text:
User:Jwood/matlab-control-tutorial

The purpose of this page is to provide a useful primer to teachers (and students) of a control theory course that requires numerical analysis using MATLAB to model complex linear systems. These systems consist of several interconnected sub-systems and are referred to as block diagrams. This primer will start with forming the most basic MATLAB system representation objects and then show how to interconnect several of them. Then a couple simulation methods will be discussed.

In linear control theory, the dynamics of the system of interest (plant) are often expressed in state-space as

$$\frac{dx}{dt} = Ax + Bu$$

$$y = Cx + Du \frac{}{}$$

where $$x \in R^n$$ is the vector of system states, $$y \in R^m$$ is the vector of system outputs, and $$u \in R^i$$ is the vector of inputs. In MATLAB a state-space object can be constructed as

$ ss(A,B,C,D);

Alternatively, a system can be represented as a transfer function. The transfer function is the system's output response to an input impulse. The output response is then

$$Y(s) = H(s)U(s) \frac{}{}$$

For one input one output (SISO), a transfer function is of the form

$$H(s) = \frac{b_{n_z-1} s^{n_z-1} + b_{n_z-2} s^{n_z-2} + ... + b_0}{a_{n_p-1} s^{n_p-1} + a_{n_p-2} s^{n_p-2} + ... + a_0}$$

In MATLAB a transfer function object can be constructed as

$ tf(num,den);

where num is the numerator polynomial and den is the denominator polynomial of the transfer function. For example, if the transfer function is

$$H(s) = \frac{1}{s+.5}$$

then you would construct the transfer function object as

$ tf([1],[1 .5]);

You can always get a state-space object from a transfer function object as

$ ss(tf(num,den));

Often a complex system can be modeled by interconnecting several sub-systems. After connecting the sub-systems, the resulting system consists of some subset of the inputs and outputs of the sub-systems. Constructing a state-space object in this manner can be done in MATLAB. To do this, you first need your sub-system state-space objects (ss1, ss2, ..., ssN). You then need to construct a system that has them independent of each other. You do this as

$ append(ss1,ss2,...,ssN);

You now need to connect inputs to outputs. To do this, define a matrix of connectivity. This matrix is of the form

$$ M = \begin{bmatrix} 1 & sysIn_{1,1} & sysIn_{1,2} & .. & sysIn_{1,N} \\ 2 & sysIn_{2,1} & sysIn_{2,2} & .. & sysIn_{2,N} \\

N & sysIn_{N,1} & sysIn_{N,2} & .. & sysIn_{N,N} \end{bmatrix} $$

In this matrix, the first column specifies the sub-system number. The elements $$sysIn_{i,j}$$ specify the outputs of which sub-systems sum together to feed into the input of the sub-system specified by the number in the first column. As a first example, consider a simple negative feedback system. You construct the feedback system by using the matrix

$$ M = \begin{bmatrix} 1 & -1 \end{bmatrix} $$

As a more complex example, if you have two sub-systems for which the output of the first sub-system sums with the negative output of the second sub-system and then feeds into the input of the second sub-system, then the matrix is

$$ M = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \end{bmatrix} $$

Notice that elements that are $$0$$ specify an empty input.

Now that the connectivity of the sub-systems is specified, you need to specify which of the sub-system inputs and outputs are the inputs and outputs of the total connected system. To do this you form vectors specifying the numbers of the inputs and outputs to be used. For example, if the inputs of sub-systems 2 and 5 are the connected system's inputs, then the input vector is

$$ in = \begin{bmatrix} 2 \\ 5 \end{bmatrix} $$

If the outputs of the sub-systems 1 and 3 are the outputs of the connected system, then the output vector is

$$ out = \begin{bmatrix} 1 \\ 3 \end{bmatrix} $$

So, after constructing the sub-system state-space objects ($$ss1$$, $$ss2$$, ..., $$ssN$$), the connectivity matrix ($$M$$), the input indices vector ($$in$$), and the output indices vector ($$out$$), and the appended system ($$sysApp$$), the final connected system is constructed in MATLAB as

$ connect(sysApp,M,in,out);

To check the stability of your connected system, one option is to plot the root loop and observe where the closed loop are. Do this in MATLAB as

$ rlocus(sys);

Now to simulate the the response of the connected system ($$sys$$), the input signals have to be specified. After specifying an arbitrary input signal-time pair ($$u$$ and $$t$$), you can simulate the response of the system as

$ lsim(sys,u,t);

The size of the input signal $$u$$ must match the length of the time vector $$t$$ and the number of system inputs. For example, if the number of inputs is $$N_i$$ and the size of $$t$$ is $$1 \times N_t$$, then the input signal must be of size $$N_i \times N_t$$. Note also that the order of the inputs is the same as the order specified when creating the connected system (i.e., check the $$in$$ vector).

An example input signal is a noisy step input. The nominal step is all ones. At each step in time add a random disturbance. The loop below illustrates the construction of this input signal.

$ dt = .01; tt = 0; $ t = zeros(1,N); u = zeros(1,N); $ for i=1:length(t) $    tt += dt; $    t(i) = tt; $    u(i) = 1+rand(-.1,.1); $ end

Hopefully this short system dynamics primer has helped jump start you in constructing block diagrams in MATLAB. The methods introduced here are very general so that arbitrary block diagrams can be constructed and arbitrary input signals can be applied to the system.

Jwood 29 November 2009

title: User:Jwood/matlab-control-tutorial


text:
User talk:Demosthenes~enwikiversity

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title: User talk:Demosthenes~enwikiversity


text:
User talk:Argentarose

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

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title: User talk:Argentarose


text:
User talk:Akohnert

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

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 30 November 2009 (UTC)

title: User talk:Akohnert


text:
New Zealand Law/Criminal/Mistake

Definition
Where no intention, knowledge or recklessness exists due to mistake (even if the mistake is not reasonable).

R v Wood

 * Facts
 * 1) D cultivated pot, claimed she thought they were tomatoes.
 * 2) Judged directed jury to use their own knowledge of how cannibus and tomatoes grown.

 * Held
 * 1) Misdirection – no evidence on matter, jury not entitled to take notice.
 * 2) Following DPP v Morgan, honest belief need not be reasonable.

R v Metuariki

 * Facts
 * 1) Defendant picked wild mushrooms, sold to undercover police officer, charged with supply of class A controlled drug.
 * 2) Defendant knew mushrooms made you high, didn’t know contained class A drug, or illegal.

 * Held
 * 1) To establish prima facie case, crown must prove defendant (i) supplied mushrooms (ii) mushrooms were a controlled drug.
 * 2) Knowledge would be presumed unless defendant introduce evidence that he honestly believed his actions were innocent.
 * 3) Ignorance of the law is no defence.
 * 4) For example, it is not a defence to claim that a person thought he was supplying a different controlled drug.
 * 5) Accused knew mushrooms had hallucinatory effects but they were growing in a public place. There was no evidence that he thought they were anything other than innocent vegetation.
 * 6) Accused entitled to acquittal unless jury convinced otherwise beyond a reasonable doubt.

title: New Zealand Law/Criminal/Mistake


text:
New Zealand Law/Criminal/SelfDefence

Requirements

 * 1) Use of force or the threat of force in the defence of self or another.
 * 2) Force that defendant believes in the circumstances to be reasonable.
 * 3) Elements of the offence must require force.
 * 4) The prosecution must first prove prima facie case involving mens rea.

R v Tavete

 * Held

 * Judge must put defence to jury which is capable of being supported by evidence, even if for tactical reasons defence did not.

R v Ranger

 * Facts

 * 1) During argument with boyfriend, defendant stabbed and killed him.
 * 2) Defence claimed provocation.
 * 3) Defence did not raise self-defence although there was evidence that defendant believed boyfriend was going for gun and would shoot her and son.

 * Held
 * 1) If the accused really believed in threat, then it would be going too far to say that a jury could not entertain a reasonable doubt as to whether a pre-emptive strike with a knife was reasonable force in the circumstances.
 * 2) It is sometimes tactically difficult for the defence to run 2 defences. It is, therefore, the responsibility of the trial judge to place before the jury any defence capable of being supported by the evidence.

R v Wang

 * Facts

 * 1) Husband passed out, D stabbed him 4 times.
 * 2) Husband had threatened to kill her and family, and was also blackmailing sister.
 * 3) On appeal, defendant said judge should have put self-defence to jury.

 * Held
 * 1) Impossible for jury to find that force reasonable in the circumstances, however a pre-emptive strike not always unreasonable.

R v Savage

 * Facts

 * 1) Savage was a patched gang member. He and others got into a fight with O’Leary.  Savage stabbed O’Leary and fled.
 * 2) Evidence was that O’Leary had been causing trouble all day.
 * 3) Savage claimed self-defence or provocation.

 * Held
 * 1) When a knife is used, the accused must be under real threat, not merely think that there might be some future danger.
 * 2) Whether a person should retreat from a fight depends on what is reasonable in the circumstances.
 * 3) It is a misdirection to say either that (i) provocation must occur immediately before killing or (ii) that the def’s action must be proportionate to the provocation.

R v Thomas

 * Facts

 * 1) Police trying to arrest suspect.
 * 2) Intoxicated defendant thought police were using excessive force, physically intervened.

 * Held
 * 1) Mistaken belief (that force used by police is unlawful) must be honest, need not be reasonable.
 * 2) Mistake of fact.
 * 3) Self defence available in defence of stranger.
 * 4) Not confined to defence against unlawful assault.

R v Kneale

 * Facts
 * 1) Defendant left daughter with ex-wife to go fishing. On return, K and ex had fight.  K assaulted wife.
 * 2) Claimed self-defence - protecting daughter from wife.

 * Held
 * 1) Judge wrong not to direct jury on self-defence but the threshold for psychological injury is high (fear, panic or grief is not enough).
 * 2) Misdirection made no difference. There was not enough evidence of threat, after all, he left daughter with ex-wife hours earlier to go fishing.

R v Howard

 * Held
 * 1) Self defence is not confined to situations involving threats of death or serious bodily harm.
 * 2) The fact that there is another motivation, eg spite or anger, does not rule out self-defence. In Howard’s case, also had fear of future assault.
 * 3) Force that is reasonable may not be proportionate (eg if accused has no real choice).

R v Bridger

 * Held
 * 1) It will often be necessary to assess whether defendant acting in self-defence by what accused did in response.
 * 2) Court declined to opine about whether the circumstances - as the defendant believes them to be - must be based on a sane belief.
 * 3) Trial judge may withdraw a defence from jury if it is impossible for a reasonable jury to regard accused’s reaction as reasonable. Withdrawal of a defence is not a denial of the right to trial by jury.

title: New Zealand Law/Criminal/SelfDefence


text: File:Home page screen.jpg

title: File:Home page screen.jpg


text:
User:SiGarb

Hi! See my userpage at English Wikipedia. SiGarb 18:31, 30 November 2009 (UTC)

title: User:SiGarb


text:
LQR Control for an Inverted Pendulum

A LQR controller is used to stabilize a cart-pole system around its unstable equilibrium with the cart at the origin and the pole in its upright position. A region of attraction for the LQR controller on the systems linearized dynamics is empirically found. Outside of this region an energy pumping scheme is derived to drive the system to its homoclinic orbit. A hybrid system is found to stabilize the cart-pole from any initial conditions by switching to the LQR policy when the system reaches its region of attraction. The cart-pole is described by the following dynamics:

title: LQR Control for an Inverted Pendulum


text:
User:Wikieat

'''EDIT MY PAGE!!!!! WTF IDIOTS WHO POST VULGAR FILFTH ARNT COOL! DUDE DONT DO THAT!'''

   Current Wikistress:

  This page has been vandalised

twenty-two times.

About Me
I am a cool cat, online I use the alias Swindle.

I am an [[ EDIT MY PAGE!!!!! WTF IDIOTS WHO POST VULGAR FILFTH ARNT COOL! DUDE DONT DO THAT!

   Current Wikistress:

About Me
I am a cool cat, online I use the alias Swindle.

I am an Bangladeshi, although I was born in the United Kingdom.

To-do List
Find something to do...

Useful Links

 * Anonymous Recent Changes
 * List of common misspellings

Useless Links

 * My Wiki Advertisment
 * The Something Awful Forums

Crazy Wikipedians
Crazy Wikipedians is a list of the results of an online personality disorder test. It's completely unscientific, but funny to see how crazy you are.

LOL, sorry Wikieat, I had to. :D It's all in good fun!

Etc
I grabbed the original layout of this page from User:AngryParsley.

simple:User:AngryParsley

Below is some vandalism I'm keeping for fun.

Jimbo the Pirate is awesome and if you disagree I will eat you.

]], although I was born in the United Kingdom.

To-do List
Find something to do...

Useful Links

 * Anonymous Recent Changes
 * List of common misspellings

Crazy Wikipedians
Crazy Wikipedians is a list of the results of an online personality disorder test. It's completely unscientific, but funny to see how crazy you are. If you want me to add your results to the list, talk to me in #wikipedia.a

Countries I've been to
{| border="1"
 * [[Image:Flag of Bangladesh.svg|40px]]
 * [[Image:Flag of Bangladesh.svg|40px]]

Etc
I grabbed the original layout of this page from User:AngryParsley.

simple:User:AngryParsley

Below is some vandalism I'm keeping for fun.

Jimbo the Pirate is awesome and if you disagree I will eat you.

title: User:Wikieat


text:
Comparative Teaching of Old Greek and Latin/Lesson 04 Part 2

Comparative Teaching of Old Greek and Latin Lesson 04 Part 2

Translated from the Greek Wikipedia|Wikiversity: "Συγκριτική διδασκαλία των κλασικών γλωσσών / (Comparative Teaching of the Classical Languages)"

'''LESSON 04  PART 2. The verb. Regular verbs. Present, imperfect, future indicative, active voice.'''

4.2.	Analyse and translate the texts:

(Each text should be separated in periods, each period in sentences and each sentence should be  analyzed grammatically  and syntacticly and after be translated.)

OLD GREEK

4.2.1.  Xenophon's Cyrus' Anabasis, 1, 2, 7-8.

'''Ἐντεῡθεν ἐξελαύνει σταθμοὺς τρεῖς παρασάγγας εἴκοσιν εἰς Κελαινὰς τῆς Φρυγίας, πόλιν οἰκουμένην, μεγάλην καὶ εὐδαίμονα. Ἐνταῦθα Κύρῳ βασίλεια ἦν καὶ παράδεισος μέγας ἀγρίων θηρίων πλήρης, ἅ ἐκεῖνος ἐθήρευεν ἀπὸ ἵππου, ὁπότε γυμνάσαι βούλοιτο ἑαυτόν τε καὶ τοὺς ἵππους. Διὰ μέσου δὲ τοῦ παραδείσου ῥεῖ ὁ Μαίανδρος ποταμός. Αἱ δὲ πηγαὶ αὐτοῦ εἰσὶν ἀπὸ τῶν βασιλείων. Ῥεῖ δὲ καὶ διὰ τῆς Κελαινῶν πόλεως. Ἔστι δὲ καὶ μεγάλου βασιλέως βασίλεια ἐν Κελαιναῖς ἐρυμνὰ ἐπὶ ταῖς πηγαῖς τοῦ Μαρσύου ποταμοῦ  ὑπὸ τῇ  ἀκροπόλει. Ῥεῖ δὲ καὶ οὗτος διὰ τῆς πόλεως καὶ ἐμβάλλει εἰς τὸν Μαίανδρον. Τοῦ δὲ Μαρσύου τὸ εὖρός ἐστιν εἴκοσι καὶ πέντε ποδῶν.'''

'''Vocabulary - comments. Analysis in detail'''

(Substantives will be given and  should  be learned with  their genitive, adjectives  and pronouns with the three genders, adverbs with the three degrees and  verbs with  their initial  tenses - in Latin  the present, the perfect, the supine and the present  infinitive.)

The text has eight periods. We first find the sentences and clauses of each period, the infinitives and the participles, if there exist any, and then the main basic terms, that is to say, the verb, the subject, the direct object, the indirect object, the predicative, provided that of course they exist. The remaining words will be determinations, depending on the kind of word and the term that it determines.

Main basic terms of the 1st period: Ἐντεῡθεν … εὐδαίμονα.

ἐξελαύνει: he goes on, he progresses, verb of the main clause, initial tenses:   ἐλαύνω, ἤλαυνον, ἐλῶ (ἐλᾷς, ἐλᾷ, ἐλῶμεν etc) ἤλασα, ἐλήλακα, ἐληλάκειν, of the same root:  ελαύνομαι, απελαύνομαι, απέλασις (banishment), παρέλασις (parade), νεήλατος, σφυρήλατος (hammered, forged) ελατήριο (spring), έλασμα, ιχνηλάτης, ποδήλατο (bicycle),  σταθμούς, παρασάγγας: stathmos = stopping, station, distance of one day's march, divided in five parasanges, object to the verb,  οἰκουμένην: being inhabited, adjectival participle, initial tenses:  οἰκοῦμαι, ᾠκούμην, οἰκήσομαι, ᾠκήθην,  ᾤκημαι, ᾠκήμην,  of the same root: οικέτης (servant), οικία (house),  οίκος (house), οἰκήτωρ, διοίκηση (administration), οικήσιμος, structures: οἰκῶ τι, οἰκῶ ἔν τινι.

Determinations of the first period:

ἐντεῦθεν: thence, from there, adverbial determination denoting motion from a place, τρεῖς, εἴκοσιν (twenty), μεγάλην (big), εὐδαίμονα, (rich, happy): adjectival determinations,   εἰς Κελαινάς: to kelaines, prepositional determination denoting motion toward a place, πόλιν: city, apposition,  τῆς Φρυγίας: of Frygia, genitive possessive.

Analysis of the second period: Ἐνταῦθα … ἵππους.

ἦν (he was), ἐθήρευεν (he hunted), βούλοιτο (he wanted): verbs of the sentence,   γυμνάσαι: to train, to exercise, infinitive, object to βούλοιτο, final infinitive (translated by να, ἵνα),  βασίλεια: palace, subject to ἦν, attic structure,  παράδεισος: garden, subject to ἦν  ἐκεῖνος: that, subject to ἐθήρευεν,  ἅ ἐκεῖνος ἐθήρευεν ἀπὸ ἵππου: subordinate relative clause,  ὁπότε γυμνάσαι βούλοιτο ἑαυτόν τε καὶ τοὺς ἵππους: subordinate time clause,  ἅ, γυμνάσαι, ἑαυτὸν (himself), ἵππους: objects,  ἐνταῦθα: there, adverbial determination denoting stopping at a place,  Κύρῳ: to Cyrus, dative personal, possessive,   μέγας (big), ἀγρίων (wild, fierce), πλήρης (full): adjectival determinations,  θηρίων: of beasts, genitive objective,  ἀπὸ ἵππου: being on the horse, prepositional determination denoting motion from a place.

Analysis of the third period: Διὰ μέσου … ποταμός.

ῥεῖ: it flows, it streams, verb of the main clause, initial tenses: ῥέω, ἔρρεον,  ῥυήσομαι, ἐρρύην,  ἐρρύηκα, ἐρρυήκειν, of the same root: ρεύμα (the draught), ροή (the flow), ῥοῦς (the flow), απόρροια (consequence, result), χείμαρρος (the torrent), ῥύμη, ρυθμός (rythm), ρύση, ρείθρο, ρεύση, structures:  ῥέω ἐπί τι, εἴς τι, διά τινος, ἔκ τινος,  ποταμός: river, subject of the verb,  Μαίανδρος: Meander, adjectival determination, διὰ τοῦ παραδείσου: through the garden, prepositional determination denoting motion through a place,  μέσου: through the middle of the river,  predicative determination.

'''Analysis of the fourth period:. Ῥεῖ δὲ … πόλεως.'''

διὰ τῆς πόλεως: through the city, prepositional determination denoting motion through a place, Κελαινῶν: apposition.

Analysis of the fifth period: Αἱ δὲ πηγαὶ … ἀπὸ τῶν βασιλείων.

ἀπὸ τῶν βασιλείων: from the palace, prepositional determination denoting motion from a place, Αἱ πηγαὶ: (the springs), subject.

Analysis of the sixth period: Ἔστι δὲ …  ἀκροπόλει.

ἔστι: is, here = there are, attic structure, the verb is in 3rd singular and the subject in neuter plural, present subjunctive:  ὦ, ᾖς, ᾖ, ὦμεν, ἦτε, ὦσι,  βασίλεια: palace, subject,  βασιλέως, ποταμοῦ: genitive possessive,  μεγάλου (big), ἐρυμνά  (fortified), Μαρσύου (of river Marsyas): adjectival determinations,   ἐν Κελαιναῖς (in Kelaines), ἐπὶ ταῖς πηγαῖς (near the springs), ὑπὸ τῇ ἀκροπόλει (beneath, under the acropolis): prepositional determinations.

Analysis of the seventh period: Ῥεῖ δὲ … τὸν Μαίανδρον.

ῥεῖ (it flows), ἐμβάλλει (it flows into): verbs of the two main clauses,  διὰ τῆς πόλεως: (through the city),  εἰς τὸν Μαίανδρον: prepositional determinations denoting motion through a place and motion toward a place.

Analysis of the eighth period: Τοῦ δὲ Μαρσύου … πέντε ποδῶν.

Μαρσύου: genitive possessive, τὸ εὖρός: the width,  εἴκοσι καὶ πέντε ποδῶν: having width twenty five feet, genitive of quality.

LATIN

4.2.2. From the old reading-book of the Latin language, 6, 9.

(The accentual marks will enter for facilitation of reading until the thirtieth fifth course.)

'''Roma’ni, po’pulus bellico’sus, armis et telis pugnant. Arma Romano’rum sunt ga’lea, scutum, lo’rica. Tela sunt pilum, gla’dius et sagi’ttae. Gla’dii Romano’rum non sunt longi, pila sunt longa. Roma’ni in bellis coti’die castra colloca’bant et vallo fossa’que firma’bant. Valla alta et fossae latae erant. Ha’nnibal ante pugnam a’nimos Poeno’rum his verbis infla’mmat: «Nisi fo’rtiter pu-gna’bitis Roma’ni po’pulos A’fricae supera’bunt et campos vico’sque pa’triae vasta’bunt. Ego post pugnam impa’vidos lauda’bo et dona’bo, igna’vos vitupera’bo et castiga’bo».'''

'''Vocabulary - comments. Analysis in detail.'''

The text has nine periods. We first find the sentences and clauses, the infinitives and the participles, if there exist any, and then the main basic terms, that is to say, the verb, the subject, the direct object, the indirect object, the predicative, provided that of course they exist. The remaining words will be determinations, depending on the kind of word and the term that it determines.

Main basic terms of the first period: Roma’ni … pugnant.

pugnant: they fight, verb of the sentence, initial tenses:  pugno, -a’vi, -a’tum, -a’re, of the same root: pugna (the fight), pugna’citas, pugna’citer, pugna’culum, pugna’tor (fighter, warrior), pugnato’rius, pugnax (quarrelsome),  po’pulus: people, apposition to Romani,  po’pulus, -i, m, (people)  popu’lus, -i, f, (poplar-tree),  bellico’sus: (warlike, bellicose, military), bellico’sus, -a, -um, adjectival determination,  armis et telis: by defensive and aggressive arms), ablatives denoting instrument.

Analysis of the second period: Arma Romano’rum … lo’rica.

sunt: they are, initial tenses: sum, fui, esse, present perfect indicative: fui, fui’sti, fuit, fu’imus, fui’stis, fuerunt, past perfect indicative: fu’eram, fu’eras, fu’erat, fuera’mus, fuera’tis, fu’erant, future perfect indicative: fu’ero, fu’eris, fu’erit, fue’rimus, fue’ritis, fu’erint,  Romano’rum: of the Romans, genitive possessive, ga’lea (helmet), scutum (shield), lo’rica (thorax): subjects of the verb,   arma: arms, predicative.

Analysis of the third period: Tela sunt … sagi’ttae.

Pilum (javelin), gla’dius  (sword), sagi’ttae (arrow): subjects of the verb sunt, tela: predicative.

Analysis of the fourth period: Gla’dii … sunt longa.

longi (long), longa (long): predicative.

Analysis of the fifth period: Roma’ni … firma’bant. colloca’bant (they founded, they established), firma’bant (they fortified): verbs of the two main clauses,  in bellis:, during the war, prepositional determination of time, bellum, -i, the war,   coti’die: every day, daily,  adverbial determination of time,  castra: military camp, object, castra, -o’rum, neuter, it is used only in plural,   vallo:  by trench, fossa:, by ditch : instrumental ablatives, vallum, -i, fossa, -ae,   que: and, copulative conjunction.

'''Analysis of the sixth period: Valla alta …. latae erant.'''

alta (high),lata (wide, broad): predicatives, altus, -a, -um (high), latus, -a, -um (wide).

Analysis of the seventh period: Ha’nnibal … infla’mmat. infla’mmat (he excites): verb of the main clause, initial tenses: infla’mmo, -a’vi, -a’tum, -a’re,  Ha’nnibal: Hannibal, Ha’nnibal, Hanni’balis, the son of Hamilcar, leader of the people of Carthage,  a’nimos:  the souls, object, a’nimus, -i,m, related to  a’nima, -ae, f, Poeno’rum: of Carthaginians, genitive possessive,  Poeni, -o’rum,   his verbis: by these words, ablative of manner, verbum, -i,  hic, haec, hoc (this),   ante pugnam, before the fight, prepositional determination of time, pugna, -ae  (fight)  ante: before, preposition structured with accusative.

Analysis of the eighth period: Nisi fo’rtiter … vasta’bunt.

supera’bunt (they will win), vasta’bunt (they will devastate), verbs of the two main clauses, nisi (unless) fo’rtiter (bravely) pugna’bitis (you will fight): subordinate if clause,   Roma’ni: subject of the two verbs,   po’pulos, campos, (the plains), vicos, (villages): objects,   nisi pugna’bitis: if you do not fight, verb of the subordinate if clause,  fo’rtiter: adverbial determination of manner,   A’fricae, pa’triae: genitives possessive.

Analysis of the ninth period: Ego post pugnam … castiga’bo.

lauda’bo (I'll praise), dona’bo (I'll give as present), vitupera’bo (I'll accuse),  castiga’bo (I'll punish): verbs of the four main clauses   impa’vidos (brave), igna’vos (timid, cowardly)): objects,   post pugnam: after the fight, prepositional determination of time,  post: after, preposition structured with accusative.

To continue look at: ../Lesson 04 Part 3/ To see the Introduction look at: ../Introduction/


text:
User:Albane69

Hi,

My name is Albane, I'm French and am currently living in Spain for my studies.

I discovered Wikiversity and the French department only some days ago. This department is really a good thing and I will try to update some lessons and add new ones.

I am also a French mentor so if you need any help with your French, don't hesitate to contact me (use my talk page at first). I can't promise you I will be available for a voice chat as I am quite busy right now but I can easily answer to mails or even chat in text form via msn or gmail.

title: User:Albane69


text:
Category talk:User bg

What language is this?
What language is bg code for? I'm trying to make boxes to explain what each of the languages are, both in English for English speakers and in the language of the category of users, and I can't for the life of me figure out what this one is. If anyone would care to illuminate me it would be much appreciated! :-) Thanks, --Trinity507 23:03, 30 November 2009 (UTC)

title: Category talk:User bg


text:
LQR Control for an Inverted Pendulum on a Cart

The purpose of this page is to show the derivation of the linearized dynamics for the highly non-linear cart-pole system and to propose an algorithm to calculate the optimal balancing input using a LQR controller. This system is classically studied in non-linear controls.

A LQR controller is used to stabilize the cart-pole system around its unstable equilibrium with the cart at the origin and the pole in its upright position.The cart-pole is described by the following dynamics:

$$(m_c + m_p) \ddot{x} + m_p l \ddot{\theta} \cos \theta - m_p l \dot{\theta}^2 \sin \theta = u$$

$$m_p l \ddot{x} cos \theta + m_p l^2 \ddot{\theta} + m_p g l \sin \theta = 0$$

This may be put into standard form with $$q = [x,\theta]^T$$:

$$\mathbf{H} (\mathbf{q}) \mathbf{\ddot{q}} + \mathbf{C} (\mathbf{q}, \mathbf{\dot{q}}) \mathbf{\dot{q}} + \mathbf{G} (\mathbf{q}) = \mathbf{Bu}$$

Where the above matrices are defined as:

The linearized dynamics are found by performing a Taylor expansion around the fixed point of interest:

With the linearized dynamics a stable LQR controller may be calculated using the cost function:

$$g(x,u) = (x^T \textbf{Q}x + u^T \textbf{R}u)$$

$$\sum^\infty_{k=0} \gamma^k g(x_k,u_k)$$

The optimal solution to this cost function may be found using techniques in dynamic programming and is of the form:

$$\mathbf{K}_{i + 1} = -(\mathbf{R} + \gamma\mathbf{B}^T \mathbf{P}_i \mathbf{B})^{-1} \mathbf{B}^T \mathbf{P}_i \mathbf{A}$$

$$\mathbf{P}_{i + 1} = \mathbf{Q} + \mathbf{K}^T_{i + 1} \mathbf{R} \mathbf{K}_{i + 1} + \gamma (\mathbf{A} + \mathbf{B} \mathbf{K})^T \mathbf{P}_i (\mathbf{A} + \mathbf{B} \mathbf{K}_{i + 1})$$

The following code may be used in MATLAB to converge on the optimal gain matrix:

For the given system, the optimal gain matrix can be found to be:

$$\mathbf{K} = \begin{bmatrix} 21.13 & -320.74 & 30.23 & -70.18 \end{bmatrix}$$

With the Qand R weights as follows:

$$\mathbf{Q} = \begin{bmatrix} 500 & 0 & 0 & 0 \\ 0 & 100 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}, R = 1$$

And the optimal control input can be calculated.

$$u(t) = \mathbf{K}x(t)$$

This policy can be empirically shown to balance the pendulum at its upright position:


text: File:Matrix def.jpg

title: File:Matrix def.jpg


text: File:Linear matrix.jpg

title: File:Linear matrix.jpg


title: File:Plot2.jpg


text:
File:Lqr code.jpg

Licensing
for school project

title: File:Lqr code.jpg


text:
User:Qianjiqianji

Using MATLAB to show the Lagrange Interpolation
The purpose of this page is to provide a simple example of the applicaiton of MATLAB in solving the problem. In the practical engineering, most of the time you can not obtain an exact equation for the problem. Then you have to use interpolation to approach the equation. Lagrange Method is the most popular way to do so. More points means more order of the Lagrange equations. Whether it will be closer to the exact equation with the increasing orders interests us. So here is an example to show that.

Given a function:
$$f(x)=$$$$1\over1+25x^2$$, the range of x is [-1,1]. Divide the range in the same distance and every point is : xi=-1+$$2i\over n$$, i=0,1,2,3,...,n The Lagrange polynomial expression is: Ln$$(x)$$=$$\sum_{i=0}^\infty \frac{1}{1+25x_i^2}$$ li(x) For $$l_i(x)$$, i=0,1,2,3,...,n is the basis of the Lagrange function. Pick up several points n=2.3...., graph the original function $$f(x)$$ and $$L_n(x)$$ in the range of [-1,1].

Solution:
The programe sentences are:

function y=lagrange(X,Y,x) n=length(X); m=length(x); for i=1:m z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n if j~=k p=p*(z-X(j))/(X(k)-X(j)); end end s=p*Y(k)+s; end y(i)=s;

1. when n=3,

x=-1:0.01:1;y1=1./(1.+25*x.^2); n=3;X=-1:2/(n-1):1;Y=1./(1.+25*X.^2);y=lagrange(X,Y,x); plot(x,y,'r.',x,y1,'g') gtext('y=1/(1+25*x^2)','fontsize',12) gtext('n=3','fontsize',12)

2. when n=10,

x=-1:0.01:1;y1=1./(1.+25*x.^2); n=10;X=-1:2/(n-1):1;Y=1./(1.+25*X.^2);y=lagrange(X,Y,x); plot(x,y,'r.',x,y1,'g') gtext('y=1/(1+25*x^2)','fontsize',12) gtext('n=10,'fontsize',12)

Run the MATLAB and we will obtain the graphs of the different conditions from which we can see that the lagrange equations is closer to the exact one with the increasing order.

title: User:Qianjiqianji


text:
User talk:Sam161998

Please try http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Science for questions about headaches upon awakening. Thank you. 99.56.137.239 04:16, 1 December 2009 (UTC)

title: User talk:Sam161998


text:
User:בינה תפארת

` ¡Shalom! Guten Tag. Lá maith. Bonjour. `

Favorites
School:Computer science `

`

Projects
http://en.wikiversity.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA/%D7%90 /b/2

http://meta.wikimedia.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wikinews.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://commons.wikimedia.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://species.wikimedia.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wikibooks.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wikiquote.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wikisource.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wiktionary.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

http://en.wikipedia.org/wiki/User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA

בינה תפארת 05:50, 1 December 2009 (UTC)

י

בינה תפארת 14:33, 1 December 2009 (UTC)

John A. Mac

בינה תפארת 07:59, 7 December 2009 (UTC)

title: User:בינה תפארת


text:
User:בינה תפארת/א

1d description string
flatland. in 3d. with time travel and alternate topology

/* story */

the common interests of the free peoples and wise despots of midgard have finally led to an end of the costly 'wars among equals'. the 'gaon lavan lubavitch of lodz' and the 'grey pilgrim' have established a strategic partnership, turning the vast armies of golems against the common enemy.

the horse-lords and the steward have pledged heavy infantry and light cavalry, and the king has offered scouts and light artillery support.

the seventeen diamonds and twenty-three emeralds have been found at last.

an intradimensional portal, located in a cave on the isen river, has been activated, but the reconnaisance team has yet to return.

the noble shimon hanasi, along with representatives of half-blood land-owners, has undertaken a dangerous and unpredictable project, which could possibly end the war.

the dark despot, along with his slaves and servants, is finally afraid.

equilateral3polyseg, grid square, elevation map
2:1 parallel-gram, icosahedron topology; square 4-angle;

` icw
http://en.wikipedia.org/wiki/Javelin

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

http://en.wikipedia.org/wiki/M1911_pistol

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

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

` ` apc/ifv
http://en.wikipedia.org/wiki/BMP-1

` ` mbt/ЗСУ
http://en.wikipedia.org/wiki/T-72

http://en.wikipedia.org/wiki/ZSU-23-4

` ` sph/mrl
http://en.wikipedia.org/wiki/2S3_Akatsiya

http://en.wikipedia.org/wiki/M270_Multiple_Launch_Rocket_System

` loh/ah1
http://en.wikipedia.org/wiki/OH-6_Cayuse

http://en.wikipedia.org/wiki/AH-1_SuperCobra

` f.w.a.;
http://en.wikipedia.org/wiki/Piper_J-3

http://en.wikipedia.org/wiki/Bell_X-1

http://en.wikipedia.org/wiki/Cessna_172

http://en.wikipedia.org/wiki/Hawker_Siddeley HS 748

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

http://en.wikipedia.org/wiki/Boeing_727

http://en.wikipedia.org/wiki/Boeing_737

http://en.wikipedia.org/wiki/F14_Tomcat

http://en.wikipedia.org/wiki/B-52_Stratofortress

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

` ffr
twin 86-round mrls, AH1-portable

3d space buildings
http://en.wikipedia.org/wiki/Proton_(rocket)

` urban geography
http://en.wikipedia.org/wiki/Architecture

` transport
http://www.orbiterwiki.org/wiki/Delta-glider

http://stargate.wikia.com/wiki/НИИ-88_Королёв

4d active remembrance
Yosef David Rachel Leah

fractal dimension

http://en.wikipedia.org/robots.txt

...

page history forest
rough draft

בינה תפארת 05:23, 1 December 2009 (UTC)

draft

בינה תפארת 10:09, 7 December 2009 (UTC)

good copy

בינה תפארת 10:14, 7 December 2009 (UTC)

page history trees
http://en.wikiversity.org/w/index.php?title=User:%D7%91%D7%99%D7%A0%D7%94_%D7%AA%D7%A4%D7%90%D7%A8%D7%AA/%D7%90&action=history

title: User:בינה תפארת/א


text:
User:Kennethyim

MBTI

title: User:Kennethyim


text:
User talk:PatrickE

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 1 December 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 1 December 2009 (UTC)

Your material
Thank you for giving permission to continue your work.

And thank you for mailing me all of your contribution drafts. There is a real holocaust going on at CP, and many people's work is in real danger of being deleted. Some of it already has been. I've been going around copying the contributions of the people who have given me permission to post here.

I can easily understand your having lost interest in this, after what happened to you, MarkGall, and KSorenson. But the introduction to general relativity article is excellent.

SamHB 00:24, 1 May 2010 (UTC)

title: User talk:PatrickE


text:
User talk:Coachpeople

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 1 December 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 1 December 2009 (UTC)

title: User talk:Coachpeople


text:
User talk:Drewjgray

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

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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:19, 1 December 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:19, 1 December 2009 (UTC)

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 * File:Lqr code.jpg

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

title: User talk:Drewjgray


text:
User:Oliverparish

This user has a page on the English wikipedia:

http://en.wikipedia.org/wiki/User:Oliverparish

title: User:Oliverparish


text:
User:בינה תפארת/Sefirot

In the esoteric teachings of the Jewish faith, there are ten numbers. There is also the concept of DA'AT.

title: User:בינה תפארת/Sefirot


text:
User:Stevenarntson/The Commons/surrealism

As we've been looking "out" at the Commons this quarter, especially when discussing artistic ideas and collage, so I'd like to end looking "in" at the ideas behind the original surrealism movement as it found itself in the 1920s.

 * Surrealism's relationship to the Enlightenment, and to Dadaism.
 * The first surrealist manifesto by Andre Breton. Shout out to Freud.
 * Games as related to the Surrealist enterprise: automatic writing and conditionals.
 * Un Chien Andalou dir. Bunuel.
 * Modern sleep research, especially with respect to the work of Stephen Laberge at Stanford.


text:
Life/What is life?

{| cellpadding="10" cellspacing="5" style="width: 900px; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #969696; border: 2px solid black; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

WHERE DOES LIFE COME FROM?

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: 9th grade:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Introduction to Biology
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: Cellular Biology
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 3-4 days
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: Imovie, Pikiwiki, Memory Miner
 * }
 * }

{| cellpadding="5" cellspacing="5" style="width: 90%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 60%; background-color: #e4e3c8; border: 2px solid #969696; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

Introduction/Specifications

 * style="width: 40%; background-color: #cdd6ee; border: 2px solid #969696; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 10px;" |

Ideas/Examples
{| cellpadding="5" cellspacing="5" style="width: 90%; background-color: inherit; margin-left: auto; margin-right: auto"
 * }
 * style="width: 40%; background-color: #e4c8cd; border: 2px solid #969696; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

Software

 * style="width: 60%; background-color: #c9e4c8; border: 2px solid #969696; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 10px;" |

Articles that Support this Lesson

 * }

title: Life/What is life?


text:
User:Reginaldnickerson

caddilac jones

title: User:Reginaldnickerson


text:
User:Brianne Martinez/WHAT WOULD YOU DO?

{| cellpadding="10" cellspacing="5" style="width: 900px; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #969696; border: 2px solid black; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

WHAT WOULD YOU DO?

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: 9th grade:
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: ADVISORY COURSE
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: FACING DILEMMAS
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 3-4 DAYS
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: Imovie, Pikiwiki, Memory Miner
 * }
 * }

{| cellpadding="5" cellspacing="5" style="width: 90%; background-color: inherit; margin-left: auto; margin-right: auto"
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Articles that Support this Lesson

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text:
User:Egm6321.f09.team4/HW7

= Homework #7: Team 4 =

Problem #1:
Lecture (37-1):

 * Given:

 * Non-polynomial homogenous solution to Legendre equation for $$n=1$$:

 * $$ \begin{align}

Q_{1}\left(x\right) = \frac{1}{2}x\ln \left( \frac{1+x}{1-x} \right) - 1 \end{align} $$

 * Find :

 * Show that this definition is equivalent to:

 * $$ \begin{align}

Q_{1}\left(x\right) = x \tanh^{-1}\left(x\right)-1 \end{align} $$

 * Solution :

 * First, remember the solution to Problem #13 in HW 6:

 * Start with the following statement:

 * $$ \begin{align}

\tanh\left( \ \tanh^{-1}\left(x\right)\ \right) &= x \\ \tanh\left(\theta\right) &= x, \quad \text{(where } \theta = \tanh^{-1}\left(x\right) \text{)} \end{align} $$

 * Insert $$\text{ } \theta$$ into the definition of $$tanh(x)$$:

 * $$ \begin{align}

\left(\frac{\exp\left(\theta\right)-\exp\left(-\theta\right)}{\exp\left(\theta\right) + \exp\left(-\theta\right)}\right) &= x \\ \exp\left(\theta\right)-\exp\left(-\theta\right) &= x\exp\left(\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right) - x\exp\left(\theta\right) &= \exp\left(-\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right)\left(1-x\right) &= \exp\left(-\theta\right)\left(1+x\right)\\ \exp\left(2\theta\right) &= \frac{1+x}{1-x} \\ 2 \theta &= \ln\left(\frac{1+x}{1-x}\right) \\ \theta &= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$

 * Since $$\theta = \tanh^{-1}\left(x\right)$$:

 * $$ \begin{align}

\tanh^{-1}\left(x\right) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$

 * Then, plug this result into the second definition for $$Q_{1}$$:

 * $$ \begin{align}

Q_{1}\left(x\right) &= x \tanh^{-1}x-1 \\ &= x \left[ \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \right] - 1 \\ &= \frac{1}{2}x\ln \left( \frac{1+x}{1-x} \right) - 1 \qquad \text{Proof.} \end{align} $$

Contributors:

 * Egm6321.f09.team4.sls 18:12, 1 December 2009 (UTC)

Problem #2:
Lecture (37-1):

 * Given:

 * The general formula for non-polynomial, homogenous solutions to the Legendre equation:

 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ \frac{2n-2j+1}{\left(2n-j+1\right)j}P_{n-j}\left(x\right) \right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$

 * Find :

 * Whether $$Q_{n}$$ is even or odd, for any $$n$$. In other words, for which $$n$$ values are $$Q_{n}$$ odd, and for which $$n$$ are $$Q_{n}$$ even?

 * Solution :

 * From HW 6, Problem 9, we know that:

 * $$ \begin{align}

&P_{2k}\left(x\right) \quad \text{ is even.} \\ &P_{2k+1}\left(x\right) \quad \text{ is odd.} \end{align} $$

 * So, this solution will involve two cases: when $$n$$ is an even number (i.e., $$n=2k$$ for some pos. integer $$k$$), and when $$n$$ is an odd number (i.e., $$n=2k+1$$ for some pos. integer k).

 * Note: values that do not depend on $$x$$ will not affect the evenness or oddness of the result. So, to simplify the derivation, the equation for $$Q_{n}$$ will be rewritten as:

 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{n-j}\left(x\right) \right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$

 * Case 1 : $$\ n=2k$$ (even #)

 * $$ \begin{align}

Q_{2k}\left(x\right) &= P_{2k}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{2k-j}\left(x\right) \right] \end{align} $$

 * $$\begin{align} P_{2k}\left(x\right) \end{align}$$ is even, and $$\tanh^{-1}\left(x\right)$$ is known to be odd. Since the product of an odd and an even function is odd:

 * $$ \begin{align}

P_{2k}\left(x\right)\tanh^{-1}\left(x\right) \Rightarrow \left[\text{even}\right]\cdot\left[\text{odd}\right] \Rightarrow \left[\text{odd}\right] \end{align} $$

 * Inside the summation, the index ($$2k-j$$) must be an odd number, since the difference between an even number ($$2k$$) and an odd number ($$j$$) is always odd. Since ($$P_{n} \Rightarrow \left[odd\right]$$) for odd numbered $$n$$, and since the sum of odd numbers is itself odd, we get the following result:

 * $$ \begin{align}

-2\sum_{j=1,3,5\dots}^{J}\left\{ C\ P_{2k-j}\left(x\right)\right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left\{-2C\ \left[\text{odd}\right] \right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left[\text{odd}\right] \Rightarrow \left[\text{odd}\right] \end{align} $$

 * Since both terms are odd, the sum of the two terms must also be odd. Therefore, we can make the following conclusion:

 * $$ \begin{align}

Q_{2k}\left(x\right) \Rightarrow \left[\text{odd}\right], \text{ for } k\geq 0, k \in \mathbb{Z} \quad \text{(even numbered } n \text{)} \end{align} $$

 * Case 2 : $$\ n=2k+1$$ (odd #)

 * $$ \begin{align}

Q_{2k+1}\left(x\right) &= P_{2k+1}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{2k+1-j}\left(x\right) \right] \end{align} $$

 * Remember, $$\begin{align} P_{2k+1}\left(x\right) \Rightarrow \left[\text{odd}\right] \end{align}$$, and $$\tanh^{-1}\left(x\right) \Rightarrow \left[\text{odd}\right]$$. Since the product of two odd functions is even, the first term is even:

 * $$ \begin{align}

P_{2k+1}\left(x\right)\tanh^{-1}\left(x\right) \Rightarrow \left[\text{odd}\right]\cdot\left[\text{odd}\right] \Rightarrow \left[\text{even}\right] \end{align} $$

 * Inside the summation, the index ($$2k+1-j$$) must be an even number, since the difference between an odd number ($$2k+1$$) and another odd number ($$j$$) is always even. Since ($$P_{n} \Rightarrow \left[even\right]$$) for even numbered $$n$$, and since the sum of even numbers is itself even, we get the following result:

 * $$ \begin{align}

-2\sum_{j=1,3,5\dots}^{J}\left\{ C\ P_{2k+1-j}\left(x\right)\right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left\{-2C\ \left[\text{even}\right] \right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left[\text{even}\right] \Rightarrow \left[\text{even}\right] \end{align} $$

 * Since both terms are even, the sum of the two terms must also be even. Therefore, we can make the following conclusion:

 * $$ \begin{align}

Q_{2k+1}\left(x\right) \Rightarrow \left[\text{even}\right], \text{ for } k\geq 0, k \in \mathbb{Z} \quad \text{(odd numbered } n \text{)} \end{align} $$

 * In summary, we have shown that the polynomial solutions to the Legendre equation always have the opposite "oddness" to the non-polynomial solutions. In other words:

 * For even numbered $$n$$: $$P_{n}$$ is even, and $$Q_{n}$$ is odd.

 * For odd numbered $$n$$: $$P_{n}$$ is odd, and $$Q_{n}$$ is even.

Contributors:
 * Egm6321.f09.team4.sls 18:32, 2 December 2009 (UTC)

Egm6321.f09.Team4.meyerat 01:59, 8 December 2009 (UTC)

Problem #3:
Lecture (37-1):

 * Given:

 * Polynomial Legendre solutions (Legendre functions of the first kind) given as:

 * $$ \begin{align}

P_{n}\left(x\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1,3,\dots,\left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$

 * Non-Polynomial Legendre solutions (Legendre functions of the second kind) given as:

 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2\sum_{j=1,3,5,\dots}^{J}\left[ \frac{2n-2j+1}{\left(2n-j+1\right)j} P_{n-j}\left(x\right)\right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$

 * Find :

 Plot $$\left\{P_{0},P_{1},P_{2},P_{3},P_{4}\right\}$$. Plot $$\left\{Q_{0},Q_{1},Q_{2},Q_{3},Q_{4}\right\}$$.

 * Solution :

 * Graphs of the Legendre polynomials of the first kind $$P_{n}$$:

 * [[Image:egm6321.f09.sls.hw7.leg_P.png|450px|$$P_{n}$$]]

 * Graphs of the Legendre polynomials of the second kind $$Q_{n}$$:

 * [[Image:egm6321.f09.sls.hw7.leg_Q.png|450px|$$Q_{n}$$]]

Contributors:
 * Egm6321.f09.team4.sls 20:46, 1 December 2009 (UTC)

Problem #4:
Lecture (37-2):

 * Given:

 * Definition of the inner product of two functions on the interval $$\left[a,b\right]$$:

 * $$ \begin{align}

\left\langle g,h \right\rangle = \int_{a}^{b}g\left(x\right)h\left(x\right)\text{d}x \end{align} $$

 * Find :

 * The inner product between the $$n^{th}$$ polynomial and non-polynomial homogenous solution to the Legendre equation, on the interval $$\left[-1,1\right]$$:

 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = \int_{-1}^{1} P_{n}\left(x\right) Q_{n}\left(x\right) \text{d}x \end{align} $$

 * Solution :

 * From Problem #3, we know that $$P_{n}$$ and $$Q_{n}$$ always have opposite "oddness". Because the product of an odd function and an even function is always odd:

 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = \int_{-1}^{1} \left[\text{odd}\right] \text{d}x \end{align} $$

 * By symmetry, we know that any integral of an odd function over a finite integral centered on the origin must equal zero. Therefore:

 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = 0 \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:36, 2 December 2009 (UTC)

Problem #5:
Lecture (37-3):

 * Given:

 * The following portion of the proof of $$\left\langle P_{n},Q_{n}\right\rangle = 0$$, for $$n\neq m$$:

 * $$ \begin{align}

\alpha = \int_{-1}^{+1} L_{m}\left[ \left(1-x^{2}\right)L_{n}'\right]'\text{d}x \end{align} $$

 * and the definition of integration by parts:

 * $$ \begin{align}

\int_{a}^{b}f\left(x\right)g'\left(x\right)\text{d}x = \left[f\left(x\right)g\left(x\right)\right]_{a}^{b} - \int_{a}^{b}f'\left(x\right)g\left(x\right)\text{d}x \end{align} $$

 * Find :

 * Show that the following is true, by performing integration by parts:

 * $$ \begin{align}

\alpha = -\int_{-1}^{+1} \left(1-x^{2}\right)L_{n}'L_{m}'\text{d}x \end{align} $$

 * Solution :

 * First, make the following definitions:

 * $$ \begin{align}

f\left(x\right) &= L_{m} \\ g\left(x\right) &= \left(1-x^{2}\right)L_{n}' \\ a = -1, b = 1 \end{align} $$

 * Then, substitute these values into the integration by parts formula:

 * $$ \begin{align}

\int_{-1}^{1} L_{m}\left[\left(1-x^{2}\right)L_{n}'\right]'\text{d}x = \left[L_{m}\left(1-x^{2}\right)L_{n}'\right]_{-1}^{1} - \int_{-1}^{1} L_{m}'\left(1-x^{2}\right)L_{n}'\text{d}x \end{align} $$

 * Examining the first term, note that $$(1-x^{2})=0$$ for both -1 and 1. So, this term evaluates to zero: $$\left[L_{m}\left(1-x^{2}\right)L_{n}'\right]_{-1}^{1} = 0$$. Simplifying:

 * $$ \begin{align}

\int_{-1}^{1}L_{m}\left[\left(1-x^{2}\right)L_{n}'\right]'\text{d}x = \alpha = -\int_{-1}^{1} L_{m}'L_{n}'\left(1-x^{2}\right)\text{d}x \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:53, 2 December 2009 (UTC)

Problem #6:
Lecture (38-2):

 * Given:

 * Points $$P$$ and $$Q$$ in spherical coordinates, such that:

 * $$ \begin{align}

P &= \left(r_{p},\phi_{p},\theta_{p}\right) \\ Q &= \left(r_{q},\phi_{q},\theta_{q}\right) \end{align} $$

 * Let the cartesian coordinates of these points be given as:

 * $$ \begin{align}

P &= \left(x_{p1}, x_{p2}, x_{p3}\right) \\ Q &= \left(x_{q1}, x_{q2}, x_{q3}\right) \end{align} $$

 * The relationship between these two coordinate systems is given as:

 * $$ \begin{align}

x_{p1} &= r_{p}\cos\left(\theta_{p}\right)\cos\left(\phi_{p}\right) \\ x_{p2} &= r_{p}\cos\left(\theta_{p}\right)\sin\left(\phi_{p}\right) \\ x_{p3} &= r_{p}\sin\left(\theta_{p}\right) \\ \\ x_{q1} &= r_{q}\cos\left(\theta_{q}\right)\cos\left(\phi_{q}\right) \\ x_{q2} &= r_{q}\cos\left(\theta_{q}\right)\sin\left(\phi_{q}\right) \\ x_{q3} &= r_{q}\sin\left(\theta_{q}\right) \end{align} $$

 * Find :

 * Convert the following equation for $$\left(r_{pq}\right)^{2}$$ to spherical coordinates, by substituting in the relationships detailed above:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} = ||\vec{PQ}||^{2} = \sum_{i=1}^{3}\left(x_{qi}-x_{pi}\right)^{2} \end{align} $$

 * Solution :

 * Start off by inserting the definitions of $$x_{pi}\ \forall i$$ and $$x_{qi}\ \forall i$$ into the formula for $$\left(r_{pq}\right)^{2}$$:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} = &\left[r_{q}\cos\left(\theta_{q}\right)\cos\left(\phi_{q}\right) - r_{p}\cos\left(\theta_{p}\right)\cos\left(\phi_{p}\right)\right]^{2} \\ &+ \left[r_{q}\cos\left(\theta_{q}\right)\sin\left(\phi_{q}\right) - r_{p}\cos\left(\theta_{p}\right)\sin\left(\phi_{p}\right)\right]^{2} \\ &+ \left[ r_{q}\sin\left(\theta_{q}\right) - r_{p}\sin\left(\theta_{p}\right)\right]^{2} \end{align} $$

 * Expanding each squared term:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)\cos^{2}\left(\phi_{q}\right) - 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{p}\right)\cos\left(\phi_{q}\right) + r_p^{2}\cos^{2}\left(\theta_{p}\right)\cos^{2}\left(\phi_{p}\right) \\ &+ r_{q}^{2}\cos^{2}\left(\theta_{q}\right)\sin^{2}\left(\phi_{q}\right) - 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\sin\left(\phi_{p}\right)\sin\left(\phi_{q}\right) + r_p^{2}\cos^{2}\left(\theta_{p}\right)\sin^{2}\left(\phi_{p}\right) \\ &+ r_{q}^{2}\sin^{2}\left(\theta_{q}\right) -2r_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \end{align} $$

 * Then, rearranging the terms for simplification:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)\left[ \sin^{2}\left(\phi_{q}\right) + \cos^{2}\left(\phi_{q}\right)\right] + r_{p}^{2}\cos^{2}\left(\theta_{p}\right)\left[\sin^{2}\left(\phi_{p}\right)+\cos^{2}\left(\phi_{p}\right)\right]\\ &+ r_{q}^{2}\sin^{2}\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \\ &- 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right) \left[ \cos\left(\phi_{p}\right)\cos\left(\phi_{q}\right) + \sin\left(\phi_{p}\right)\sin\left(\phi_{q}\right) \right] \\ &- 2r_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) \end{align} $$

 * Using the trig. identities ($$\sin^{2}\left(x\right) + \cos^{2}\left(x\right)=1$$) and ($$\cos\left(x\right)\cos\left(y\right)+\sin\left(x\right)\sin\left(y\right)=\cos\left(x-y\right)$$), we can further simplify the equation:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)+r_{p}^{2}\cos^{2}\left(\theta_{p}\right)+r_{q}^{2}\sin^{2}\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \\ &- 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{q}-\phi_{p}\right) \\ &- 2r_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) \end{align} $$

 * Performing another regrouping allows further simplification using the first trig identity:

 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{p}^{2}\left[\sin^{2}\left(\theta_{p}\right) + \cos^{2}\left(\theta_{p}\right)\right] + r_{q}^{2} \left[\sin^{2}\left(\theta_{q}\right)+\cos^{2}\left(\theta_{q}\right)\right] \\ &- 2r_{p}r_{q}\left[\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right) \cos\left(\phi_{q}-\phi_{p}\right) + \sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right)\right] \end{align} $$

 * After performing the last simplification, the final form of the solution is reached.

 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ r_{p}^{2}+r_{q}^{2}-2r_{p}r_{q}\left[\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{q}-\phi_{p}\right) + \sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right)\right] \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 19:49, 2 December 2009 (UTC)

Problem #7:
Lecture (38-4):

 * Given:

 * The binomial theorem, generalized to real values of $$r$$ (as opposed to integers):

 * $$ \begin{align}

\left(a+b\right)^{r} &= \sum_{k=0}^{\infty}\left( \begin{array}{c} r \\ k \end{array} \right) a^{r-k}b^{k} \\ &\text{ where } \left( \begin{array}{c} r\\ k \end{array} \right) = \frac{r\left(r-1\right)\dots\left(r-k+1\right)}{k!} \end{align} $$

 * Find :

 * Show that the following equation is true:

 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\alpha_{i}x^{i} \\ &\text{ where } \alpha_{i} = \frac{1 \cdot 3 \cdot \ldots \cdot \left(2i-1\right)}{2 \cdot 4 \cdot \ldots \cdot \left(2i\right)} \end{align} $$

 * Solution :

 * Define the following values:

 * $$ \begin{align}

&a = 1 \\ &b = -x \\ &r = -\frac{1}{2} \\ &k = i \end{align} $$

 * Substitute these values into the binomial theorem:

 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\frac{\left(-0.5\right)\left(-0.5-1\right)\cdots \left(-0.5-i+1\right)}{i!} \left(1\right)\left(-1\right)^{i}\left(x\right)^{i} \\ &= \sum_{i=0}^{\infty}\frac{\left(0.5\right)\left(1+0.5\right)\left(2+0.5\right)\cdots\left(i-1+0.5\right)}{\left(1\right)\left(2\right)\left(3\right)\cdots\left(i\right)}x^{i} \end{align} $$

 * Now, multiply both the top and the bottom of the fraction by ($$2^{i}$$):

 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\frac{\left(1\right)\left(2+1\right)\left(4+1\right) \cdots \left(2i-1\right)}{\left(2\right)\left(4\right)\cdots\left(2i\right)}x^{i} \\ &= \sum_{i=0}^{\infty}\frac{\left(1\right)\left(3\right)\left(5\right)\cdots\left(2i-1\right)}{\left(2\right)\left(4\right)\cdots\left(2i\right)}x^{i} \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 20:24, 2 December 2009 (UTC)

Egm6321.f09.Team4.meyerat 02:07, 8 December 2009 (UTC)

Problem #8:
Lecture (39-2):

 * Given:

 * The following result from the generalized binomial theorem (see last problem):

 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\alpha_{i}x^{i} \\ &\text{ where } \alpha_{i} = \frac{1 \cdot 3 \cdot \ldots \cdot \left(2i-1\right)}{2 \cdot 4 \cdot \ldots \cdot \left(2i\right)} \end{align} $$

 * The primary definitions for the Legendre polynomials given in Problem #5 of HW 6:

 * $$ \begin{align}

P_{n}\left(\mu\right) &= \sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ \mu^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} \\ &\text{where } \left\lfloor n/2\right\rfloor = \ \text{floor of } n/2 \end{align} $$

 * OR

 * $$ \begin{align}

P_{n}\left(\mu\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ \mu^{n-2i} \left(-1\right)^{i}\ \frac{1,3,\dots,\left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$

 * The following alternative definition for the Legendre polynomials $$P_{n}\left(\mu\right)$$:

 * $$ \begin{align}

\left(1 - 2\mu\rho + \rho^{2}\right)^{-\frac{1}{2}} = \sum_{n=0}^{\infty}P_{n}\left(\mu\right)\rho^{n} \end{align} $$

 * Find :

 * Use the alternate definition of the Legendre polynomials to find $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$. Compare these results to those obtained from one of the primary definitions.

 * Solution :

 * Using the primary definition, we have the following values for $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$:

 * $$ \begin{align}

P_{3}\left(\mu\right) &= \frac{1}{2}\left(5\mu^{3}-3\mu\right) \\ P_{4}\left(\mu\right) &= \frac{1}{8}\left(35\mu^{4}-30\mu^{2}+3\right) \\ P_{5}\left(\mu\right) &= \frac{1}{8}\left(63\mu^{5}-70\mu^{3}+15\mu\right) \end{align} $$

 * Continuing the power series developed in class, we obtain:

 * $$\begin{align}

\frac{1}{\sqrt{1-2\mu p+p^2}}=\alpha_0+\alpha_1(2\mu p-p^2)+\alpha_2(2\mu p-p^2)^2+\alpha_3(2\mu p-p^2)^3 \\ +\alpha_4(2\mu p-p^2)^4+\alpha_5(2\mu p-p^2)^5+... \end{align} $$

\begin{align} =\alpha_0+\alpha_1(2\mu p-p^2)+\alpha_2(4\mu^2p^2-4\mu p^3+p^4)+\alpha_3(8\mu^3p^3-12\mu^2p^4+6\mu p^5-p^6) +\alpha_4(16\mu^4p^4-32\mu^3p^5+24\mu^2p^6-8\mu p^7+p^8)\\ +\alpha_5(32\mu^5p^5-80\mu^4p^6+80\mu^3p^7-40\mu^2p^8+10\mu p^9-p^{10})+... \end{align} $$

 * The terms can be grouped according to $$p^n$$ as:

\begin{align} \frac{1}{\sqrt{1-2\mu p+p^2}}=\alpha_0p^0+(2\mu \alpha_1)p^1+(4\mu^2 \alpha_2 -\alpha_1)p^2+(8\mu^3 \alpha_3-4\mu\alpha_2)p^3+(16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2)p^4\\ +(32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3)p^5+... \end{align} $$

 * Where,

 * $$P_0\left(\mu\right)=\alpha_0$$

 * $$P_1\left(\mu\right)=2\mu \alpha_1$$

 * $$P_2\left(\mu\right)=4\mu^2 \alpha_2 -\alpha_1$$

 * $$P_3\left(\mu\right)=8\mu^3 \alpha_3-4\mu\alpha_2$$

 * $$P_4\left(\mu\right)=16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2$$

 * $$P_5\left(\mu\right)=32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3$$

 * The values for $$\alpha_i$$ can be determined from:

 * $$\alpha_i=\frac{1\cdot 3 \cdot(...)\cdot(2i-1)}{2\cdot4\cdot (...)\cdot(2i)}$$

 * As such, we obtain:

 * $$\alpha_0=1$$

 * $$\alpha_1=\frac{1}{2}$$

 * $$\alpha_2=\frac{3}{8}$$

 * $$\alpha_3=\frac{5}{16}$$

 * $$\alpha_4=\frac{35}{128}$$

 * $$\alpha_5=\frac{63}{256}$$

 * Substituting in yields:

 * $$P_0\left(\mu\right)=\alpha_0=1$$

 * $$P_1\left(\mu\right)=2\mu \alpha_1=\mu$$

 * $$P_2\left(\mu\right)=4\mu^2 \alpha_2 -\alpha_1=\frac{1}{2}(3\mu^2-1)$$

 * $$P_3\left(\mu\right)=8\mu^3 \alpha_3-4\mu\alpha_2=8\mu^3\left(\frac{5}{16}\right)-4\mu\left(\frac{3}{8}\right)=\frac{1}{2}(5\mu^3-3\mu)$$

 * $$P_4\left(\mu\right)=16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2=16\mu^4 \left ( \frac{35}{128} \right )-12\mu^2\left ( \frac{5}{16} \right )+\left ( \frac{3}{8} \right )=\frac{1}{8}(35\mu^4-30\mu^2+3)$$

 * $$\begin{align}

P_5\left(\mu\right)=32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3=32\mu^5\left ( \frac{63}{256} \right )-32\mu^3\left ( \frac{35}{128} \right )+6\mu\left ( \frac{5}{16} \right ) &= \\ \frac{63}{8}\mu^5-\frac{35}{4}\mu^3+\frac{15}{8}\mu &= \\ \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \end{align}$$

 * Therefore the alternative definition yields the same values for $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$

Contributors:

Egm6321.f09.team4.palubin.d 17:48, 6 December 2009 (UTC)

title: User:Egm6321.f09.team4/HW7


text:
Vertical Height and Horizontal Length and Slope

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Grade Level: 8th Grade
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 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Algebra I
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 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: An investigation between steepness and the slope of a line
 * }

 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 2 - 4 days
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 * {| cellpadding="4" cellspacing="0" width="350px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used:
 * YouTube
 * World's Largest Skateboard Ramp
 * Ski Jumping Video

Mathwarehouse (algebra)
 * Interactive software to explore slopes of lines
 * }
 * }

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Learning Goals
Lesson / Content Skills

Students will be able to recognize applications of slope and see the relationship between the dependent and independent variables. Specifically students will focus on slope within the context of a rate of change, where students will learn how to find the slope of a line, given the coordinates of two points on a line.

Specifically students:
 * 1) Investigate and make sense of the relationship between the vertical heights to the horizontal length.
 * 2) Find the slope of a line given the coordinates of two points on the line.
 * 3) Connect the learning's of slope to real life.

Mathematical Proficiency Skills

There are five strands of Mathematical Proficiency that are essential for students to develop math skills that can be extended across many facets. Students are exposed to the skills of all five strands which are listed below.

 * Conceptual Understanding: Comprehension of mathematical concepts, operations, and relations.
 * Procedural Fluency: Skill in carrying out procedures flexibly, accurately, efficiently, and appropriately.
 * Strategic Competence: Ability to formulate, represent, and solve mathematical problems.
 * Adaptive Reasoning: Capacity for logical thought, reflection, explanation, and justification.
 * Productive Disposition: Habitual inclination to see mathematics as sensible, useful, and worthwhile, coupled with a belief in diligence and one's own efficacy.

21st Century Skills

The buzzword in education today is 21st Century Skills. At a time when increasingly advanced skills are required for success in life and work, people of all ages are seeking a diverse range of learning experiences to inspire, guide, and enhance their personal and professional lives. This lesson provides students the opportunity to develop a subset of the 21st Century Skills.

Specifically students develop:
 * Critical Thinking and Problem Solving Skills
 * Communication and Collaboration Skills

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CA Content Standards
The lesson addresses California Algebra I standard 7.0 (for 8th - 12th grade) - "Students verify that a point lies on a line, given an equation of a line. Students are able to derive linear equations by using the point-slope formula."

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Lesson Flow
Prerequisite

It is assumed that students have prior knowledge in coordinate planes, including graphing and locating points and addition and subtraction of integers. Exposure to ratios and the relationship between the numerator and denominator and how it affects the magnitude is helpful, but not essential.

Introduction

The subject of slope is introduced using videos from youtube or a video of your choice. Two videos from youtube are provided. The first, lets the students think about slopes in the context of the world's largest skateboard ramp. The second, lets the students consider slopes viewing ski jumping. You can show one of these videos, both of them, or one of your choice, based on your class interests.

skate board ramp: http://www.youtube.com/watch?v=3_1Y8UoLlu4

ski jumping: http://www.youtube.com/watch?v=6ZeLbTYeNk4

A key connection between either of these videos and the topic of slope is steepness and how it relates to speed and distance. Allow students to consider different aspects of the ramp with minimal guidance from the teacher. Have the teacher inject comments and direction as needed. The objective is to promote the student's observations and thoughts to transition into the remainder of the lesson plan.

This introduction is intended to "wow" the students and stimulate their interest in learning about slopes, its relationship to steepness and the real world. It assumes the students have an interest in skate boarding or skiing or TBD which could be indirectly applied.

Instructional

The instruction of slope is divided into three parts.

The "first part" the teacher and the students engage in the idea of slope in a hands on manner. Students are provided the opportunity to play with cars and a ramp to assess the effect of the steepness on speed and distance.

This activity requires the students to divide into groups of 2-3. Each group is provided a piece of cardboard (or other fairly rigid flat material) of the same length between 12 - 20 inches, a matchbox car or other, a ruler, means to measure time and a sheet to record data. Students will have the opportunity to vary the height and assess its affect on the speed of the car and the distance it travels. Students record the speed and distance of the car at a measured vertical height and horizontal length. As the students explore the concept of slope, the teacher should wander the classroom and observe and probe the groups on observations they are seeing.

Points the teacher might want to incorporate into the discussion are:
 * 1) If speed is important, what is the relationship between the vertical height to the horizontal length?
 * 2) If the distance you travel is important, what is the relationship between the vertical height to the horizontal length?
 * 3) As the vertical height increases, what happens to the horizontal length?
 * 4) How might the coefficient of friction play a factor on speed and distance the car travels?
 * 5) What other outside factors might influence speed and distance?
 * 6) Where might we place the origin of a coordinate plane in relation to the line of the ramp in order to make calculations to find a slope?
 * 7) What real-world situations incorporate the concept of slope(s)? Examples might be, construction and architecture.

The "second part" of the instructional activity is to scaffold the students into the calculation of slope using coordinates and the coordinate plane. Students interactively learn about slopes using the computer. The program from math warehouse allows students to learn about slopes in relation to the coordinate plane, how to calculate slope, and manipulate points to visually watch the slope of a line change along with the slope value. At the completion of the discussion of slopes, the students will be able to practice calculating on the computer.

http://www.mathwarehouse.com/algebra/linear_equation/slope-of-a-line.php

http://www.mathwarehouse.com/algebra/linear_equation/interactive-slope.php

NOTE: If the teacher does not have access to a computer for every student, they can use one computer and lead the activity with the class, otherwise have the students learn about slopes with their partner(s) and the teacher assist students as needed.

The "third part" of the instructional activity connects part one and part two. Discuss the following prompts with the students:
 * 1) If speed is important, what is the relationship between the vertical height and the horizontal length?
 * 2) If the distance you travel is important, what is the relationship between the vertical hight to the horizontal length?
 * 3) How does the slope change when considering speed vs distance?
 * 4) As the vertical height increases, what happens to the horizontal length?
 * 5) Where might we place the origin of a coordinate plane in relation to the line of the ramp in order to make calculations to find a slope?

Close

At the end of the class, have the students complete three practice problems about slope and write a reflection based on provided prompts. After the students finish, have students share responses with the class for further discussion and clarification. The students work becomes an exit slip for the teacher to use for student assessment.

Materials

The following materials will be used during the lesson:
 * 1) YouTube Video: World's Largest Skateboard Ramp, Ski Jumping
 * 2) TBD ramps, cars and rulers for group manipulative activity
 * 3) Means to measure time (watch, calculator, phone, wall clock...)
 * 4) Slope discussion using interactive software from math warehouse
 * 5) computer (at least one), projector
 * 6) Handouts for students

Assessment Strategy

This lesson can be assessed with class participation and exit slip.

Class Participation

There will be ample opportunity to assess the students during the following activities:
 * 1) Conceptual discussion of slope and steepness during the group manipulative activity and as a class (using suggested prompts or others of your choice)
 * 2) Extension of the manipulative activity to the group interactive computer activity and exercise.

Exit Slip

Here the teacher can assess each students' individual understanding. Students are asked questions pertaining to slope along with a written reflection based on prompts provided. This exit slip is a direct reflection of the main lesson goals.

Optional Assessment Strategy

Studies have shown that one of the most effective ways for students to learn and for teachers to assess if students understood the material covered is to have students explain the concepts to other students. "The best way to learn something is to have to teach it." In this assessment the students, working with their same groups during the lesson, generate a video explaining the concept of slope. It is up to the students to determine how they want to explain this concept. The students will show their videos to the class for feedback and then, with teacher assistance, place on the school website, you tube, teacher tube, or TBD location. The intent is to place the videos in a location such that students from the school and outside of the school can view when they need help in understanding slopes.

Students can use the technology of their choice to make their videos. The website for JingProject is listed below. This website can be downloaded and allows students to easily produce videos, capturing both sound and video.

http://www.jingproject.com

 * }

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Related Articles
Dewey, J. (1916). "Democracy and education: An introduction to the philosophy of education" (pp. 179 - 192). New York, NY: The Macmillan Company.

Dewey emphasizes the theoretical idea that "good habits of thinking" should be fostered in school. However, actual practice does not support this theory of developing students ability to think. Typical classroom environments support knowledge, which "develops the poison of conceit" rather than "further growth in the grace of intelligence." Intelligent learning promotes thinking, which uses and rewards the mind. Dewey defines the essential methods of thinking as:
 * 1) The student environment promotes a genuine experience, where they can develop an interest for "its own sake";
 * 2) Students engage in a genuine-problem that stimulates thought;
 * 3) Students are able to posses or are provided the information to work the problem;
 * 4) Students are able to develop and suggest solutions; and
 * 5) Students have the opportunity to test and validate their ideas, thus allowing meaning to become clear.

In this lesson plan students learning (thinking) progression of slopes evolves in the following manner:
 * 1) Wow students with skateboard video and the concept of steepness and speed (develop interest);
 * 2) Engage students by allowing them to explore steepness using cars and a ramp, hands on (problem to stimulate thought);
 * 3) Allow students to interactively learn about slopes on the computer (information to work problems);
 * 4) Students connect interactive with hands on activities (develop suggestions to solutions); and
 * 5) Students reflect on big ideas of the lesson (test & validate ideas).

Sigler, J.W., & Hiebert, J. (2009). The teaching gap: Best ideas from the world's teachers for improving education in the classroom (pp. 85 - 101). New York, NY: Free Press.

This article emphasizes the concept that teaching is a cultural activity, that evolves over long periods of time in ways that are consistent with the beliefs and assumptions that are part of the culture. A comparison is made between the culture in the United States versus Japan. The typical U.S. lesson follows the belief that the mathematics is a set of procedures that emphasizes students developing skills. In Japan, lessons follow the belief that mathematics is a set of relationships between concepts, facts, and procedures, that emphasize students seeing relationships between mathematical ideas. This lesson was developed with the Japanese culture in mind. The students are provided the opportunity to explore, make mistakes and reflect on the mathematical concept of slope. The students are not shown procedures and equations to calculate slopes, rather they explore the underlying meaning of slope and how two variables can deliver different outcomes.

Gee, J. (2008). Games for learning institute. Retrieved from: http://vimeo.com/4513412

This video highlights how video games can organize deep conceptual understanding, as well as a variety of 21st century skills, that are rarely offered in schools today. An important concept Gee talks about is "situated meaning". Typical classrooms use verbal (words thru words), rather than the desired situated (image, action, dialogue, words) which is more suited for problem solving. The most effective way for a student to learn is thru action and experience and then read the text (instructions). Providing students with equations without mental images of what they represent is an ineffective way of learning. It is preferable to provide the student with a situated meaning and then provide the equations that describe that situation. In this lesson, even though the initial work is hands on and not thru a video game, the students are provided the opportunity to play with the idea of slopes, speed and distance prior to learning the equation for slope. When students have the opportunity to interactively learn the equation for slope, they now have the mental model of how vertical / horizontal changes effect the slope and ultimately speed and distance.

Chi, M.T.H., De Leeuw, N., Chiu, M.-H., & Lavancher, C. Eliziting self-explanation improves understanding. Cognitive Science, 18(3), 439-477.

Chi talks about the importance of self-explanation in promoting student learning. The article discusses prompts that the students considered while reading about a subject matter. This philosophy was used in this lesson by having the teacher provide prompts for the students to consider at the start of the instructional activities, along with the middle and the end. It is the hopes that these prompts promote student thinking on a deeper level and not just memorizing a formula. In addition, these prompts should provide connection a theme throughout the lesson.

Barab, S., Thomas, M., Dodge, T., Carteaux, R., & Tuzun, H. (2005). Making learning fun: Quest Atlantis, a game without guns. "Educational Technology Research and Development", 53(1), 86-107.

Even though this lesson does not employ computer games, it does apply some of the key concepts that the authors note are important for student learning. First, this lesson uses more than one form of media that connects and provides meaning to students as they participate in the lesson. Students work in groups of 2-3, thus they are participating in social engagement that taps on their experiences and ideas as they progress thru the lesson. The authors derived their educational games from Vygotsky where children's play creates a zone of proximal development for the child. This lesson is designed for understanding using a participatory framework that stresses hands-on action and reflection as components central to the learning process. The authors emphasize that learners should be involved in doing domain-related activities, not simply receiving the results of someone else's activities as summarized in texts or heard in lectures. This lesson does not use a textbook or lecture. Students learn best when the learning process involves inquiry as opposed to the memorization of the facts and principals that were generated from someone else's inquiry. In this activity the students explore the relationship of vertical height and horizontal length in regards to speed, distance, and steepness, before they even learn the formula to calculate the slope.

Robinson, K. (2006) Schools kill creativity [TED talk]. Retrieved from http://www.ted.com/talks/lang/eng/ken_robinson_says_schools_kill_creativity.html

Robinson notes that education is supposed to take us into the future that we cannot grasp at this time. The best way to approach the unknown is thru creativity. He notes that students have extraordinary capacity for creativity and talent that we squash with the current education environment. The classroom should allow students to explore, take chances, and not be frightened of being wrong. In fact, that is when students can be the most creative. Schools are educating students out of their creativity, as emphasis is placed more on the right answer. Students are now afraid to be wrong. The only hope for the future is to adopt a new ecology on education, recognize the gift of human imagination. This lesson does not ask students to show their creativity by inventing new things in many different facets, however, it does allow students to inquire about the concept of slope and how it might apply to everyday life. The topic is math, but aspects of art, social science, and English can be incorporated by each student, depending on their interests and desires.

Answers to the three main questions of the course:

How does learning occur?

Leaning occurs thru the use of manipulatives, computer interactive tools, and the social engagement of inquiring prompts.

In what environment can technology promote learning?

Technology is being used in the classroom as the students engage in an interactive math program on slopes. The students can work alone or in groups and change two points ultimately changing the slope of the line. During this activity the students can visually see the slope of a line change and the slope formula calculate the slope at the same time.

What is the process by which technology enhances learning?

In this lesson students can develop a better understanding of what the slope equation represents. Before the students learn the equation, they develop a mental model of what slope represents and how vertical height and horizontal length change this value. This visual representation of slope is interactive by the students. Now as they calculate the slope of a line, they have a deeper understanding of the relationship between the x and y coordinates.

LINKS

Educational Technologies: http://en.wikiversity.org/wiki/Educational_Technology

Distance Education Lesson Plans: http://en.wikiversity.org/wiki/Distance_Education_Lesson_Plans

http://en.wikiversity.org/wiki/Vertical_Height_and_Horizontal_Length_and_Slope

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Appendix - Possible Exit Slip
Possible exit slip with prompts for teacher to use:

Find the slope of the line that passes through each pair of points.
 * (1,2), (4,3)
 * (7,2), (3,5)
 * 1) (-3,-2), (4,-2)

Prompts
 * 1) What happens to the slope of a line if your increase the vertical height and keep the horizontal length fixed?
 * 2) What happens to the slope of a line if you fix the vertical height and increase the horizontal length?
 * 3) Comparing the slope from question 1 versus the slope from question 2, which would be larger
 * 4) You want to design two skateboard ramps.  The first is for you and would allow you to go as fast as possible.  The second is for your friend who does not want to go fast.  Describe using the vocabulary that discussed in this lesson, what the two ramp designs might look like.  Provide and example of what each of the ramp's slopes might look like.

 * }


text:
User:John Bessa/Capital Structure/Notes

The contents of this page was simply inserted into the main text under Article Development, as the ideas developed in this page concluded the contextual setting phase.

=To Do=
 * Inserting the new ideas into their proper places in the Setting
 * Filling-in and refining the structure
 * Collecting citation material
 * Mumford
 * Durant
 * Spring
 * Wiki articles
 * Constructivists and Synergists
 * Evolutionists, both pro-Darwin and anti

This final supporting task will attempt to introduce annotated bibliographies as a form of mediated support, that can contain various interpretations of cited text. Annotations from cited material that have differing critical approaches will allow back-linking to other threads of research,as the differing approaches will link to the material they support. This allows for "conceptual complexity" in the purely hierarchical format of these Wikis. Multiple threads of research, of course, allow for protecting research freedom, and prevent the types of conflict that plagues WP, something we want to avoid here.

title: User:John Bessa/Capital Structure/Notes


text:
Talk:Introduction to Orthodox Judaism

looks good
good page בינה תפארת 19:22, 1 December 2009 (UTC)

title: Talk:Introduction to Orthodox Judaism


text:
User:Alphagirl74

AlphaGirl

title: User:Alphagirl74


text:
User talk:Alphagirl74

In search of adult learning statistics and ideas to enhance projects and training efforts at my company--Alphagirl74 19:29, 1 December 2009 (UTC)

title: User talk:Alphagirl74


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