 9 January 2014 (UTC)


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User talk:Dross4565/enes100/My Work 3

Appreciate the help on the project also nice find on the walkthrough--Garcia123 (talk) 03:41, 23 April 2012 (UTC)

Week 1 p3 Comment Kim DelVillar Thanks for helping me understand anything and everything that isn't clear to me, I feel like we got a good start on this project.

week 1 p3 grading 1sfoerster (talk) 19:59, 24 April 2012 (UTC)
 * 0/0/0 .. nothing to grade

week 2 p3 grading 1sfoerster (talk) 17:08, 1 May 2012 (UTC)
 * 10/10/15 ... need pictures of the product, not pictures from someone else's work

week 3 p3 grading 1sfoerster (talk) 13:31, 8 May 2012 (UTC)
 * 10/10/30 ... why reverse the picture of the glue bottle? why put up a picture of a glue bottle? Why not a picture of the gun and video of it working? Why not description of all the decisions, little frustrations encountered assembling? I am looking forward to seeing the final product.

Week 4 p3 Comment Kim DelVillar Wow good job on getting the rubber band gun to work! Even though you told me about your troubles going through it, at the end you made it all come together nicely.

week 4 p3 grading 1sfoerster (talk) 20:46, 15 May 2012 (UTC)
 * 10/10/80 ... good job ...

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

title: User talk:Dross4565/enes100/My Work 3


text:
User talk:Kdelvill4474/enes100/My Work 3

Appreciate the effort and help on the project--Garcia123 (talk) 03:43, 23 April 2012 (UTC)

week 1 p3 grading
 * 10/10/50 .. good job

Week 1 p3 Comment David Ross
 * Love the pictures and amount of detail you put into your narration

week 2 p3 grading 1sfoerster (talk) 17:06, 1 May 2012 (UTC)
 * 10/5/15 .. good, detailed narrative ... better if had pictures of what you cut

Week 2 p3 Comment David Ross
 * Nice pictures of your work and good detailed description of what you precisely did. I like your use of the jewelers saw, I used a jigsaw and had to smooth out the edges.

Week 3 p3 Comment David Ross
 * Again nice pictures and details on the work you did over the week. I know, first hand, that cutting the wood out was a little difficult so good job on getting the pieces almost precise and using the sandpaper to smooth them out.

week 3 p3 grading 1sfoerster (talk) 13:27, 8 May 2012 (UTC)
 * 10/5/50 .. good work, lots of pictures .. looking forward to finished product

Week 4 p3 Comment David Ross
 * Great that u got the gun finished. sorry that the sliding bolt was a little off, i fell into that same problem the first time.

week 4 p3 grading 1sfoerster (talk) 20:44, 15 May 2012 (UTC)
 * 10/10/30 ... why didn't you bring to class ... leave in the glass room .. shoot video of it working?


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User talk:Amirreza Saharkhiz/My Third Work

Great job working on the gimbal!! its coming along great! Jromanch0088 (talk) 04:08, 23 April 2012 (UTC)

Yeah! It's lookin' good! I'm a little confused though. Were the motors supposed to stay on so that the gimbal's axes could spin automatically, or are they just for smooth motion? Also, I think you need to move this page into your "userspace." Shatton8111 (talk) 17:09, 23 April 2012 (UTC)


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User:Egm4313.s12.team8.tclamb/R7-1

=R7.2 - Plot wave equation solution=

Problem Statement
Plot the truncated-series:
 * $$u(x,t)\,=\,\sum\limits_{j=1}^n a_j \cos(c \omega_j t) \sin(\omega_j x)$$

with $$n=5$$, and for:
 * $$t\,=\,\alpha p_1\,=\,\alpha \frac{2\pi}{c \omega_1}\,=\,\alpha\frac{2L}{c}$$
 * $$\alpha\,=\,0.5,\,1,\,1.5,\,2$$

and:
 * $$\omega_j\,=\,j\frac{\pi}{L}$$
 * $$a_j\,=\,\begin{cases} 0 \text{ for } j=2m \text{ (even)} \\ -4 / (\pi^3 j^3) \text{ for } j = 2 m + 1 \text{ (odd)} \end{cases}$$

Solution
The plots are generated with the following Mathematica code:
 * [[Image:egm4313.s12.team8.tclamb.r7.2.src.gif]]

 * [[Image:egm4313.s12.team8.tclamb.plot1.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot2.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot3.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot4.png]]

title: User:Egm4313.s12.team8.tclamb/R7-1


text: File:Egm4313.s12.team8.tclamb.r7.2.src.gif


title: File:Egm4313.s12.team8.tclamb.plot1.png


title: File:Egm4313.s12.team8.tclamb.plot2.png


title: File:Egm4313.s12.team8.tclamb.plot3.png


text:
File:Egm4313.s12.team8.tclamb.plot4.png

Summary
ʀ

title: File:Egm4313.s12.team8.tclamb.plot4.png


text:
User:Egm4313.s12.team8/R7

=R7.1 - Verify the functions=

Problem Statement
$$ \displaystyle \text{Verify (1)-(2) }\ $$ ((4) and (5) p19.9, respectively)

$$ \displaystyle \text{Given:} $$

Verifying (1)
$$ \displaystyle \text{From equation } \, $$($$) $$ \displaystyle \, \text{ we have } $$ $$ \displaystyle \langle \bar f, \bar g \rangle := \int_0^L \bar f(x) \, \bar g(x) \, dx, \text{ where we can plug in } \phi \, \text{ from } \, $$ ($$) $$ \displaystyle \, \text{ to get} $$

$$ \displaystyle \text{From } $$ ($$), $$ \displaystyle \text{ and using equation } \, $$ ($$), $$ \displaystyle \, \text{ we get}$$

$$ \displaystyle \text{or} $$

$$ \displaystyle \text{From equation } \, $$ ($$), $$ \displaystyle \text{ for all constants of } \omega_i \text{ and } \omega_j \text{ yield } 0 \text{ as all values of } \sin (\pi) \text{ equal zero. Thus,} $$ $$ \displaystyle \langle \phi_i, \phi_j \rangle = 0 \ \text{ for } \ i \ne j $$

Verifying (2)
$$ \displaystyle \text{Given equation } \, $$($$) $$ \displaystyle \, \text{ and from } $$ ($$), $$ \displaystyle \, \text{ we have } $$

$$ \displaystyle \text{Plugging in the identity, and substituting } L=\pi \text{ from } [0,\pi] \text{ gives } $$

$$ \displaystyle \text{or} $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{1}{2} \left [ x - \frac{1}{2 \omega_j} \sin(2 \omega_j x) \right ]_0^\pi $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{1}{2} \left [ \pi - \frac{1}{2 \omega_j} \sin(2 \omega_j \pi) \right ] $$ $$ \displaystyle \text{Thus} $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{\pi}{2} = \frac{L}{2} \text{ for } i=j. $$

=R7.2 - Plot wave equation solution=

Problem Statement
Plot the truncated-series:
 * $$u(x,t)\,=\,\sum\limits_{j=1}^n a_j \cos(c \omega_j t) \sin(\omega_j x)$$

with $$n=5$$, and for:
 * $$t\,=\,\alpha p_1\,=\,\alpha \frac{2\pi}{c \omega_1}\,=\,\alpha\frac{2L}{c}$$
 * $$\alpha\,=\,0.5,\,1,\,1.5,\,2$$

and:
 * $$\omega_j\,=\,j\frac{\pi}{L}$$
 * $$a_j\,=\,\begin{cases} 0 \text{ for } j=2m \text{ (even)} \\ -4 / (\pi^3 j^3) \text{ for } j = 2 m + 1 \text{ (odd)} \end{cases}$$

Level 1 Solution
The plots are generated with the following Mathematica code:
 * [[Image:egm4313.s12.team8.tclamb.r7.2.src.gif]]

 * [[Image:egm4313.s12.team8.tclamb.plot1.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot2.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot3.png]]
 * [[Image:egm4313.s12.team8.tclamb.plot4.png]]

=R7.3 - Dot products, magnitudes, and angles between functions=

Problem Statement
Find: (a) The scalar product $$\displaystyle < f,g> .$$ (b) The magnitude of $$\displaystyle f $$ and $$\displaystyle g $$. (c) The angle between $$\displaystyle f $$ and $$\displaystyle g $$. For: (1) $$\displaystyle f(x)=cos(x), g(x)=x $$ for $$\displaystyle -2\leq x\leq 10 $$. (2) $$\displaystyle f(x)=\frac{1}{2}(3x^2-1), g(x)=\frac{1}2{(5x^2-3x)} $$ for $$\displaystyle -1\leq x\leq 1 $$.

Part 1
(a) We know that $$\displaystyle < f,g>=\int_{a}^{b}f(x)g(x)dx $$ Plugging in our given $$\displaystyle f $$ and $$\displaystyle g $$ functions: $$\displaystyle \int_{-2}^{10}xcos(x)dx $$    (7.3.1) Integrating (7.3.1) by parts where: $$\displaystyle du=cos(x),v=x,u=sin(x),dv=dx $$ We find: $$\displaystyle \int vdu=vu-\int udv $$ $$\displaystyle \int xcosx=xsinx-\int sinxdx $$ This leads us to the solution: $$\displaystyle =xsinx+cosx+C     (7.3.2)$$ where C=0.

(b) We know that the magnitude of a function is obtained as follows: $$\displaystyle \parallel f\parallel ==[\int_{a}^b{f^2(x)dx}]^{1/2} $$ Plugging our given values in and using trigonometric identities: $$\displaystyle \parallel f(x)\parallel =[\int_{-2}^{10}{cos^2(x)dx}]^{1/2}=[\int_{-2}^{10}{(\frac{1}{2}(cos(2x))+\frac{1}{2})dx}]^{1/2} $$    (7.3.3) Using u substitution on (7.3.3) to integrate, where: $$\displaystyle u=2x,du=2dx $$ We find: $$\displaystyle \frac{1}{4}\int_{-2}^{10}cos(u)du=\int_{-2}^{10}\frac{1}{2}dx $$ This leads to: $$\displaystyle [\frac{1}{2}(sinxcosx+x)\mid _{-2}^{10}]^{1/2} $$ And allows us to solve for our answer: $$\displaystyle \parallel f(x)\parallel =2.47 $$ Doing the same process (without needing to integrate by parts) for our given g function: $$\displaystyle \parallel g(x)\parallel =[\int_{-2}^{10}x^2dx]^{1/2} $$ $$\displaystyle \parallel g(x)\parallel =[\frac{x^3}{3}\mid _{-2}^{10}]$$ Giving us our final magnitude of: $$\displaystyle \parallel g(x)\parallel =18.33 $$

(c) We know that: $$\displaystyle cos(\theta )=\frac{}{\parallel f\parallel \parallel g\parallel }    (7.3.4)$$ Plugging in solved values shows: $$\displaystyle cos(\theta )=\frac{-7.68}{(2.478)(18.33)} $$ Leading us to: $$\displaystyle \theta =1.74 radians $$

Part 2
(a) We know that $$\displaystyle < f,g>=\int_{a}^{b}f(x)g(x)dx $$ Plugging in our given $$\displaystyle f $$ and $$\displaystyle g $$ functions: $$\displaystyle \frac{1}{2}\int_{-1}^{1}(3x^2-1)(5x^3-3x)dx $$    (7.3.5) Integrating (7.3.5), we find: $$\displaystyle \frac{1}{2}[\frac{15}{6}x^6-\frac{14}{4}x^4+\frac{3}{2}x^2\mid _{-1}^{1}]$$ This leads us to the solution: $$\displaystyle =0 $$

(b) We know that the magnitude of a function is obtained as follows: $$\displaystyle \parallel f\parallel ==[\int_{a}^b{f^2(x)dx}]^{1/2} $$ Plugging our given values in and using trigonometric identities: $$\displaystyle \parallel f(x)\parallel =[\int_{-1}^{1}(\frac{3}{2}x^2-\frac{1}{2})^2dx]^{1/2}=[\frac{9}{20}x^5-\frac{x^3}{2}+\frac{x}{4}\mid _{-1}^{1}]^{1/2}$$    (7.3.7) This allows us to solve for our answer: $$\displaystyle \parallel f\parallel =.632456$$

We now obtain the magnitude of the g function in the same way, as follows: $$\displaystyle \parallel g(x)\parallel =[\int_{-1}^{1}(\frac{5}{2}x^3-\frac{3}{2}x)^2dx]^{1/2}=[\frac{25}{28}x^7-\frac{3}{2}x^5+\frac{3}{4}x^3]\mid _{-1}^{1}]^{1/2}$$ Which leads up to our final magnitude: $$\displaystyle \parallel g(x)\parallel = .5345 $$

(c) We know that: $$\displaystyle cos(\theta )=\frac{}{\parallel f\parallel \parallel g\parallel }    (7.3.4)$$ Plugging in (7.3.8) and (7.3.9) shows: $$\displaystyle cos(\theta )=\frac{0}{(.632456)(.5345)} $$ Leading us to: $$\displaystyle \theta =\frac{\pi}{2} $$ radians

=R7.4 - Fourier series of |x|=

Problem Statement
K 2011 page 482 # 6, 9, 12, 13

-for 6 and 9 graph f(x) on $$ \displaystyle -\pi $$ to $$ \displaystyle \pi $$

-for 12 and 13 find the Fourier series of the given function f(x) and sketch or graph the partial sums up to that including $$ \displaystyle \cos(5x) $$ and $$ \displaystyle \sin(5x) $$.

Problem 6
$$ \displaystyle f(x)=|x| $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=abs(x)

>> plot(x,y)

9.
$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi,0) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (0,\pi) $$

Matlab Code:

>> x1=linspace(-pi,0)

>> y1=x1

>> x2=linspace(0,pi)

>> y2=pi-x2

>> plot(x1,y1,x2,y2)

Problem 12
$$ \displaystyle f(x)=|x| $$

This function is an even function.

This is proven by the fact that $$ \displaystyle f(-x)=f(x) $$

Since this is an even function and of period $$ \displaystyle 2\pi $$ its Fourier series reduces to the Fourier cosine series shown here:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty} a_n\cos(nx) $$

Where the coefficients are equal to:

$$ \displaystyle a_0=\frac{1}{\pi}\int_{0}^{\pi}f(x)dx $$

$$ \displaystyle a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle a_0=\frac{1}{\pi}\int_{0}^{\pi}xdx $$

$$ \displaystyle a_0=\frac{\pi}{2} $$

$$ \displaystyle a_n=\frac{2}{\pi}\int_{0}^{\pi}x\cos(nx)dx $$

Using integration by parts:

$$ \displaystyle a_n=\frac{2}{\pi}(\frac{nx\sin(nx)+\cos(nx)}{n^2})$$ evaluated on x=$$ \displaystyle (0,\pi) $$

Evaluating, reduces to:

$$ \displaystyle a_n=\frac{2}{\pi}(\frac{\cos(\pi n)-1}{n^2}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\frac{\pi}{2} + \sum_{n = 1}^{\infty} \frac{2}{\pi}(\frac{\cos(\pi n)-1}{n^2}) \cos(nx) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{\pi}{2} - \frac{4}{\pi}(\cos x+ \frac{1}{9}\cos3x+ \frac{1}{25}\cos5x+...) $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=pi/2-(4/pi)*(cos(x)+(1/9)*cos(3*x)+(1/25)*cos(5*x))

>> plot(x,y)

Problem 13
$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi,0) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (0,\pi) $$

Since the function has a period of $$ \displaystyle 2\pi $$ its Fourier series is:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty}(a_n\cos(nx)+ b_n\sin(nx)) $$

Where the coefficients equal:

$$ \displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx $$

$$ \displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx $$

$$ \displaystyle a_0=\frac{1}{2\pi}(\int_{-\pi}^{0}xdx+ \int_{0}^{\pi}(\pi-x)dx) $$

$$ \displaystyle a_0=\frac{1}{2\pi}(\frac{-\pi^2}{2}+\pi^2-\frac{\pi^2}{2}) $$

$$ \displaystyle a_0=0 $$

$$ \displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle a_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\cos(nx)dx + \int_{0}^{\pi}(\pi-x)\cos(nx)dx) $$

$$ \displaystyle a_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\cos(nx)dx + \int_{0}^{\pi}\pi\cos(nx)dx - \int_{0}^{\pi}x\cos(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle a_n=\frac{1}{\pi}(\frac{2-2\cos(\pi n)}{n^2})=\frac{2-2\cos(\pi n)}{\pi n^2} $$

$$ \displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle b_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\sin(nx)dx + \int_{0}^{\pi}(\pi-x)\sin(nx)dx) $$

$$ \displaystyle b_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\sin(nx)dx + \int_{0}^{\pi}\pi\sin(nx)dx - \int_{0}^{\pi}x\sin(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle b_n=\frac{1}{\pi}(\frac{\pi-\pi cos(\pi n)}{n})=\frac{1-cos(\pi n)}{n} $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty}(\frac{2-2\cos(\pi n)}{\pi n^2}\cos(nx)+ \frac{1-cos(\pi n)}{n}\sin(nx)) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{4}{\pi}(\cos x+\frac{1}{9}\cos3x+\frac{1}{25}\cos5x) + 2(\sin x + \frac{1}{3}\sin3x + \frac{1}{5}\sin5x) $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=(4/pi)*(cos(x)+(1/9)*cos(3*x)+(1/25)*cos(5*x))+2*(sin(x)+(1/3)*sin(3*x)+(1/5)*sin(5*x))

>> plot(x,y)

= R7.5 - Exact integration and numerical confirmation =

Problem Statement
Part 1: Find the exact integration of $$ \int_{0}^{p} \sin (j \omega x) \sin (k \omega x) dx$$ Where $$j \not\equiv k$$ and $$ j,k = 1,2,... $$

Given: $$ p=2 \pi, j=2, k=3$$

Part 2: Confirm the result with MATLAB's trapz command for the trapezoidal rule.

Part 1
Substituting in the provided data gives us:

$$ \int_{0}^{2 \pi} \sin (2 \omega x) \sin (3 \omega x) dx$$

Apply the following trigonometric identity:

$$ \sin(a) \sin(b)=\frac{1}{2} ( \cos(a-b)- \cos(a+b))$$ Where $$ a=2 \omega x, b=3 \omega x$$

The integral becomes:

$$ \int_{0}^{2 \pi} \frac{1}{2} ( \cos(\omega x)- \cos(5 \omega x)) dx$$

$$ \frac{1}{2 \omega} \int_{0}^{2 \pi} \cos(u)du- \frac{1}{10 \omega} \int_{0}^{2 \pi} \cos(v)dv$$

Where $$ u= \omega x, du= \omega dx, v=5 \omega x, dv=5 \omega dx$$

The finite integral becomes:

$$[ \frac{1}{2 \omega} \sin(\omega x)- \frac{1}{10 \omega} \sin(5 \omega x)]_{0}^{2 \pi}$$

We know from the book that $$\omega = \frac{2 \pi}{p}$$ Therefore $$\omega = \frac{2 \pi}{2 \pi} = 1$$

So our finite integral is now:

$$[ \frac{1}{2} \sin(x)- \frac{1}{10} \sin(5x)]_{0}^{2 \pi}=0$$

Part 2
The following is MATLAB code used to determine the definite integral from 0 to $$2 \pi$$ using a step of 0.05:

>> x=0:0.05:2*pi; y=sin(2*x).*sin(3*x); z=trapz(x,y)

z = -1.5564e-004

As we decrease the size of the step and increase the accuracy the result continues to get smaller:

>> x=0:0.01:2*pi; y=sin(2*x).*sin(3*x); z=trapz(x,y)

z = -3.8317e-007 >> x=0:0.001:2*pi; y=sin(2*x).*sin(3*x); z=trapz(x,y)

z = -1.9803e-010

The values are essentially zero and therefore, they confirm the result from integration.

=Team Contributions=

title: User:Egm4313.s12.team8/R7


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Instructional design/Summary and References

← Go back to ID Page

Summary
Congratulations, you have successfully finished this online course of how to generate PBL problems. In this online course, we have talked about three important features of PBL problems, which are:
 * Problems should enhance students’ motivation.
 * Problems should enhance students’ critical thinking skills.
 * Problems should enhance students' collaboration.

If you have chance to use PBL method in your future teaching, please remember to consider the features listed above to generate good problems for students to work on.

Books
1. Bloom, B.S., Engelhart, M. D., Furst, E. J., Hill, W. H., & Krathwohl, D. R.(1956). Taxonomy of educational objectives: the classification of educational goals; Handbook I: Cognitive Domain New York, Longmans, Green, 1956. 2. Johnson, D., Johnson, R.& Holubec, E. (1998). Cooperation in the classroom. Boston: Allyn and Bacon.

Journals
1. Chin, C. and Chia, L.G. (2006) Problem-Based Learning: Using Ill-Structured Problems in Biology Project Work. Science Education, 90, 44-67.

Websites
1. Lady Caitriona & looney_tunes.(2012). Quizzes - Architecture.Retrieved April,20,2012, from source Website: http://www.funtrivia.com/quizzes/humanities/architecture.html 2. Wikipedia.(2012). Ultraviolet-Subtypes.Retrieved April,20,2012, from source Website: http://en.wikipedia.org/wiki/Ultraviolet#Subtypes 3. Wilder, P.(2012). Assessing Student Interests and Strengths. Retrieved April,18,2012, from source Website: http://www.readwritethink.org/professional-development/strategy-guides/assessing-student-interests-strengths-30100.html?main-tab=1#main-tabs

Video Clips
1. BIEPBL. (2010, December 9). Project Based Learning: Explained. Retried April, 18, 2012, from http://www.youtube.com/watch?v=LMCZvGesRz8 2. DragonflyNoir10. (2009, December 9). Mona Lisa Smile Part 1/12. Retrieved April, 18, 2012, from http://www.youtube.com/watch?v=1w9QYDmV3MU 3. Facdent1981. (2011, September 30). Introduction to Problem-Based Learning at the HKU Faculty of Dentistry. Retried April, 18, 2012, from http://www.youtube.com/watch?v=lnAq0oX8zWs 4. Making Learning Real (2010, October, 26). Student Insights. Retrieved April, 18, 2012, from http://www.youtube.com/watch?v=OWkAVSRkDgo 5. Making Learning Real (2010, October, 26). Problem-Based Case Learning Why it Works. Retrieved April, 18, 2012, from http://www.youtube.com/watch?v=cT8_YEtSFQQ 6. 1234rockford (2011, August, 17). Geometry Meets Architecture - Applying Math in a Real World Project. Retrieved April, 19, 2012, from http://www.youtube.com/watch?v=QbeQpSkQUnw


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CIVICS/Lagodekhi Case Study

Lagodekhi is a town of about 7,000 inhabitants in Georgia's Kaheti province. It borders Azerbeijan in the East, and the Russian territory of Dagestan in the North. Lagodekhi is home to the oldest national parks in Georgia. 12 villages are located at the so-called buffer zone of the national park, including about 3000 households.

Information on Lagedekhi National Park and Conservation Managed Area

 * NP between poaching and tourism, Worldbank Article from 2005
 * Marina Metreveli & Avtandil Abuli (2009) Book on Kakheti Conservation: Vashlovani & Lagodekhi (Overview)
 * The present reference book deals with the customs and traditions of the gateway communities of Vashlovani and Lagodekhi Protected Areas founded under the auspices of the law passed by the Parliament of Georgia in 2003 with facilitation of “Georgia’s Protected Areas Development Project” supported by the World Bank and Global Environment Facility.

Development strategy 2009-2014 for Kakheti Province

 * supported by UNDP: online available
 * the text below consists mainly of selected direct quotes from the document. The quotes are indicated by Italics. In the source text (editing view), the extent of the single quotes can be seen. Omissions are indicated by (...) elipses. Bold fact added for emphasis.

Main Vision

 * Kakheti Region will become recognized by the world as a “motherland” region of wine, with growing wine clusters and successful exports. and
 * By 2014, Kakheti will become a leading region in Georgia’s cultural and economic development, supported by a hospitality infrastructure – hotels, restaurants, networks of café-bars, tour operators, as well as regional and international events.

Region description and SWOT analyis

 * The Kakheti lowlands are not rich in forest resources. Forest areas are becoming more extensive towards Caucasus and Tusheti. Timber production is growing in certain areas of Kakheti, such as Akhmeta; effective use of forest resources is vital. Uncontrolled use of forest resources, especially at the foothills of the Caucasus Mountains, has increased the risks of floods. This is especially true of the Alazani River Basin, where severe results are already felt
 * Kakheti has the highest ratio of fertile agriculture lands to the entire land resources and area per population in the country. Kakheti has also traditionally been the country’s cereal production region. It is responsible for half of the total lands under wheat and the largest share of the wheat produced in the country.  Beekeeping may be considered one of the region’s priorities, because it is rich with forests, flowers and fields under crops, which provide it with adequate resources.
 * Kakheti is famous for its protected territories, Vashlovani, Batsara-Babaneuri, Lagodekhi and Chachuna, which were created to preserve the unique flora and fauna of the region. Vashlovani and Tusheti National Parks as well as Lagodekhi Reserve play a critical role in developing its tourism industry.
 * Today, around 65–70 percent of all vineyards in Georgia are concentrated in Kakheti Region. Production vineyards giving the best wines are located in the basins of Alazani and Iori Rivers, at 400-700 m altitude.

Lagodekhi mentioned

 * Most of the irrigated areas are thus used for melons and vegetables production, with Signagi, Sagarejo, Kvareli and Lagodekhi being the leading producers. Lagodekhi, Signagi and Gurjaani are the main vegetable-producing municipalities due to their climatic conditions and the availability of irrigation water.
 * Cattle: .Sagarejo, Lagodekhi and Akhmeta are the leading municipalities within Kakheti in this regard. Lagodekhi, Sagarejo and Akhmeta are the largest milk producers in the Kakheti Region (...). However, milk yield remains low at 1.2 tonnes per cow per year. Animal productivity could be increased through improved feed management and quality veterinary services, breed improvement and calving management to smoothen the seasonality of milk production.

 * Pigs: according to 2005 data, the leading producers were Kvareli, Telavi, Akhmeta and Lagodekhi. In these four municipalities, forest resources are used for swine feed. (...) In 2007, the number of pigs significantly decreased in Kakheti due to the spread of Montgomery’s disease.
 * Kakheti has a long tradition of 'sheep raising, and (..) it is the region with the largest number of sheep in the country. Favourable conditions for sheep production in Kakheti are the availability of winter and summer pasturelands. Sheep and goats are present in all municipalities (...). Akhmeta and Sagarejo are first among Kakheti Municipalities in producing lamb, sheep milk and wool, emphasizing the importance of this sector to the local economy.
 * Sagarejo and Lagodekhi Municipalities are the leading municipalities of the region in the poultry industry.
 * Viticulture The highest yields of Saperavi were in Akhmeta, Telavi, Gurjaani, Sagarejo and Kvareli, and the lowest in Lagodekhi and Signagi. The estimated area under the new vineyards is around 5,000 ha with estimated investments of US$40–55 million, assuming that 1 ha requires around US$8,000–11,000. This process has not been significant in Akhmeta, Signagi, Sagarejo, Lagodekhi and Dedoplistskaro Municipalities, where vine growing is not a leading sector.
 * As of 1 August 2007, the total capacity of hotels in support territories of the Vashlovani, Tusheti, Batsara-Babaneuri and Lagodekhi was only 237 places, which was far below demand. In general, Kakheti Region has the lowest gross and per capita turnover of hotels and restaurants. The same is true for the growth rate of the hospitality sector. There is also some potential, though underused and fragmented, in Tusheti and Lagodekhi and Vashlovani Reserves.
 * ANNEX: Lagodekhi Municipality: Baisubani Wine Factory, large-scale, old factory.

"Priotity Directions" & "Priorities"

 * "Priotity Directions" (Environment) 2. Preventing uncontrolled usage of the forest resources for the Caucasian foothills and seeking funding for Duruji River bank reconstruction.

 * "Overall Priotities":
 * Setting up food processing enterprises: Processing of fruits and vegetables; setting up milk collection centres and cheese and butter production plants; ensuring accessibility of financial resources (low interest rate and subsidized loans, grants) for small entrepreneurs.
 * Supporting agriculture development: Supporting the wine and vine sectors, organizing agricultural extension services, facilitating farmers’ cooperation, providing support to the animal husbandry sector (breed improvement, feed production, veterinary services, among others).
 * Drinking Water
 * Developing tourism
 * Garbage and Waste Management

Current Projects

 * New (2012) Worldbank Project on Kakheti Region (75m USD), Focus on General water, sanitation and flood protection sector (60%), Urban Transport (30%), Sub-national government administration (8%), Voc ational training (2%) (http://www-wds.worldbank.org/external/default/WDSContentServer/WDSP/ECA/2012/02/08/D4D4F4302F82D69B8525799E00561037/1_0/Rendered/PDF/P1260330PID0Ap008201201328715602188.pdf online)

People of the buffer zone
Lagodekhi has no traditional buffer zone, but a de facto one. 12 villages with about 3000 households are located in this zone. Prestudy visits have shown that the region has a better infrastructure and utility grid support, than Lake Arpi and Samtskhe-Javakheti even so poverty is high. Mobility connection is quite adequate and there are possibilities to find jobs outside agriculture. Traditional activities in Lagodekhi are corn production and animal husbandry, but families do also often have home gardens, where fruits and nuts are produced. Very little communal pasture land is existing in the region and farmers see their chances for investing limited. Visits have shown furthermore that wives are often not living with their families and are working aboard for support.

title: CIVICS/Lagodekhi Case Study


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User talk:Garcia123/ENES100/My Work 3

Week 1 p3 Comment Kim DelVillar Good job on getting your gun started and having the pieces cut out!

week 1 p3 grading 1sfoerster (talk) 19:57, 24 April 2012 (UTC)
 * 10/10/50 ... good job

Week 1 p3 Comment David Ross
 * Great job getting the pieces cut out already and glued together. You got the hard part out of the way very quickly.

week 2 p3 grading 1sfoerster (talk) 17:07, 1 May 2012 (UTC)
 * 0/0/0 .. nothing to grade

week 3 p3 grading 1sfoerster (talk) 13:27, 8 May 2012 (UTC)
 * 0/0/0 .. nothing to grade

Week 4 p3 Comment Kim DelVillar Thanks for all the help that you gave me.

week 4 p3 grading 1sfoerster (talk) 20:44, 15 May 2012 (UTC)
 * 0/0/0 .. nothing to grade

title: User talk:Garcia123/ENES100/My Work 3


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New Zealand Pilot License/Meteorology/Private/Subject 1

Linear actuators

 * 1. open google chrome or mozilla fire fox.
 * 2. go to www.google.com
 * 3. Type "linear actuators"
 * 4. I click on the 8th link. http://www.linak.com/products/Linear-Actuators.aspx
 * 5. I first click on the "LA 12" http://www.linak.com/products/Linear-Actuators.aspx?product=LA12

:2. It does not hold 1000N :1 It has a rod.
 * Problems encounter with this Linear actuator:
 * Good this about it:
 * 6. Click Back
 * 7. Then click on the "LA 22" http://www.linak.com/products/Linear-Actuators.aspx?product=LA22

:2. It hold 400N :1 A rod
 * Problem encounter with LA 22 Actuator:
 * Good this about it:
 * 8 Click Back
 * 9 Click on the "LA 27" http://www.linak.com/products/Linear-Actuators.aspx?product=LA27

:1. It hold more than 1000N :2. A rod
 * Problems encounter with the LA 27
 * Good things about the "LA 27"


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User:Shatton8111/enes100/my work 3

Helicopter Project, Part 3

Write problem/project Goal
To create an Arduino-based quadcopter that flies itself without a remote control.

My First Task
I'm working on stabilizing the thrusts of the Hacker A10-13L motors on the seesaw apparatus.

Summary of actual work over first weekend
I managed to control the individual thrusts of the motors and show tuned reactions to changes in orientation. This involves the motors receiving control signals from the Thunderbird 9 ESC's, which are sent variable pulse widths from the Arduino Mega 2560. The Arduino runs behavior-controlling code that responds to filtered changes of orientation from the 5DOF SparkFun gyro-accelerometer sensor. The control code in the Arduino calculates response speeds for the individual motors using math based on complementary angles.

Week1 Narrative
In class, I've been focused on editing the behavior-controlling code that is in the Arduino. What is currently working is that one motor attempts to counteract forces on its side until the angle measured (by the Inertial Measurement Unit or 5DOF sensor) is within the definition of a level range. However, once this side's motor is done, the other responds exaggeratedly and doesn't stop until power is removed (in most cases).

I have yet to discover why one motor appears to behave appropriately while its complement is responds aggressively. There are several possibilities for why this is occurring. Problems with the IMU sensor's drift may be changing the definition of what is a stable and level position. After several tests, the calibrated "zenith" appears to shift from ~0.61 to ~0.18. Having a dirty electrical current supplied to the sensor may also induce drift. One motor may be more powerful than the other. They are the same model, but manufacturing methods aren't perfect. Because one motor is occasionally overheating, the "wear and tear" on it might reduce it to being very weak compared to the other motor.

However, despite these problems, it should be possible to account for the differences in thrusts. One possibility I've been told about is using a PID controller. While it would add more complexity and code to the program, if I'm able to understand and implement it successfully, it may resolve thrust balancing issue.

Counteracting the IMU sensor's drift shouldn't be difficult either. However, my attempts haven't worked so far. I've tried recording the initial filtered orientation readings and using them to offset the later readings for calibration. As I tried this, I discovered that such a method is futile since the Kalman filter in use calibrates the readings using multiple readings during the main "run loop" (the "loop" function) of the executing program. As such, it appears that either something else is the problem and the sensor is behaving properly or the calibration would need to watch for the readings to become stable during the run loop's execution (which would be tricky). Hopefully this drift problem is not the case, but I should be able to discover whether or not the problem exists once the PID controller is soundly balancing the seesaw. Erratic behavior across tests would most likely be caused by drift or power supply issues.

My Second Task
Unless I get to it today, I'm going to work on implementing PID controllers in order to control the balancing of the seesaw.

Summary of actual work over second weekend
I'm trying to redesign may "seesaw balancing code" to use PID controllers. The current PID controller implementation I'm using is the third-party PID library for the Arduino.

Week2 Narrative
I've added one PID controller to my code, and now a single side responds to counteract angles measured to be below a certain set point. One problem is that it responds very slowly. It's possible that this delay could be decreased by tuning the P, I, and D parameters to the feedback system. That will most likely be done manually or with an automatic tuning system.

I'm interested in adding the second motor to the system. I also need to redesign the code to clearly show and separate the parts involved in:

 * holding the desired input or control values for speed and angles,
 * ramping up the speeds of the motors gradually from zero to desired values,
 * introducing the corrective actions of the PID controllers,
 * keeping the produced values within safe and usable ranges, and finally...
 * giving the resulting speed values to the ESC's in order to change the motors' speeds.

To do this, I'm trying to design the conceptual outline for how the software would interact with other components. Here's the current idea:

Angle Sensor                                    Desired speed _______                                        (base speed)              Motor |      |                     Desired angle        _______                 (ESC) |      |                            |            |       |                  _  |_______|                            V            |       |                 | | PID Controller     |_______|               __|_|__ |                           ________             |                  |       |      |                           |        |            V                  |       | `> |        | -> + ---> |       |                                  |________|                               |_______|

This is how I'm thinking of redesigning the software to interact with different variables and parts:

Control Inputs Desired angle (X) base speed                                                      | X angle                                                         V  Y angle                                 Angle sensor ---> PID controller |                                                                  V                                 .---> + ---> ESC ---> Motor |                                |                           Desired angle (X) |                                |                                 |                                 V                                 |        Angle sensor ---> PID controller (reversed) |                                |                                 |                                 V                                 |---> + ---> ESC ---> Motor |                                |                           Desired angle (Y) |                                |                                 |                                 V Base speed ---> -|        Angle sensor ---> PID controller |                                |                                 |                                 V                                 |---> + ---> ESC ---> Motor |                                |                           Desired angle (Y) |                                |                                 |                                 V                                 |        Angle sensor ---> PID controller (reversed) |                                |                                 |                                 V                                 `---> + ---> ESC ---> Motor

As I test the code I currently have in simulation, I'm watching a single motor vary its output depending on how far above or below the horizon it is. When it is above the horizon, it slowly decreases its output down to the base speed, where it stays. When it's below the horizon, it gradually increases its speed until it reaches, and stays at, the maximum limit I set.

There are two problems I'm seeing. First, the delay between updates of the PID controller are fairly long. I think it will need to respond quicker for actual use. Second, once the PID controller's internal value is past one of the boundaries, it takes a very long time to reach it again. This is because it doesn't have internal limits (that I know about) and the delay is too long. I'm going to try tuning the values in order to have faster responses.

I tested some adjustments to the P, I, and D parameters. After testing through the Arduino editor's Serial Monitor with the full system (without power for the motors attached), I was able to find some tuning parameters that seemed to respond quickly enough for the seesaw to react quickly: 2, 100, 1; for P, I, and D, respectively.

From testing with the motor (with power), there seems to be a problem: the side that is being controlled is backwards. This leads to slightly confusing behavior. I'm going to try fixing it by setting the PID controller to be "reverse acting." This is done by changing the last initial parameter of the PID function from "DIRECT" to "REVERSE." For testing safely, I'm also going to change the P, I, and D values to 2, 10, and 1, respectively.

The test went relatively well. The motor turned on and spun. It wasn't really easy to tell if it was slowing down or speeding up. I'm going to retry with a servo "speed" of 65 instead of 70.

The test was the same, but the speed seemed slightly lower, based on the pitch of the sound. I'm going to retry with a speed of 60.

It seemed the same. One interesting thing is that when the Arduino starts the motors for the first time after being reprogrammed, it sets the motor to a very high speed. It's very dangerous, and annoying. I'm not sure why it does that. I'm trying again with a speed of 50.

From testing, I found out that the ESC attached to the motor on that side seems to only update its speed when power is first applied to it. That would mean that it becomes inoperable once the Arduino is attempting to update it. The difference between that and the other side is that it is missing the smaller 5 volt and grounding wires to the ESC. I'm going to try giving it those to see if it improves its responsiveness.

That did it. Now it works and is responsive to changes. Now I'm going to add the other side's motor with a "direct acting" (using the "DIRECT" parameter).

I simulated with only the Serial Monitor's output, and it looks good. I'm now trying with power. It works! It doesn't quite balance, but the seesaw is moving back and forth. I'm going to use PID values of 2, 20, & 1 to see if it will respond more quickly. I'm also going to use "somePID.SetSampleTime(15)" to make it update the controllers faster.

It didn't really help that much. I was able to almost balance it in my hands, but not on its own. I'm going to try it without the "safety boundaries" I had set up.

It didn't help much either. But, I did find out that it's basically fine to run the system without the limiters. Although, it might work better to limit within five or ten values of the desired range.

I looked through the ArduPilot Mega code and wiki for ideas. Eventually, I came across the wiki page for PID controller tuning. I found the explanations of PID control in the section "A bit more about PID loops" to be really helpful for understanding tuning. I tested in class today with P, I, and D values of 0.5, 0.5, and 1. The seesaw balances! Now I need to tune it to respond quickly and with little overcompensation, instead of slowly with a little wobbling.

Video: Slowly Balancing Seesaw

My Third task
Record what you are planning on doing for the team during the third weekend.

Summary of actual work over third weekend
How is what you did different than what you planned?

Week3 Narrative
I got the seesaw basically balancing, and then I decided to make the PID controller sample every millisecond instead of ever 15 milliseconds. That completely changed the behavior, making it respond much faster but with stronger motions. This meant that I had to adjust the tuning parameters again.

Here's the code at this point: TiltControl code

My Fourth task
''Record what you are planning on doing for the team during the final weekend of project work. ''

Summary of actual work over fourth weekend
How is what you did different than what you planned?

Week4 Narrative
I've been trying to balance the seesaw by experimenting with the PID parameters. I'm also experimenting with the sampling rates of the two PID controllers and the use of an initial "autotuning" system for finding tunings. Not much progress is being made unfortunately.

From talking to people about the current design, I think it would be good to try using a single PID controller per axis (controlling two motors). I've been worrying that having a single feedback loop for two motors of varying output will lead to instability, due to the different thrusts of the two motors. However, it might make sense, as temperature control systems appear to use a single feedback loop to control both the heating and cooling systems for buildings. I would suspect that each of these parts have varying output, just like the motors I'm using. I haven't researched or verified any of this, but I'll try using a single PID controller.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).

title: User:Shatton8111/enes100/my work 3


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User talk:Jromanch0088/enes100/My Work Heli3

Great work on the seesaw. Amirreza Saharkhiz (talk)

Yeah, really awesome job with the seesaw's wiring and construction. The Bluetooth stuff looks like it's coming along too! Shatton8111 (talk) 17:02, 23 April 2012 (UTC)

week 1 p3 grading 1sfoerster (talk) 20:18, 24 April 2012 (UTC)
 * 10/10/200 great job ...

week 2 p3 grading 1sfoerster (talk) 17:41, 1 May 2012 (UTC)
 * 0/0/0 .. nothing to grade

Thanks for helping me to attach gimbal on the base.Amirreza Saharkhiz (talk)

week 3 p3 grading 1sfoerster (talk) 13:49, 8 May 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 4 p3 grading 1sfoerster (talk) 21:04, 15 May 2012 (UTC)
 * 0/0/0 ... nothing to grade


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Template:Substub/doc

This template is placed on the top of pages when they are a substub (about the length of a YouTube video description or a Twitter tweet). Any pages should be deleted or replaced with a better page.

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User:Elijahignatieff

Elijah Ignatieff is a researcher and designer of educational tools to assist individuals, teams, organizations, and communities in the transition from the "old paradigm" economic system to a "new paradigm" economic system.

Elijah has designed a business operating system called the Inflow Matrix. This operating system was created by placing concepts on sacred geometry. It is also fractally organized.

The Inflow Matrix can be used to design any job, any organization, and any community.

The Inflow Matrix acts as an operating system to run either an individual job, or an organization or a "Shared Knowledge Community". A Shared Knowledge Community is a "New Paradigm" economic organizational structure.

The Inflow Matrix as as the operating system to connect a series of maps, card sets, gameboards, processes and software called the New Paradigm Toolkit.

The online location to learn about how the Inflow Matrix works, how to Build a Shared Knowledge Community, and how to use the New Paradigm Toolkit is the School of Conscious Communication.

There are currently two research groups looking at building a Shared Knowledge Community in Canada. Elemental Soulutions in Saskatchewan and Vancouver Knexus in Vancouver. Lori Ann Bandura is organizing the founding group in Saskatchewan and Zamir and Raamayan Dhanji are organizing the founding group in Vancouver.

title: User:Elijahignatieff


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text: File:2012-04-23 14.28.37.jpg

title: File:2012-04-23 14.28.37.jpg


text:
MakerBot/How to download 123D

Purpose
The purpose of this tutorials is show all the steps required to successfully download 123D software

Steps

 * 1) Go to http://www.123dapp.com/
 * 2) Click on the link that says 123d
 * 3) On the next page there is download button "Download 123D" Click on it to go to the next pageStep 2 123D.jpg
 * 4) A new window would appear with download details and instructions, read through the instructions to make sure that your computer meets the requirnments to run the 123D application

System Requirements:
 * A) Windows 7 (32-bit or 64-bit), Windows Vista® (32-bit or 64-bit), Windows® XP Professional or Home Edition (SP3), or Windows XP Professional x64 Edition (SP2), Windows Server 2008
 * B) Intel® Pentium® 4 or AMD Athlon® 64 or later with 2 GHz or faster processor; or compatible
 * C) 1+ GB RAM (512 MB minimum)
 * D) 1.5+ GB free disk space (for installation)
 * E) Direct3D® 9 or 10 graphics support with 64+ MB
 * F) Internet connection

 * 5) On this page options for operating systems is given, one is for 32bit OS and the other is for the 62 bit OS. Choose the one which is appropriate and begin to download.32or64bit.png
 * 6) Next, it would ask for the account information, if you dont have an account, you have to make one. just provide with a vaild email address and create a password. Account information.jpg
 * 7) After creating an account and successfully loging in, it would ask you to accept terms of service,check the "I Agree" check box and click on the continue botton.

 * 8) Now you can download the 123D software,save it on the computer, desktop would be a nice place to save it since it is easier to find. The downloading process may take several minutes depeding on the internet connection.
 * 9)Double click on 123d file to install the software package.123 software.jpg

title: MakerBot/How to download 123D


text:
User talk:Shatton8111/enes100/my work 3

Nice job working on motor programming. Amirreza Saharkhiz (talk)

Good job getting the code to work (kind of) for the accelerometer! Jromanch0088 (talk) 09:52, 24 April 2012 (UTC)

week 1 p3 grading 1sfoerster (talk) 20:17, 24 April 2012 (UTC)
 * 10/10/50 .. need links to the arduino code ... the "pid" needs links ... a "pic" is just another cpu like the amtel in the arduino

Great work to keep a sea-saw balancing Amirreza Saharkhiz (talk)

 * 10/0/150 ... great job with the software

Nice job working on sea-saw and keep it balance Amirreza Saharkhiz (talk)

week 3 p3 grading 1sfoerster (talk) 13:53, 8 May 2012 (UTC)
 * 10/0/50 .... good job tuning the code and posting a link to it ... keep up the work!

week 4 p3 grading 1sfoerster (talk) 21:07, 15 May 2012 (UTC)
 * 10/0/30 ... still need the pid software you are using and modifying linked up

title: User talk:Shatton8111/enes100/my work 3


text:
User:Egm6322.s12.team1.zheng.zx/R14

Problem 5: Derive the Christoffel-Darboux identity for Legendre polynomials
 Solved without assistance

Given:
The 3-term recurrence relation for any sequence of orthogonal polynomials;
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$
 * }
 * }
 * }

for n=0,1,2,..., and $$\,\,p_{-1}(x)=0\,\,$$
 * {| style="width:100%" border="0" align="left"

\displaystyle A_n\,,B_n\,,C_n \in \mathcal R\,,with\, A_{n-1}A_nC_n > 0 $$
 * }
 * }
 * }

If the highest coefficient of $$\,\,p_n(x)\,\,$$ is kn, then
 * {| style="width:100%" border="0" align="left"

\displaystyle A_n=\frac{k_{n+1}}{k_n} $$
 * }
 * }
 * }

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

\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
 * }
 * }
 * }

Find:

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

\displaystyle \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k}=\frac{p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)}{A_nh_n(x-y)} $$
 * }
 * }
 * }

Solution:
From the three term recurrence relation
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$
 * }
 * }
 * }

multiply both sides by pn(y):
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)p_n(y)=(A_nx+B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x) $$ $$
 * $$\displaystyle (Equation\;5.1 )
 * }
 * }

And for the symmetry between the factors
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(y)p_n(x)=(A_ny+B_n)p_n(x)p_n(y)-C_np_n(x)p_{n-1}(y) $$ $$
 * $$\displaystyle (Equation\;5.2 )
 * }
 * }

Substract equation 5.2 from equation 5.1:
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)=A_n(x-y)p_n(x)p_n(y)-C_n[p_n(y)p_{n-1}(x)-p_n(x)p_{n-1}(y)] $$
 * }
 * }
 * }

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

\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
 * }
 * }
 * }

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

\displaystyle \begin{align} & \frac{p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)}{A_nh_n(x-y)} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_n(x)p_{n-1}(y)-p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)h_{n-1}} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_{n-1}(x)p_{n-1}(y)}{h_{n-1}}+\frac{p_{n-1}(x)p_{n-2}(y)-p_{n-1}(y)p_{n-2}(x)}{A_{n-2}(x-y)h_{n-2}} \\ & \vdots \\ =& \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k} \end{align} $$
 * }
 * }
 * }

title: User:Egm6322.s12.team1.zheng.zx/R14


text:
User:1sfoerster/enes100/Tri-BotHack/MotorTest

/* BallBot Exercise This is test code to exercise the WowWee Tri-Bot capabilities Have to push down the head to get the robot to move This example code is in the public domain. */

// constants won't change. They're used here to // set pin numbers: const int buttonPin = 3;    // the number of the pushbutton pin const int ledPin = 13;      // the number of the LED pin

// variables will change: int buttonState = 0;        // variable for reading the pushbutton status int headPush=0; //if 0 means execute any instructions and wait, if 1 then do and check if headpushed regularly

void setup { // initialize the digital pin as an output. pinMode(12, OUTPUT); //yellow wire to motor enable 1 pinMode(11, OUTPUT); //teal wire to motor enable 2 pinMode(10, OUTPUT); //blue wire to motor enable 3 pinMode(9, OUTPUT); // white wire to motor direction 1 pinMode(8, OUTPUT); //gray wire to motor direction 2 pinMode(7, OUTPUT); //purple wire to motor direction 3 pinMode(6, INPUT); //orange encoder from motor 1 pinMode(5, INPUT); //red encoder from motor 1 pinMode(4, INPUT); //brown encoder from motor 1 pinMode(ledPin, OUTPUT);  // initialize the LED pin as an output: pinMode(buttonPin, INPUT); // initialize the pushbutton pin as an input: }

void loop{ while (headPush == 0) { // wait for head push before starting buttonState = digitalRead(buttonPin); if (buttonState == HIGH) { // turn LED on: digitalWrite(ledPin, HIGH); headPush = 1; } } //  oneMotorAtATime; // spinTightCCW; oneMotorAtATime; delay(1000); spinTightCCW; delay(1000); spinTightCW; delay(1000); gradualTurn; delay(1000); forWard; delay(1000); backWard; delay(1000); rightForward; delay(1000); rightBackward; delay(1000); headPush=0; }

void oneMotorAtATime {

//motors turning individually delay(1000);             // wait for a second digitalWrite(12, HIGH);  // set the LED on  delay(2000);              // wait for a second digitalWrite(12, LOW);   // set the LED off delay(2000);             // wait for a second digitalWrite(11, HIGH);  // set the LED on  delay(2000);              // wait for a second digitalWrite(11, LOW);   // set the LED off delay(2000);             // wait for a second digitalWrite(10, HIGH);  // set the LED on  delay(2000);              // wait for a second digitalWrite(10, LOW);   // set the LED off headPush=0; }

void spinTightCCW { delay(1000); digitalWrite(9, HIGH);  // reverse motor 1 digitalWrite(8, HIGH);  // reverse motor 2 digitalWrite(7, HIGH);  // reverse motor 3 delay(100); digitalWrite(12, HIGH);  // turn motor 1 on   digitalWrite(11, HIGH);   // turn motor 1 on  digitalWrite(10, HIGH);   // turn motor 3 on  delay(840); digitalWrite(12, LOW);  // turn motor 1 off digitalWrite(11, LOW);  // turn motor 2 off digitalWrite(10, LOW);  // turn motor 3 off digitalWrite(9, LOW);  // normal motor 1 digitalWrite(8, LOW);  // normal motor 2 digitalWrite(7, LOW);  // normal motor 3 headPush=0; }

void spinTightCW { delay(1000); digitalWrite(12, HIGH);  // turn motor 1 on   digitalWrite(11, HIGH);   // turn motor 1 on  digitalWrite(10, HIGH);   // turn motor 3 on  delay(840); digitalWrite(12, LOW);  // turn motor 1 off digitalWrite(11, LOW);  // turn motor 2 off digitalWrite(10, LOW);  // turn motor 3 off headPush=0; }

void forWard { delay(1000); digitalWrite(12, HIGH);  // turn motor 1 on   digitalWrite(8, HIGH);   // reverse direction of motor 2 digitalWrite(11, HIGH);  // turn motor 2 on  delay(2000); digitalWrite(12, LOW);  // turn motor 1 off digitalWrite(8, LOW);  // make direction of motor 2 normal digitalWrite(11, LOW);  // turn motor 2 off headPush=0; }

void backWard { delay(1000); digitalWrite(9, HIGH);  // reverse direction of motor 1 digitalWrite(12, HIGH);  // turn motor 1 on     digitalWrite(11, HIGH);   // turn motor 2 on  delay(2000); digitalWrite(12, LOW);  // turn motor 1 off digitalWrite(9, LOW);  // make direction of motor 2 normal digitalWrite(11, LOW);  // turn motor 2 off headPush=0; }

void rightForward { delay(1000); digitalWrite(11, HIGH);  // turn motor 1 on   digitalWrite(7, HIGH);   // reverse direction of motor 2 digitalWrite(10, HIGH);  // turn motor 2 on  delay(2000); digitalWrite(11, LOW);  // turn motor 1 off digitalWrite(7, LOW);  // make direction of motor 2 normal digitalWrite(10, LOW);  // turn motor 2 off headPush=0; }

void rightBackward { delay(1000); digitalWrite(11, HIGH);  // turn motor 1 on   digitalWrite(8, HIGH);   // reverse direction of motor 2 digitalWrite(10, HIGH);  // turn motor 2 on  delay(2000); digitalWrite(11, LOW);  // turn motor 1 off digitalWrite(8, LOW);  // make direction of motor 2 normal digitalWrite(10, LOW);  // turn motor 2 off headPush=0; }

void gradualTurn { delay(1000); digitalWrite(12, HIGH);  // turn motor 1 on   digitalWrite(8, HIGH);   // reverse direction of motor 2 digitalWrite(11, HIGH);  // turn motor 2 on  delay(200); digitalWrite(10,HIGH); // turn motor 3 on delay(200); digitalWrite(10,LOW); // turn motor 3 on delay(200); digitalWrite(10,HIGH); // turn motor 3 on delay(200); digitalWrite(10,LOW); // turn motor 3 on delay(200); digitalWrite(10,HIGH); // turn motor 3 on delay(200); digitalWrite(10,LOW); // turn motor 3 on delay(200); digitalWrite(10,HIGH); // turn motor 3 on delay(200); digitalWrite(10,LOW); // turn motor 3 on delay(200); digitalWrite(12, LOW);  // turn motor 1 off digitalWrite(8, LOW);  // make direction of motor 2 normal digitalWrite(11, LOW);  // turn motor 2 off headPush=0; }


text:
User:Egm4313.s12.team8.colocar/R7.1

Problem Statement
$$ \displaystyle \text{Verify (1)-(2) }\ $$ ((4) and (5) p19.9, respectively)

$$ \displaystyle \text{Given:} $$

Verifying (1)
$$ \displaystyle \text{From equation } \, $$($$) $$ \displaystyle \, \text{ we have } $$ $$ \displaystyle \langle \bar f, \bar g \rangle := \int_0^L \bar f(x) \, \bar g(x) \, dx, \text{ where we can plug in } \phi \, \text{ from } \, $$ ($$) $$ \displaystyle \, \text{ to get} $$

$$ \displaystyle \text{From } $$ ($$), $$ \displaystyle \text{ and using equation } \, $$ ($$), $$ \displaystyle \, \text{ we get}$$

$$ \displaystyle \text{or} $$

$$ \displaystyle \text{From equation } \, $$ ($$), $$ \displaystyle \text{ for all constants of } \omega_i \text{ and } \omega_j \text{ yield } 0 \text{ as all values of } \sin (\pi) \text{ equal zero. Thus,} $$ $$ \displaystyle \langle \phi_i, \phi_j \rangle = 0 \ \text{ for } \ i \ne j $$

Verifying (2)
$$ \displaystyle \text{Given equation } \, $$($$) $$ \displaystyle \, \text{ and from } $$ ($$), $$ \displaystyle \, \text{ we have } $$

$$ \displaystyle \text{Plugging in the identity, and substituting } L=\pi \text{ from } [0,\pi] \text{ gives } $$

$$ \displaystyle \text{or} $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{1}{2} \left [ x - \frac{1}{2 \omega_j} \sin(2 \omega_j x) \right ]_0^\pi $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{1}{2} \left [ \pi - \frac{1}{2 \omega_j} \sin(2 \omega_j \pi) \right ] $$ $$ \displaystyle \text{Thus} $$ $$ \displaystyle \langle \phi_j, \phi_j \rangle := \frac{\pi}{2} = \frac{L}{2} \text{ for } i=j. $$

title: User:Egm4313.s12.team8.colocar/R7.1


text: File:2012-04-23 15.25.55.jpg

title: File:2012-04-23 15.25.55.jpg


text:
User talk:Jasper Deng

I couldn't undo the Spam on the Colloquium. It said there was a conflict of edits. Maybe its your area of expertise! Brett Johnston (talk) 05:24, 24 November 2012 (UTC)
 * Which spam? I revert lots of spam here!--Jasper Deng (talk) 00:20, 26 November 2012 (UTC)
 * You do indeed; please keep it up. Thanks for the note about spammers.--Collingwood (talk) 08:06, 15 December 2012 (UTC)

About your comment
I saw you oppose me, I said that we were resting the case. And yes, I am mature enough for the job, I have created a lot of educational pages and supported educational pages. Bye! --Draubb (discuss • contribs) 19:50, 24 May 2013 (UTC)
 * To be frank, your comments did not give me the impression that you could act like a full adult. That is an important prerequisite for any user right on a Wikimedia wiki.--Jasper Deng (talk) 06:28, 25 May 2013 (UTC)

 * Jasper Deng, I am trying to hold my aggression, but clearly, have you heard the phrase "Do not judge a book by it's cover" so even thought my comments (as you said!) childish. I am really a full and mature person. Thank you. --Draubb (discuss • contribs) 23:58, 25 May 2013 (UTC)
 * And here's what I have to say to you: You yourself are never a good judge of your own maturity and experience. For all I know, you could be a full-grown adult. However, the only evidence I have is those comments and they indicate immaturity. If you're really grown-up, act like it. (and if you act aggressively to such criticism, I understand, but it also further convinces me you don't have the patience that I would expect from a custodian candidate) --Jasper Deng (talk) 00:14, 26 May 2013 (UTC)

Okay Jasper Deng - I am very sorry for being annoying to you. I promise this unfriendly discussion will never happen again. Sorry about that. See you around Jasper Deng! --Draubb (discuss • contribs) 00:25, 26 May 2013 (UTC)

Vote
Please Jasper Deng! I do not want to get into a disturbing discussion. I am not voting twice and what was the reason for crossing the line out! Yingshong did not ask you to do that (Yingshong is my brother's account). Jasper Deng, I am not voting twice! I am setting up a trial for myself and what I mean is to start fresh and easy! Please let me be able to do it. --Draubb (discuss • contribs) 19:43, 27 May 2013 (UTC)
 * At this point, you're simply being disruptive (and I doubt your claims about not committing sockpuppetry, given your apparent connection to m:User:Extinct Birds).
 * In that case, Yingshong had added two votes, so there was no ambiguity, sockpuppetry or not, about crossing it out.
 * You aren't getting custodianship anytime soon, period. In fact, you're closer to a block from editing than custodianship. You've already been judged, so nothing much will really change if you start another request (which you didn't do properly, and are in no position to do).--Jasper Deng (talk) 19:48, 27 May 2013 (UTC)

You know what!You and Titodutta are right. I do not have any experience. I am sorry to everyone. I am so sorry! --Draubb (discuss • contribs) 19:52, 27 May 2013 (UTC)

Candidates for Custodianship for Draubb
I have but a ❌ sign to show that I am not ready for custodianship. A new upload of my custodianship request will be on June 4, 2013. --Draubb (discuss • contribs) 20:53, 29 May 2013 (UTC)

Aaqib A.
Thanks for the article, but I am not Aaqib A. --The Gir’s and Sing 18:11, 24 August 2013 (UTC)
 * You basically admitted otherwise at Candidates for Custodianship/Draubb. Given that you've had a history of deception with sockpuppets on this wiki, I'm not inclined to believe anything you say.--Jasper Deng (talk) 00:23, 28 August 2013 (UTC)

Children
Hi Jasper Deng, I would like to see your opinion on children on the Wikimedia Projects? Thank you. --Goldenburg111 20:27, 9 January 2014 (UTC)
 * The one condition everyone absolutely has to meet is maturity. Based on his edits on other projects, the locked account whose talk page I'm commenting on is a sock of a user who obviously doesn't have that maturity. See Stewards' noticeboard/Archives/2012-05 for more information.--Jasper Deng (talk) 20:34, 9 January 2014 (UTC)
 * Did you know, in 2011, that Wikiversity was "attacked" (Most of the custodians at that time stated) where by sockpuppets controlled by a 7 year old. Did the research, and done. --Goldenburg111 20:38, 9 January 2014 (UTC)

Candidates for Custodianship/Jasper Deng
Just wondering if you still would like to be considered for probationary custodianship? All you have to do is transclude the request onto the main Candidates for Custodianship page, though I note that Wikiversity's admin activity is a bit short lately. TeleComNasSprVen (discuss • contribs) 10:48, 17 March 2014 (UTC)
 * At this time, I don't feel like the discussion on that page is worth anything anymore, so out of likelihood I'd instead file a new request if I wanted it.--Jasper Deng (talk) 06:15, 18 March 2014 (UTC)

Hey Jasper Deng,

I want to talk to you!

--CarrotLoveForever (discuss • contribs) 17:38, 18 January 2016 (UTC) (Edit my homepage with what ever you like!)

title: User talk:Jasper Deng


text:
Instructional design/Reducing cognitive load in multimedia instruction/Find and Fix Answers

Below are four issues identified. Feel free to edit this page and add any you noticed that don't appear below. Also include anything you feel would enhance this presentation even if it does not fall into any of the overload categories we discussed.

Issue 1:

Weird Background Music. The background music on this video is loud, distracting, and odd. The music and the spoken words seem to compete for the listener's attention. The music leads to incidental and extraneous processing. (Essential processing + incidental processing (caused by extraneous material) > cognitive capacity--one or both channels overloaded by essential and incidental processing (attributable to extraneous material))

Solution:

Utilize the principal of Coherence and eliminate the music.

Issue 2:

Too many visuals -- boxes, words, pictures, etc. (Essential processing + incidental processing (caused by extraneous material) > cognitive capacity One or both channels overloaded by essential and incidental processing (attributable to extraneous material)

Solution: Again use Coherence and weed out the pictures from the presentation. Learner’s will be able to create their own mental model of an eye, ear, sounds, words and images.

Issue 3: There seems to be too much going on at once – hard to track. Essential processing (in both channels) > cognitive capacity-- Both channels are overloaded by essential processing demands.

Solution: Use the principals of Segmentation and/or Pretraining Cut up the instruction over several slides and then bring the whole theory together at the end. Pretain on key terms used in the presentation.

Issue 4: Strange use of signaling – magnification creates a confusing presentation. The magnification makes holding a representation difficult because all the parts of the system blend in together. Essential processing + representational holding > cognitive capacity -- One or both channels overloaded by essential processing and representational holding. Solution: Use the Principal of Temporal and Spatial Contiguity and Coherence (improve signaling). Synchronize by presenting the model parts simultaneously rather than successively. Weed out the pictures and utilize different signaling than magnification such as arrows or circles, our highlight as each section is explained. Once the explanation is finished for a particular section, remove the signal from that section and, move the signal to the next section.

Other Things You Would Like to Add
--- Edit this section to add your ideas.

Click Next to conclude this lesson.


text: File:Foam wings.jpg

title: File:Foam wings.jpg


text: File:Glued Wing.jpg

title: File:Glued Wing.jpg


text:
User:Jasper Deng

w:en:User:Jasper Deng On this wiki, my content work mainly is in chemistry-related pages.

__NOINDEX__

title: User:Jasper Deng


text:
User:Egm4313.s12.team11.Suh/R1.6

Problem Statement
 Problem 1.6  For each Ordinary Differential Equations, determine the order, linearity, and show whether the principal of superposition can be applied.

Order of ODEs
To find the order of Ordinary Differential Equations, (ODEs) simply check the differential order of the dependent variable. If the dependent variable is differentiated once, then it is a first order ODE. If it is differentiated twice, it is a second order ODE.

Linearity
To find the linearity of the ODE, check the power of the dependent variable. If the dependent variable is raised to the power of anything one than 1, then it is not linear. If the dependent variable is raised to the power of 1, then it is linear.

Superposition Principle
For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.  The function satisfies the superposition principle if the sum of the homogeneous solution and the particular solution is equal to $$\overline{y}(x)\!$$, with $$\overline{y}=(y_{h}+y_{p})\!$$ $$\overline{y}(x):=y_{h}(x)+y_{p}(x)\!$$

1.6a

 * $$y''=g=const \!$$

 * Check superposition principle:
 * $$y''=g\!$$
 * $$y_{h}''=0\!$$
 * $$y_{p}''=g\!$$
 * $$(y_{h}+y_{p})\equiv\overline{y}''\!$$

Order of ODE: 2nd Order Linearity: Linear Superposition Principle: Yes

1.6b

 * $$mv'=mg-bv^2 \!$$

 * Check superposition principle:
 * $$mv'+bv^2=mg\!$$
 * $$mv_{h}'+bv_{h}^2=0\!$$
 * $$mv_{p}'+bv_{p}^2=mg\!$$
 * $$m(v_{h}'+v_{p}')+b(v_{h}^2+v_{p}^2)\not\equiv m\overline{v}'+b(\overline{v}^2)\!$$

Order of ODE: 1st Order Linearity: Non-linear Superposition Principle: No

1.6c

 * $$h'=-k\sqrt{h} \!$$

 * Check superposition principle:
 * $$h'+kh^{1/2}=0\!$$
 * $$h_{h}'+kh_{h}^{1/2}=0\!$$
 * $$h_{p}'+kh_{p}^{1/2}=0\!$$
 * $$(h_{h}'+h_{p}')+k(h_{h}^{1/2}+h_{p}^{1/2})\not\equiv\overline{h}'+k\overline{h}^{1/2}\!$$

Order of ODE: 1st Order Linearity: Non-linear Superposition Principle: No

1.6d

 * $$my''+ky=0 \!$$

 * Check superposition principle:
 * $$y''+\frac{k}{m}y=0\!$$
 * $$y_{h}''+\frac{k}{m}y_{h}=0\!$$
 * $$y_{p}''+\frac{k}{m}y_{p}=0\!$$
 * $$(y_{h}+y_{p})+\frac{k}{m}(y_{h}+y_{p})\equiv\overline{y}''+\frac{k}{m}\overline{y}\!$$

Order of ODE: 2nd Order Linearity: Linear Superposition Principle: Yes

1.6e

 * $$y''+\omega _{o}^{2}y=\cos (\omega t)\!$$

 * Check superposition principle:
 * $$y''+\omega _{o}^{2}y=\cos (\omega t)\!$$
 * $$y_{h}''+\omega _{o}^{2}y_{h}=0\!$$
 * $$y_{p}''+\omega _{o}^{2}y_{p}=\cos (\omega t)\!$$
 * $$(y_{h}+y_{p})+\omega _{o}^{2}(y_{h}+y_{p})\equiv\overline{y}''+\omega _{o}^{2}\overline{y}\!$$

Order of ODE: 2nd Order Linearity: Linear Superposition Principle: Yes

1.6f

 * $$LI''+RI'+\frac{1}{C}I=E' \!$$

 * Check superposition principle:
 * $$I''+\frac{R}{L}I'+\frac{1}{LC}I=\frac{E'}{L}\!$$
 * $$I_{h}''+\frac{R}{L}I_{h}'+\frac{1}{LC}I_{h}=0\!$$
 * $$I_{p}''+\frac{R}{L}I_{p}'+\frac{1}{LC}I_{p}=\frac{E'}{L}\!$$
 * $$(I_{h}+I_{p})+\frac{R}{L}(I_{h}'+I_{p}')+\frac{1}{LC}(I_{h}+I_{p})\equiv\overline{I}''+\frac{R}{L}\overline{I}'+\frac{1}{LC}\overline{I}\!$$

Order of ODE: 2nd Order Linearity: Linear Superposition Principle: Yes

1.6g

 * $$EIy^{iv}=f(x) \!$$

 * Check superposition principle:
 * $$y-\frac{1}{EI}y=0\!$$
 * $$y_{h}-\frac{1}{EI}y_{h}=0\!$$
 * $$y_{p}-\frac{1}{EI}y_{p}=0\!$$
 * $$(y_{h}'+y_{p}')-\frac{1}{EI}(y_{h}+y_{p})\equiv\overline{y}-\frac{1}{EI}\overline{y}\!$$

Order of ODE: 4th Order Linearity: Linear Superposition Principle: Yes

1.6h

 * $$L\theta''+g\sin\theta=0 \!$$

 * Check superposition principle:
 * $$\theta''+\frac{g}{L}\sin\theta=0\!$$
 * $$\theta_{h}''+\frac{g}{L}\sin\theta_{h}=0\!$$
 * $$\theta_{p}''+\frac{g}{L}\sin\theta_{p}=0\!$$
 * $$(\theta_{h}+\theta_{p})+\frac{g}{L}(\sin\theta_{h}+\sin\theta_{p})\not\equiv\overline{\theta}''+\frac{g}{L}\sin\overline{\theta}\!$$

Order of ODE: 2nd Order Linearity: Non-linear Superposition Principle: No

1.6i

 * $$y_{1}^{'}=ay_{1}-by_{1}y_{2} \!$$
 * $$y_{2}^{'}=ky_{1}y_{2}-ly_{1} \!$$

Order of ODE: 1st Order Linearity: Non-linear Superposition Principle: N/A

title: User:Egm4313.s12.team11.Suh/R1.6


text:
User:Egm4313.s12.team11.Suh/R2.5

Problem Statement
 Problem 2.5  Find an Ordinary Differential Equation. Use $$y''+ay'+by=0\!$$ for the given basis.

ODE Solutions
Two Real Roots: $$y(x)=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}\!$$ Double Real Roots: $$y(x)=c_{1}e^{\lambda_{1}x}+c_{2}xe^{\lambda_{2}x}\!$$

Reverse Engineering
$$e^{\lambda x}(\lambda^2+a\lambda+b)=0\!$$ Case 1: $$a^2-4b>0\!$$ $$\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!$$ $$y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!$$ Case 2: $$a^2-4b=0\!$$ $$\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!$$ $$y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!$$

R2.5 K2011 p.59 pbs.16
$$e^{2.6x}, e^{-4.3x}\!$$ $$(\lambda-2.6)(\lambda+4.3)\!$$ $$\lambda^2+1.7\lambda-11.18\!$$

ODE Form: $$y''+1.7y'-11.18=0\!$$

R2.5 K2011 p.59 pbs.17
$$e^{-\sqrt{5}x}, xe^{-\sqrt{5}x}\!$$ $$(\lambda+\sqrt{5})^2\!$$ $$\lambda^2+2\sqrt{5}\lambda+5\!$$

ODE Form: $$y''+2\sqrt{5}y'+5y=0\!$$

title: User:Egm4313.s12.team11.Suh/R2.5


text:
User:Egm4313.s12.team11.Suh/R3.4

Problem Statement
Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for $$y''-3y'+2y = 4x^{2}-6x^{5}\!$$ is of the form $$\sum_{j=0}^{n}c_{j}x^{j}\!$$, with n = 5.

Finding the Particular Solution with Basic Rule and Sum Rule
We know that in standard form, for a particular solution, $$y''_{p}+ay'_{p}+by_{p}=r(x)\!$$ $$y_{p}=y_{p1}+y_{p2}\!$$

Using Basic Rule (1) and Sum Rule (3), we know that $$y''_{p}+ay'_{p}+by_{p}=\sum_{i}r_i(x)\!$$

Additionally, we choose $$r_1(x)\rightarrow y_{p1}(x)\!$$ $$r_2(x)\rightarrow y_{p2}(x)\!$$

Solving for Particular Solution 1
Using the Method of Undetermined Coefficients, we find that $$y_{p1}=C_{2}x^{2}+C_{1}x+C_{0}\!$$ $$y'_{p1}=2C_{2}x+C_{1}\!$$ $$y''_{p1}=2C_{2}\!$$

Plugging $$y_{p1}\!$$ into $$y''-3y'+2y=4x^{2}\!$$ gives $$(2C_{2})-3(2C_{2}x+C_{1})+2(C_{2}x^{2}+C_{1}x+C_{0})=4x^{2}\!$$

Solving for coefficients $$2C_{2}=4\!$$ $$-6C_{2}+2C_{1}=0\!$$ $$-3C_{1}+2C_{0}+2C_{2}=0\!$$

Results in $$C_{2}=2\!$$ $$C_{1}=6\!$$ $$C_{0}=7\!$$

$$y_{p1}=2x^{2}+6x+7\!$$

Solving for Particular Solution 2
Using the Method of Undetermined Coefficients, we find that $$y_{p2}=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!$$ $$y'_{p2}=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\!$$ $$y''_{p2}=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!$$

Plugging $$y_{p2}\!$$ into $$y''-3y'+2y=-6x^{5}\!$$ gives $$(20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=-6x^{5}\!$$

Solving for coefficients results in $$K_{5}=-3\!$$ $$K_{4}=-22.5\!$$ $$K_{3}=-105\!$$ $$K_{2}=-337.5\!$$ $$K_{1}=-697.5\!$$ $$K_{0}=-708.75\!$$

$$y_{p2}=-3x^{5}-22.5x^{4}-105x^{3}-337.5x^{2}-697.5x-708.75\!$$

Combining Particular Solutions
Plugging $$y_{p1}\!$$ and $$y_{p2}\!$$ into $$y_{p}=y_{p1}+y_{p2}\!$$ gives

$$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$

Comparing to Summation Form
The particular solution in the summation form is $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$

if n=5, then $$y_{p}=\sum_{j=0}^{5}c_{j}x^j\!$$ $$y'_{p}=\sum_{j=0}^{5}c_{j}*j*x^{(j-1)}\!$$ $$y''_{p}=\sum_{j=0}^{5}c_{j}*j*(j-1)*x^{(j-2)}\!$$

Plugging the particular solution into $$y''-3y'+2y=-6x^{5}\!$$, the final solution gives $$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$ (check Problem 3.5 for explanation on how to get this answer)

Result
As you can see, the two particular solutions of $$y_{p}\!$$ are equal. Thus, using the Basic Rule (1) and Sum rule (3) does give you the correct particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$, in the form of $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$

title: User:Egm4313.s12.team11.Suh/R3.4


text:
User:Egm4313.s12.team11.Suh/R4.2

Problem Statement
Using the particular solution from Table 2.1 on Kreyszig, find the overall solution $$y(x)\!$$, and plot it compared with $$y_{n}(x)\!$$ for n = 3,5,9

$$y''-3y'+2y = r(x)\!$$ $$r(x)=sin(x)\!$$ Initial Conditions: $$y(0)=1\!$$, $$y'(0)=0\!$$

Homogenous Solution
To find $$y_{h}(x)\!$$, $$y''-3y'+2y = r(x)\!$$ $$\lambda^{2}-3\lambda+2=0\!$$ $$(\lambda-2)(\lambda-1)=0\!$$ $$\lambda_{1}=2, \lambda_{2}=1\!$$ $$y_{h}(x)=c_{1}e^{2x}+c_{2}e^{x}\!$$

Particular Solution
To find $$y_{p}(x)\!$$, Using Table 2.1, we find that for $$r(x)=sin(x)\!$$, $$y_{p}=Kcos (\omega x)+Msin (\omega x)\!$$ $$y'_{p}=-Ksin (\omega x)+Mcos (\omega x)\!$$ $$y''_{p}=-Kcos (\omega x)-Msin (\omega x)\!$$

Plugging it back into the equation, $$(-Kcos (x)-Msin (x))-3(-Ksin (x)+Mcos (x))+2(Kcos (x)+Msin (x)) = sin(x)\!$$ $$(3K+M)sin(x)+(K-3M)cos(x) = sin(x)\!$$

Solving for coefficients, $$(3K+M)sin(x) = 1\!$$ $$(K-3M)cos(x) = 0\!$$ $$K=\frac{3}{10}\!$$ $$M=\frac{1}{10}\!$$ $$y_{p}(x)=\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$

Overall Solution
$$y(x)=y_{h}(x)+y_{p}(x)\!$$ $$y(x)=c_{1}e^{2x}+c_{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$

Solving with initial conditions, $$y(x)=c_{1}e^{2x}+c_{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$ $$y'(x)=2c_{1}e^{2x}+c_{2}e^{x}-\frac{3}{10}sin(x)+\frac{1}{10}cos(x)\!$$

$$1=c_{1}+c_{2}+\frac{3}{10}\!$$ $$0=2c_{1}+c_{2}+\frac{1}{10}\!$$ $$c_{1}=\frac{-4}{5}\!$$ $$c_{2}=\frac{3}{2}\!$$

$$y(x)=\frac{-4}{5}e^{2x}+\frac{3}{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$ $$\!$$

Plot
Plotting in matlab yields: As shown on the graph, the two plots are exactly the same.

Created by [| Daniel Suh] 20:57, 7 February 2012 (UTC)

title: User:Egm4313.s12.team11.Suh/R4.2


text:
User:Egm4313.s12.team11.Suh/R5.7

Problem Statement
$$v=4e_{1}+2e_{2}=c_{1}b_{1}+c{2}b_{2}\!$$

$$b_{1}=2e_{1}+7e_{2}\!$$

$$b_{2}=1.5e_{1}+3e_{2}\!$$ 1. Find the components $$c_{1},c_{2}\!$$ using the Gram matrix. 2. Verify the result by using $$b_{1}\!$$ and $$b_{2}\!$$, and rely on the non-zero determinant matrix of $$b_{1}\!$$ and $$b_{2}\!$$ relative to the bases of $$e_{1}\!$$ and $$e_{2}\!$$.

Gram Matrix
$$ T(b_{1},b_{2})=\begin{bmatrix}  &  \\  &  \end{bmatrix}$$ $$ = \!$$ Thus,

$$ =  = <(2)(2) + (7)(7)> = 53\!$$

$$ =  = <(1.5)(1.5) + (3)(3)> = 11.25\!$$

$$ =  =  = <(2)(1.5) + (7)(3)> = 24\!$$

$$ T(b_{1},b_{2})=\begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}$$

$$\Gamma = det[T] = (53)(11.25)-(24)(24) = 20.25\!$$

Defining c
Define: $$c =\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$$ $$d = \begin{bmatrix}  \\  \end{bmatrix}= \begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}\!$$

If $$\Gamma \neq 0 \!$$, then $$\Gamma^{-1}\!$$ exists

$$c = \Gamma^{-1}d\!$$

Finding c
$$\Gamma = 20.25 \neq 0 \!$$ thus, $$\Gamma ^{-1}\!$$exists

$$d =\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix} = \begin{bmatrix} (2)(4)+(7)(2) \\ (1.5)(4)+(3)(2) \end{bmatrix} = \begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}^{-1}\begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} -2 \\ 5.33 \end{bmatrix}\!$$

Part 2 Solution
$$v = 4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = (-2)(2e_{1}+7e_{2}) + (5.33)(1.5e_{1} + 3e_{2})\!$$

$$c_{1}b_{1} + c_{2}b_{2} = -4e_{1} - 14e_{2} + 8e_{1} + 16e_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = 4e_{1} + 2e_{2}\!$$

$$4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

Thus, the solution is correct.

title: User:Egm4313.s12.team11.Suh/R5.7


text:
User:Egm4313.s12.team11.Suh/R6.1

Problem Statement

 * Find the fundamental period of $$\cos (n\omega x) \!$$ and $$\sin (n\omega x) \!$$
 * Show that these functions also have period $$p\!$$.
 * Show that the constants $$a_{0}\!$$ is also a periodic function with period $$p\!$$.

Solution

 * Find the fundamental period, and show that these functions have a period of $$p\!$$.

$$p=2L=2\frac{\pi}{\omega}\!$$

$$L = \frac{\pi}{\omega}\!$$

$$f(x+np) = f(x)\!$$

$$f(x) = cos (n\omega x)\!$$

$$f(x+n\frac{2\pi}{n\omega }) = cos (n\omega (x+\frac{2\pi}{n\omega }))\!$$

$$f(x+\frac{2\pi}{\omega}) = cos (n\omega x+2\pi)\!$$

$$f(x+2L) = cos (n\omega x+2\pi)\!$$

$$p=2L\!$$ is the fundamental period for $$cos (n\omega x)\!$$.

$$f(x+np) = f(x)\!$$

$$f(x) = sin (n\omega x)\!$$

$$f(x+n\frac{2\pi}{n\omega }) = sin (n\omega (x+\frac{2\pi}{n\omega }))\!$$

$$f(x+\frac{2\pi}{\omega }) = sin (n\omega x+2\pi)\!$$

$$f(x+2L) = sin (n\omega x+2\pi)\!$$

$$p=2L\!$$ is the fundamental period for $$sin (n\omega x)\!$$.

 * Show that the constants $$a_{0}\!$$ is also a periodic function with period $$p\!$$.

From Fourier Series, $$a_{0}=\frac{1}{2L}\int_{-L}^{L}f(x)dx\!$$

$$f(x) = \cos (n\omega x)\!$$

$$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}cos (n\omega x)dx\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{sin (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{2sin (n\pi)}{n\omega}]\!$$

$$a_{0}=0\!$$

$$f(x) = \sin (n\omega x)\!$$

$$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}sin (n\omega x)dx\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{cos (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$$

$$a_{0}=\frac{\omega}{2\pi}[\frac{-cos{n\pi}+cos{n\pi}}{n\omega}]\!$$

$$a_{0}=0\!$$

title: User:Egm4313.s12.team11.Suh/R6.1


text:
User:Egm4313.s12.team11.Suh/R6.7

Report 6, Problem 6.7
Solved by Solved by Daniel Suh

Problem Statement
Find the separated Ordinary Differential Equations for the heat equation.

$$\frac{\partial u}{\partial t}=\kappa \frac{\partial^2 u}{\partial x^2} \cdots (1)\!$$

$$u(x,t) = F(x)\cdot G(t)\!$$

Solution
$$\frac{\partial u}{\partial t} = F(x)\cdot \dot{G}(t) \cdots (2)\!$$

$$\frac{\partial^2 u}{\partial x^2} = F''(x)\cdot G(t) \cdots (3)\!$$

Combine $$(2)\!$$ and $$(3)\!$$ into $$(1)\!$$

$$F(x)\cdot \dot{G}(t) = \kappa F''(x)\cdot G(t)\!$$

$$\frac{\dot{G}(t)}{\kappa G(t)} = \frac{F''(x)}{F(x)} = c\!$$ (constant)

Separate

$$F''(x)-cF(x) = 0\!$$ $$G'(t)-\kappa cG(t) = 0\!$$

title: User:Egm4313.s12.team11.Suh/R6.7


text:
User:ChimaEzen/My work for Putt Putt

Putt Putt

Write problem/project Goal
Get the ink head to move back and forth

My First Task
Understand the applications of the Coanda effect.

Summary of actual work over first weekend
How is what you did different than what you planned?

Week1 Narrative
I was working with the ink jet printer head, and tried to follow Dr. Foester's tutorial for how to move the ink head back and forth here's a video of just the volts being changed. video after that we hooked up the all the wires to the arduino and got it to move back and forth video and now I'm study the encoder and how it works so we can mount it on the pendulum.

My Second Task
work the incode so it moves as far as we want it too

Summary of actual work over second weekend
How is what you did different than what you planned?

Week2 Narrative
First week I was working with the ink jet printer head, and tried to follow Dr. Foester's tutorial for how to move the ink head back and forth here's a video of just the volts being changed. video after that we hooked up the all the wires to the arduino and got it to move back and forth video and now I'm study the encoder and how it works so we can mount it on the pendulum.

I dot yet have to wood for the hanging structure now, but i already have the basic design, i started cleaning out my garage and closet looking for anything that i could use to help build this structure

My Third task
Record what you are planning on doing for the team during the third weekend.

Summary of actual work over third weekend
How is what you did different than what you planned?

Week3 Narrative
So i made the shelf, for the print head to sit next to the pendulum with two shelf supports

I used the dremel to cut the excess part of the printer keeping the print head away from the pendulum

My Fourth task
Record what you are planning on doing for the team during the final weekend of project work.

Summary of actual work over fourth weekend
How is what you did different than what you planned?

Week4 Narrative
I found this piece to an old guitar stand i want to use it ass a hook for the print head to connect with the pendulum i can cut it an connect the cut end to the print head and slip the circular part of the pendulum

Complete Project Page
Follow the "project done" format.

title: User:ChimaEzen/My work for Putt Putt


text:
User:Egm4313.s12.team11.Suh/R7.5

Report 7, Problem 7.5
Solved by Solved by Daniel Suh

Problem Statement
Consider the following, $$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\phi _{2j-1}(x) \cdot \phi _{2k-1}(x)dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {j\omega x} \cdot \sin {k\omega x} \; dx\!$$

with $$j\neq k\; and\; j,k = 1,2,...\!$$, and $$p=2\pi, j=2, k=3\!$$

1. Find the integration with the given data.

2. Confirm the results with Matlab's trapz command for the trapezoidal rule.

Trigonometric Identities
Angle Sum and Difference Identities

$$(1) \cos{(a+b)} = \cos{a}\cos {b} - \sin {a}\sin{b}\!$$ $$(2) \cos{(a-b)} = \cos{a}\cos {b} + \sin {a}\sin{b}\!$$

Rearrange

$$(1) \sin {a}\sin{b} = \cos{a}\cos {b}-\cos{(a+b)}\!$$ $$(2) \cos{a}\cos {b} = \cos{(a-b)} - \sin{a}\sin {b}\!$$

Substitute and Combine

$$\sin {a}\sin{b} = \cos{(a-b)} - \sin{a}\sin {b} - \cos{(a+b)}\!$$ $$2\sin {a}\sin{b} = \cos{(a-b)} - \cos{(a+b)}\!$$ $$\sin {a}\sin{b} = \frac{1}{2}\cos{(a-b)} - \frac{1}{2}\cos{(a+b)}\!$$

Utilize Trig Identities
$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {jwx} \cdot \sin {kwx} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\sin {2\omega x} \cdot \sin {3\omega x} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\frac{1}{2}\cos {(2\omega x - 3\omega x)} - \frac{1}{2}\cos{(2\omega x + 3\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\int_{0}^{2\pi}\cos {(-\omega x)} \; dx - \frac{1}{2}\int_{0}^{2\pi}\cos{(5\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\sin {(-\omega x)}|_{0}^{2\pi} - \frac{1}{2}\sin{(5\omega x)}|_{0}^{2\pi}\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= (0-0) - (0-0) \!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= 0 \!$$

title: User:Egm4313.s12.team11.Suh/R7.5