text:
SPIR608 Political Simulations and Gaming/2011/Week 6

Friday 25th February

 * 'Week 6 Discussion of Serious Games, Vietnam 1955.

* Is the design of the game's mechanics (board, pieces, cards, etc.) fit for purpose? The game was comprised of just an introductory booklet and A4 sheets with victory points. There was no board or pieces. Perhaps it would have been easier to play if poker chips were used to keep track of the game score. A map of 1950s Vietnam would have helped. The lack of a rule book was a problem at some points in the game. Vietnam 1955 was more of a interactive simulation than a game. It seemed more like reality as the rules could be changed to fit the progress of the game. However, if we played it again, we could learn how to"game the game". It was unclear whether the way to win Vietnam 1955 was gaining more victory points or acheiving certain goals. Heading at top of A4 sheets with victory points made people think they needed to fufill the conditions at the top. If you wanted to simulate history more closely, people should be forced to play more in character.

* Is the game enjoyable and sociable to play? We loved it! Vietnam 1955 was even better than Comrade Koba.

* What techniques does the game use to model its chosen subject? Vietnam 1955 was uses the free kriegsspiel system. Perhaps the game is too free sometimes. Umpire was invigilating and interacting with players in order to manipulate the game in certain directions where necessary. It showed that Russell is very experienced at running from his work with the NHS. The umpire having so much poweer can sometimes seem to be unfair. But this does work as a technique to make the simulation more realistic. Vietnam 1955 works by using the living labour of an umpire rather than the dead labour of a designer "congealed" in a board game.

* How does the game combine abstraction and realism in its workings? The biggest problem was that choices do not have real-life consequences so that threatening a nuclear war was an acceptable risk in the game. It was realistic that we did not know what the other players' victory conditions were. Vietnam 1955 was historically realistic in how the different players divided into the two Cold War blocs during the game.

* How accurately does the game simulate the decision-making processes faced by the real-life protagonists of its chosen subject? Vietnam 1955 is a bit like Origins of World War II. However it did seem strange that the French Indo-China and Vietminh players could work together so easily. Victory points do ensure that there are areas where people must confront each other, and other areas where they need to co-operate. As a participatory event, Vietnam 1955 was a bit like Rousseau's theory of the festival.

* What political lessons can people learn by playing the game? Vietnam 1955 teaches its players how diplomacy works and the need to create alliances. You need to know how to protect your own interests and predict what other people are striving for. The umpire should have deducted victory points when players were breaking out of their roles.

* How would you improve the structure and mechanics of the game?
 * Include name badges and country flags in game equipment;
 * Have the players in separate rooms so they would only meet when they want to meet;
 * Vary victory conditions if game played again;
 * Losing victory points for breaking out of role.
 * Have a giant map of Vietnam to orientate the players geographically;
 * Use chips to keep track of gaining for losing victory points;
 * Make the room layout more specific to the game, such as having negotiating areas;
 * Have a player representing the world media who issues press releases and news bulletins to give the game more flavour.

Links

 * Photos of Week 6 game playing
 * Vietnam 1955 Scenario
 * Wargame Developments


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SPIR608 Political Simulations and Gaming/2011/Week 7

Friday 4th March

 * Week 7 Discussion:.

Links

 * Photos of Week 7 game playing.
 * "I can recommend this game to any Revolutionary looking for multiple strategies." Giant Fire Breathing Robot Review

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


text:
SPIR608 Political Simulations and Gaming/2011/Week 8

Friday 4th March

 * Week 8 Discussion of Martin Wallace, London; and Klaus Teuber, The Settlers of Catan.

Links

 * Photos of Week 8 game playing.


text: SPIR608 Political Simulations and Gaming/2011/Week 9


text:
SPIR608 Political Simulations and Gaming/2011/Week 10

Favourite game

 * Vietnam 1955: 5
 * Settlers of Catan: 3
 * Origins of World War II: 2
 * War on Terror: 2
 * Red Guard!: 1
 * 1776: 1
 * Liberté: 1

Links
Photos of Week 10 play-testing.


text:
SPIR608 Political Simulations and Gaming/2011/Week 11

Links
Photos of Week 11 play-testing.


text:
SPIR608 Political Simulations and Gaming/2011/Week 12

Links
Photos of Week 12 final assessment of Prototype Political Simulations.


text:
User:Egm6341.s11.team4.BM/HW2

Given
Given the remainder term after the first integration by parts upon a function

 * $$ \int_{x_{0}}^{x} (x-t) f^{(2)}(t) dt \!$$

Objectives
1.    Repeat the integration by parts method used for 3 more degrees of the Taylor Series Approximation

 * $$ \frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0}) +   \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   \!$$

2. Use the Intermediate Mean Value Theorem to express the remainder in terms of

 * $$ f^{(5)}(\xi) \!$$
 * $$ \xi \epsilon [x_{0},x] \!$$

3. Do integration by parts a fourth time on the equation to verify the Taylor Series remainder equations for (n+1) expansion with $$ R_{n+2}(x) $$.

4. Use the Intermediate Mean Value Theorem on the Remainder term to show

 * $$ R_{n+1}(x) = \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(\xi) \!$$

Solution
1. First we will start with the "remainder" term from the Taylor series expansion as derived in class lecture [].

 * $$ R_{1}=\int_{x_{0}}^{x}(x-t)f^{(2)}(t)dt \!$$

The same integration by parts method used in class will be used on this remainder to further expand the Taylor series.

 * $$\int v(t) u'(t)\, dt = v(t) u(t) - \int v'(t) u(t)\, dt\!$$

Here we will now define the parts in our integration by parts scheme.

 * $$ u^{\prime}=(x-t) \!$$

 * $$ v=f^{(2)}(t) \!$$

 * $$  u=(xt-\frac{t^{2}}{2})       \!$$

Now, substituting into our integration by parts equation.

 * $$  R_{1}=\left [ (xt-\frac{t^{2}}{2}) f^{(2)}(t)\right ]_{t=x_{0}}^{t=x}-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt       \!$$

Expanding out the term in brackets we obtain

 * $$ R_{1}=(\frac{1}{2}x^{2})f^{(2)}(x)-(xx_{0}-\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0}) -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt  \!$$

 * $$+ [ (\frac{1}{2}x^{2})f^{(2)}(x_{0})-(\frac{1}{2}x^{2})f^{(2)}(x_{0}) ]     \!$$

The second line in this equation sums to zero and is added so that we can rearrange the resulting equation into an integral term and a non-integral term.

 * $$     R_{1}=[ \frac{1}{2}x^{2}f^{(2)}(x)-\frac{1}{2}x^{2}f^{(2)}(x_{0}) ] +(\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0})  -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                     \!$$

The bracketed term can be re-expressed as an integral. Additionally, the term $$ (\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2}) \!$$ can be expressed as $$ \frac{1}{2}(x-x_{0})^{2} \!$$. This gives us the following.

 * $$   R_{1}= \int_{x_{0}}^{x}\frac{1}{2}x^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                                 \!$$

Now regrouping the integral terms under one integral we obtain the following.

 * $$     R_{1}=  \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})                       \!$$

This can be grouped back into our Taylor series expansion to give us the following.

 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+\int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t) $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

Our new remainder term, the term in the integral, will now be further expanded.

 * $$      R_{2} =   \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)                       \!$$

Our integration by parts terms are now the following.
 * $$          u'=  \frac{1}{2}(x-t)^{2}                  \!$$
 * $$             v=f^{(3)}(t)                  \!$$
 * $$             u=\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2}                 \!$$

Substituting into our integration by parts equation we obtain the following.

 * $$ R_{2} =   [(\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2})f^{(3)}(t)]_{t=x_{0}}^{t=x} - \int_{x_{0}}^{x}  (\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2} )  f^{(4)}(t) dt            \!$$

For clarity, the term in brackets will now be expressed as $$ \alpha \!$$. Expanding the terms in bracket we obtain the following.
 * $$  R_{2} =(\frac{1}{6}x^{3}) f^{(4)}(x)- (\frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2}) f^{(4)}(x_{0}) - \alpha             \!$$

 * $$ +(\frac{1}{6}x^{3}) f^{(4)}(x_{0}) - (\frac{1}{6}x^{3}) f^{(4)}(x_{0})                    \!$$

The second line of the equation adds to zero as before and is added to help rearrange the expressions into an integral later on.

Now we can group together the terms appropriately so that we can simplify the equation.

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt   + (\frac{1}{6}x^{3} - \frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2})f^{(3)}(x_{0}) - \alpha                 \!$$

Pulling out the fractional coefficients of the terms outside of the integrals.

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!}(x^{3}-3x_{0}^{2}x+3x_{0}x^{2}-x_{0}^{3})  f^{(3)}(x_{0}) -\alpha                           \!$$

We can further simplify the term outside of the integral to be

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0}) -\alpha                           \!$$

Now combining the terms under the integrals to eliminate our $$ \alpha \!$$ expression.

 * $$  R_{2} =  \int_{x_{0}}^{x} (\frac{1}{3!}x^{3}-\frac{1}{6}t^{3}-\frac{1}{2}t^{2}x+\frac{1}{2}tx^{2})f^{(4)} + \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})                           \!$$

Further simplifying, this becomes

 * $$ R_{2} =  \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                         \!$$

Substituting this back into our original equation we obtain the following.

 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                                    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

Now we can use the same method for the next term in the Taylor series.

 * $$       \int_{x_{0}}^{x} \frac{1}{3!}(x-t)^{3}f^{(4)}(t)dt                             \!$$
 * $$      u'=     \frac{1}{3!}(x-t)^{3}                   \!$$
 * $$       v= f^{(4)}(t)                       \!$$
 * $$       u=(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )                      \!$$

Now substituting into our integration by parts equation
 * $$   R_{3}= [(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} ) f^{(4)}(t)  ]_{x_{0}}^{x} -\int_{x_{0}}^{x}  (-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )  f^{(5)}(t)  dt                           \!$$

Expanding the bracket and substituting $$ \alpha \!$$ for the integral term, we obtain the following equation.

 * $$ R_{3}= (-\frac{x^{4}}{24}+\frac{1}{6}x^{4}-\frac{1}{4}x^{4}+\frac{1}{6}x^{4} )f^{(4)}(x)  -(-\frac{x_{0}^{4}}{24}+\frac{1}{6}x_{0}^{3}x-\frac{1}{4}x_{0}^{2}x^{2}+\frac{1}{6}x_{0}x^{3} )f^{(4)}(x_{0}) -\alpha                            \!$$
 * $$ +\frac{1}{24}x^{4}  f^{(4)}(x_{0})-    \frac{1}{24}x^{4}   f^{(4)}(x_{0})                         \!$$

The second line sums to zero and is added in order to re-arrange the terms into an integral.

Now we can express the equation as the following.

 * $$ R_{3}=   \int_{x_{0}}^{x}   \frac{1}{24}x^{4}f^{(5)}(t)dt -\alpha +[\frac{1}{24}x^{4} +\frac{x_{0}^{4}}{24}-\frac{1}{6}x_{0}^{3}x+\frac{1}{4}x_{0}^{2}x^{2}-\frac{1}{6}x_{0}x^{3}  ]             \!$$

Now, combining the integral term with our $$ \alpha \!$$ term and then simplifying the terms in brackets we can express the equation as follows

 * $$R_{3}=  \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  \!$$

Substituting this into our Taylor series polynomial expression, we obtain

 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})        \!$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ +  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  $$

$$ 2.
 * $$\displaystyle
 * style= |
 * }.
 * }.

First we state the Intermediate Mean Value Theorem (IMVT)
 * $$ \int_{a}^{b} w(t)f(t)dt = f(\xi) \int_{a}^{b} w(t)dt \!$$

This was previously proven in assignment 1 [] Now, we know the remainder for our equation is the following
 * $$ R_{4}=\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt \!$$

Now, if we identify w(t) and f(t) we can use the IMVT for our problem
 * $$ W(t)= \frac{1}{4!}(x-t) \!$$
 * $$ f(t) = f^{(5)}(t) \!$$

Since we are integrating from our constants of integration x and $$ x_{0} $$ we can use the IMVT, substituting our equations, to obtain the following.

 * {| style="width:70%" border="0" align="center"

$$\displaystyle \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt = f^{(5)}(\xi)\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4} dt    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

Where

$$ \xi \in [x_{0},x] \!$$

3.

Here, we must do integration by parts for a fourth time to verify equations 3 and 4 on lecture slide 3-3 for (n+1) expansion with $$ R_{n+2}(x) $$. In essence, this requires us to see if

 * $$ R_{n+2}(x) = \frac{1}{(n+1)!} \int_{x_{0}}^{x} (x-t)^{(n+1)} f^{n+2}(t)dt \!$$

Where, in our case, n=4.

We will start with the remainder we obtained in part 1.
 * $$ R_{5}(x)=\frac{1}{4!}\int_{x_{0}}^{x} (x-t)^{4} f^{(5)}(t)dt \!$$

Now identifying our integration by parts terms.

 * $$  u'=\frac{(x-t)^{4}}{4!}       \!$$
 * $$   v=f^{(5)}(t)      \!$$
 * $$    u=(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!})     \!$$

Substituting into our integration by parts equation.

 * $$ R_{5}(x)= [(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) (f^{(5)}(t) )]_{x_{0}}^{x} - \int_{x_{0}}^{x}  (\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) f^{(6)}(t) dt     \!$$

Now referring to the term to the integral term as $$ \alpha $$ as in the previous several iterations, expanding the bracketed term as before, and adding two terms that sum to zero in order to aid in re-arranging the equation

 * $$ R_{5}(x)=(\frac{x^{5}}{5!}-\frac{x^{5}}{4!}+\frac{x^{5}}{12}-\frac{x^{5}}{12}+\frac{x^{5}}{4!}) f^{(5)}(x)   - (\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}x^{4}}{4!}) f^{(5)}(x_{0})   -\alpha \!$$
 * $$ +\frac{x^{5}}{5!}f^{(5)}(x_{0}) - \frac{x^{5}}{5!}f^{(5)}(x_{0})        \!$$

The second line of the above equation sums to zero and is added so that we can rearrange the equation into a simpler form.

 * $$  R_{5}(x)= [\frac{x^{5}}{5!}f^{(5)}(x) -\frac{x^{5}}{5!}f^{(5)}(x_{0}) ]_{x_{0}}^{x} -(\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}^{4}x}{4!}-\frac{x^{5}}{5!}) f^{(5)}(x)  -\alpha    \!$$

Now we can express the bracketed term as an integral. Meanwhile, the complex polynomial simplifies to $$ \frac{(x-x_{0})^{5}}{5!} $$.
 * $$  R_{5}(x)= \int_{x_{0}}^{x}\frac{x^{5}}{5!}f^{(6)}(t)dt+\frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0})-\alpha      \!$$

Now, combining the integral term with the $$ \alpha $$ term we end up with the following equation

 * $$ R_{6}= \frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0}) +  \int_{x_{0}}^{x}\frac{(x-t)^{5}}{5!}f^{(6)}(t)dt     \!$$

Since n=4 in our case, this verifies the expansion with $$ R_{n+2}(x) $$

4.

We can use the IMVT on the Remainder term, given as equation 4 lecture slide 3-3  in order to show equation 5 lecture slide 3-3. The IMVT can be stated as follows.

 * $$  \int_{a}^{b}w(t)g(t)dt=g(\xi)\int_{a}^{b}w(t)dt       \!$$

When applied to equation 4, the Taylor series remainder, we can evaluate the individual terms of the IMVT to be
 * $$  a=x_{0}          \!$$
 * $$  b=x        \!$$
 * $$  w(t)= \frac{(x-t)^{n}}{n!}       \!$$
 * $$     g(t)=f^{(n+1)}(t)       \!$$

Simple substitution of these equations into the IMVT gives us the following

 * $$  \int_{x_{0}}^{x}\frac{(x-t)^{n}}{n!}f^{(n+1)}(t)dt=f^{(n+1)}(\xi)  \int_{x_{0}}^{x}  \frac{(x-t)^{n}}{n!} dt        \!$$

Now, evaluating the integral present on the right hand side, we can further simplify the remainder to the following
 * $$   R_{n+1}(x) = \frac{(x-x_{0})^{(n+1)}}{(n+1)!}f^{(n+1)}(\xi)      \!$$

And this result is simply equation 5, so we have effectively shown that equation 5 arises naturally out of equation 4 in conjunction with the Intermediate Mean Value Theorem.

Given
The function f(x)
 * $$ f(x)=sin(x) \!$$

and the span of x
 * $$ x \epsilon [0, \pi] \!$$

Objective
Evaluate the Taylor Series about the point
 * $$ x_{0}=\frac{3\pi}{8} \!$$

For Taylor Series degrees of
 * $$ n=0,1,2,...,10 \!$$

We must then plot these series for each value of n in order to visually see how the Taylor Series improves with each degree of approximation.

Next we must estimate the error in our solution at the point $$ x= \frac{3\pi}{4} \!$$ for each degree of Taylor Series approximation.

Solution
Evaluating the Taylor Series of the sin(x) is a simple matter of following the general Taylor Series expansion formula for the case where f(x) is sin(x).

The general Taylor Series expansion as presented in the lecture slides is

 * $$ P_{n}(x) = f(x_{0}) + \frac{(x-x_{0})}{1!} f^{(1)}(x_{0}) +...+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) \!$$

For the case of sin(x), the equation then becomes

 * $$ f(x) = sin(x_{0}) + \frac{(sin(x)-sin(x_{0}))^{1}}{1!}cos(x_{0}) - \frac{(sin(x)-sin(x_{0}))^{2}}{2!}sin(x_{0}) - ...\!$$

The periodicity of the sine function shows up in our evaluation of the Taylor Series and can be capitalized upon. Since the derivative of sine is the sine function itself every 4 derivatives evaluating the nth derivative is a simple matter of determining what the remainder is when n is divided by 4 and matching the derivative with the appropriate trigonometric function based on what the remainder is.

This method is implemented in code about the point $$ x_{0} = \frac{3\pi}{8}$$ up to degree n=10 and then plotted. The result is a plot that graphically shows how the Taylor Series comes closer to the true function it evaluates with each successive degree evaluated.

The maximum error of the Taylor Series can be expressed mathematically using equation 5 lecture slide 3-3
 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!}(\alpha ) \!$$

Where $$ \alpha \!$$ represents the maximum of the maximum value of the n+1 derivative over the relevant range.

 * $$ \alpha = max \left | f^{(n+1)}(t) \right | \!$$
 * $$ t \epsilon [x_{0},x] \!$$

Now we will apply these equations to our specific function; the sine function and an evaluation point of $$ x_{0} = \frac{3\pi}{8} $$ and measurement point of $$ x = \frac{3\pi}{4} $$. First, we can say that

 * $$ \alpha = max \left | f^{(n+1)}(t) \right | = 1 \!$$
 * $$ t \epsilon [x_{0},x] \!$$

This arises because of the nature of the sine function. Every nth order derivative of sine is going to be positive or negative sine or cosine. Sine and cosine functions are bounded above and below by 1 and -1. This means that the maximum absolute value of any derivative of sine is 1.

Now we can use this result to simplify our remainder equation.

 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!} \!$$

Since, for the problem we are evaluating $$ x_{0} = \frac{3\pi}{8} $$ and $$ x = \frac{3\pi}{4} $$ we can say that $$ (x-x_{0})=\frac{3\pi}{8} $$

 * $$ \left | R_{n+1}(\frac{3\pi}{4}) \right | \leq \frac{(\frac{3\pi}{8})^{n+1}}{(n+1)!} \!$$

Now we have an upper bound on the error for any degree Taylor Series approximation of our function. This equation is included in the code used to plot the sine function and evaluated for all degree Taylor Series approximations from 0 to 10.

Matlab Code:

The code as printed above gives us several outputs, the plot of the Taylor Series and the error for each degree of the Taylor series about the point $$ \frac{3\pi}{4} $$.

Plots:

As can be seen, the higher order Taylor Series functions do an excellent job of approximating the sine function. In fact, every approximation above the 5th degree approximation cannot even be seen in the plot since they coincide with and are obscured by the plotting of the true value of sin(x). Thus, we would expect to see a very small error associated with these higher order approximations. The code used above gives us the following error estimations.

 * $$ \left | R_{1}(\frac{3\pi}{4}) \right | \leq 1.1781 \!$$
 * $$ \left | R_{2}(\frac{3\pi}{4}) \right | \leq 0.6940 \!$$
 * $$ \left | R_{3}(\frac{3\pi}{4}) \right | \leq 0.2725 \!$$
 * $$ \left | R_{4}(\frac{3\pi}{4}) \right | \leq 0.0803 \!$$
 * $$ \left | R_{5}(\frac{3\pi}{4}) \right | \leq 0.0189 \!$$
 * $$ \left | R_{6}(\frac{3\pi}{4}) \right | \leq 0.0037 \!$$
 * $$ \left | R_{7}(\frac{3\pi}{4}) \right | \leq 6.2494*10^{-4} \!$$
 * $$ \left | R_{8}(\frac{3\pi}{4}) \right | \leq 9.2030*10^{-5} \!$$
 * $$ \left | R_{9}(\frac{3\pi}{4}) \right | \leq 1.2047*10^{-5} \!$$
 * $$ \left | R_{10}(\frac{3\pi}{4}) \right | \leq 1.4192*10^{-6} \!$$
 * $$ \left | R_{11}(\frac{3\pi}{4}) \right | \leq 1.5200*10^{-7} \!$$

As expected, the higher order approximations from about degree six onward produce an almost negligible error.

title: User:Egm6341.s11.team4.BM/HW2


text: File:HW2.9 1.jpg

title: File:HW2.9 1.jpg


text:
User:Eml5526.s11.team7.jin/HW2

=Problem 2.3=

Given
Use $$\left\{ \right\} $$as basis, let :{| style="width:100%" border="0" $$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{\mathbf{w}} = \sum\limits_i {{\mathbf{a}}_i}

$$     (Eq 3.1) Then
 *
 * }
 * {| style="width:100%" border="0"

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

\begin{array}{ccccccccccccccc} {{\mathbf{w}} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall {\mathbf{w}} \in {\mathbb{R}^n}} \end{array}

$$     (Eq 3.2) is equal to
 *
 * }
 * {| style="width:100%" border="0"

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

\begin{array}{ccccccccccccccc} {{\mathbf{w}} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall \left\{ {{\beta _1}, \ldots ,{\beta _n}} \right\} \in {\mathbb{R}^n}} \end{array}

$$     (Eq 3.3) Such that $${\mathbf{w}} = \sum\limits_i $$
 *
 * }

Find
Show Equation 3 is equivalent to
 * {| style="width:100%" border="0"

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

\begin{array}{ccccccccccccccc} {{{\mathbf{a}}_i} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0}&{\forall i = 1, \ldots ,n} \end{array}

$$     (Eq 3.4)
 *
 * }

Solution
Since$$ \left\{ {{\beta _1}, \ldots ,{\beta _n}} \right\} $$are arbitrary, let
 * {| style="width:100%" border="0"

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

{\beta _1} = 1,{\beta _2} = {\beta _3} = \cdots  = {\beta _n} = 0

$$     (Eq 3.5) Then
 *
 * }
 * {| style="width:100%" border="0"

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

{\mathbf{w}} = {{\mathbf{a}}_1}

$$     (Eq 3.6) Equation 3 can be written as
 *
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{a}}_1} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0

$$     (Eq 3.7) Similarly, let
 *
 * }
 * {| style="width:100%" border="0"

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

{\beta _1} = {\beta _2} = \cdots  = {\beta _{n - 1}} = 0,{\beta _n} = 1

$$     (Eq 3.8) Then
 *
 * }
 * {| style="width:100%" border="0"

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

{\mathbf{w}} = {{\mathbf{a}}_n}

$$     (Eq 3.9) Equation 3 can be written as
 *
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{a}}_n} \cdot {\mathbf{P}}\left( {\mathbf{v}} \right) = 0

$$     (Eq 3.10) Thus, Equation 3 is identical to Equation 4.
 *
 * }

title: User:Eml5526.s11.team7.jin/HW2


text:
Wikimedia experience

Please help extend this page!

SPIR608 Evaluation
SPIR608 is a course being run at the University of Westminster. The first cohort will being doing the course Jan- April 2011. At the beginning of the course a survey has been carried out amongst the students as regards how much the use Wikimedia sites. The results are quite interesting. From a survey of 17
 * All had visited Wikipedia
 * 3 (18%) had edited wikipedia. None had a user account. None had edited any of the other wikis.
 * 4 (24%) had visited wikipedia, 2 (12%) had visited Wiktionary, Wikiquotes, Wikibooks or Wiki Commons, 1 (6%) had visited wikispecies or another mediawiki, and none had visited wikinews or wikisource.

I would be interested to know if any other people on wikiversity are gathering this sort of information? I would be interested in developing a way of collecting shared date (or sharing collected date!). Also there are probably ways in which the evaluation and guidance used on this occasion can be improved. I intend to re-survey the students at the end of the course to see to what extenttheir use of wikimedia sites has changed.Leutha 17:57, 29 January 2011 (UTC)
 * see also more at Colloquium

title: Wikimedia experience


text:
University of Canberra/RCC2010

RecentChangesCamp2010 at the University of Canberra (UCRCC) was an unconference focused on wikis and online collaborative practices. UCRCC was organised by University of Canberra staff in the Faculty of Health, following Recent Changes Camps held in other parts of the world. The main characteristics of the event were that it was free to attend, open to anyone, with a program determined on the day by the participants. The aim of the event was to draw together people with interest in wikis and online collaborative practices, to discuss, question and share knowledge in a friendly face to face setting.

Sessions
The sessions below were determined on the morning by attendees at the 10 - 11:30 Welcome and agenda building.

Welcome The human timeline activity worked quite well. Everyone coordinated with each other to arrange themselves in a standing order based on how long people had been editing wikis. Once that was done, each person talked briefly about their first wiki experience. From that we started to get a bit of an idea of what sessions whould be proposed...

Agenda building After the human time line, we created a list of possible sessions to discuss. People who suggested a session were expected to lead the session. Then everyone placed their names next to the sessions they wanted to attend.

Notes on sessions
 * Wikiversity
 * Wiki community and usability
 * Wikis for applied learning
 * Conversations behind wikis
 * Wikis for Community Consultation

Recordings

 * Twitter stream
 * Flickr photos
 * commons:Category:RecentChangesCamp 2010, Canberra - Media on Commons
 * Video doco on Youtube

What people have said

 * Yesterday was a great event - the dialogue was thought provoking, interesting and inspiring. I'm looking forward to sharing it with my colleagues and the documentation from everyone really helps with that. Kirsty Sharp. Resource Development Manager Tasmanian Polytechnic

 * It was great to catch up with old friends and meet new ones. I'd like to see the VET, University, not-for-profit and other sectors do this sort of thing more often. There's so much to learn with and from each other. And of course, the unconference context means that we leave our occupational/professional hats at the door and collaborate, argue and opine freely and openly. A great learning environment and a worthwhile day. Thanks Leigh for organising it. Rose Grozdanic, Australian Flexible Learning Framework

 * A tremendously edifying experience. It's a pity it wasn't a two day event. I especially enjoyed the 'real world' tales of planning and integration that the unconference participants brought with them and shared with the rest of us. The event re-shaped my thinking about open content and its place in formal education Peter Shanks, Skills Tasmania

 * The day captured such wide-ranging discussion from so many viewpoints. Very refreshing! Great to see so many newcomers not only to the format but to the discussion of wikis. I'd love to see more events of this kind permeate the often 'dry' walls of our educational sites... Why not a student based forum of this kind too? Marg O'Connell, Canberra Institute of Technology

 * Having people from the community come to the University of Canberra is a very good thing. In many ways, here at this University, we're developing a key group of people who are very innovative in their approach to teaching and open education, and I think on a National scale we're really starting to lead the pack in regards to some of the approaches to teaching... a real kick start for UC becoming a leader in this area. Well done... Michael de Percy. Lecturer in Government-Business Relations, UC

 * As newcomers to the Wiki sphere, we found the unconference very helpful and informative - it was fantastic to talk with people using this platform in their work and opened our eyes to the current and future possibilities of Wiki. It's great to see Wikis being used as educational tools for social and collaborative learning in universities. We hope one day they will be as prolific in the health sector. Thank you very much for hosting such a valuable workshop. Christine Vuletich and Alice Winter-Irving, Cancer Council Australia

 * A great opportunity to meet all of those wiki people and to understand some of the challenges that they're facing. A lot of the people here are from the educational realm, and a lot of their challenges are the same challenges they we're facing at Wikimedia Projects... its been a really great experience. Andrew Garrett, Wikimedia Foundation

 * Past events
 * 2010, Montreal, Quebec
 * 2009, Portland, Oregon
 * Pictures on Flickr tagged rcc09

 * 2008, Palo Alto, California
 * Wikimedia Commons category of images

 * 2007, Montréal, Quebec (aka RoCoCo)
 * Video from RoCoCo

 * 2007, Portland, Oregon
 * 2006, Portland, Oregon
 * http://recentchangescamp.org/

RCC events bring together people for a common purpose: discussing the past, present, and future of the technology and collaborative method that is wiki. RCC is a chance for everyone in the wiki community, to meet and have a fun, and have productive conversation about our passion for wikis of all stripes.
 * What is Recent Changes Camp, anyway?

At RCC, everybody is welcomed. You don't need to be an expert on anything, and you certainly don't need to consider yourself a geek. Collaboration thrives on diversity! All you need to bring is an open mind, and a willingness to participate, whether by teaching or by taking an active role in discussions. And, don't forget, an unconference is what we make it, so let's make it enlightening and fun.

 * Sessions covering an array of interests

 * Cross-Cultural Collaboration (RoCoCo, 2007 Montréal)
 * Throwing and Catching Swords - Hypersensory Awareness for Better Community (2007 Portland)
 * The Evolutionary Edge of Software Development (2006 Portland)
 * Using wikis for government processes (2008 Palo Alto)
 * Wine tasting (2008 Palo Alto)
 * Nonlinear conversations from the Glass Plate Game (2006 Portland)
 * Who edits Wikipedia? By Ed Chi of PARC's Augmented Social Cognition research group

 * Remember that the schedule will be set during the opening circle.

title: University of Canberra/RCC2010


text: University of Canberra/RCC/Navigation

title: University of Canberra/RCC/Navigation


text: User:Taichitanian

title: User:Taichitanian


text:
University of Canberra/RCC2012

20, 21, 22 January 2012 - INSPIRE Centre, University of Canberra

RecentChangesCamp2012 is an OpenSpace unconference focused on wikis and online collaborative practices. It has been held in Canberra twice so far, and held several times in other parts of the world. The main characteristics of the event are that it is free to attend, open to anyone, with a program determined on the day by those who take part. The aim of these events are to draw together people interested in worldwide iterative knowledge involvement or wikis, to discuss and share knowledge, and eat and socialise in a friendly face to face setting.

Planning
We're in the planning stages of RecentChangesCamp 2012! If you can help, please comment on the talk page! So far the dates and venue are confirmed. 20, 21, 22 of January at the INSPIRE Centre at the University of Canberra.

Sponsorship
We are looking for sponsors. If you are interested in sponsoring RecentChangesCamp, you'd get the following:
 * Your logo on any promotional material we use offline,
 * Your logo and a link to your site on RecentChangesCamp's page,
 * Table and wall space to put promotional materials on at RecentChangesCamp, and
 * A thank you and mention of your organisation during opening and closing circle at RecentChangesCamp.

Current sponsors: The University of Canberra who are providing the space, and Wikimedia Australia, who are providing travel grants for Australians and Kiwis who want to be able to attend but could not otherwise afford to travel. See wmau:Meeting:Committee_(2011-10-16)

Pre-camp dinner, Thursday 19 Jan
Join us for dinner at an Ethiopian restaurant in Dickson (not far from Civic), 7pm the night before RCC starts.

Details are at the Thursday dinner page. Please add your name to that page if you're coming.

Daily schedule
Turn up at 9am on Friday to mingle (and hopefully we'll have tea and coffee to mingle over). We'll have the first day's opening circle at 10am. Details for Saturday's start will be worked out on Friday.

 * Fri & Sat: 9am-5pm.
 * Sun: 10am-2pm (approx). We'll either arrange lunch at the venue, or go out for lunch.

Session schedule
Being a BarCamp style event, the sessions depend on all of us, as participants. We start each day with an opening circle where we begin to set the topics for the day. There are no set topics beforehand (though there are some session ideas).

Images

 * [Wikimedia Commons Category:RecentChangesCamp_2012,_Canberra]

Outcomes

 * NEXT

 * Past events
 * 2011, Canberra, Australia
 * 2010, Canberra, Australia
 * 2010, Montreal, Quebec
 * 2009, Portland, Oregon
 * Pictures on Flickr tagged rcc09

 * 2008, Palo Alto, California
 * Wikimedia Commons category of images

 * 2007, Montréal, Quebec (aka RoCoCo)
 * Video from RoCoCo

 * 2007, Portland, Oregon
 * 2006, Portland, Oregon
 * http://recentchangescamp.org/

RecentChangesCamp events bring together people for a common purpose: discussing the past, present, and future of the technology and collaborative method that is wiki. RCC is a chance for everyone in the wiki community, to meet and have a fun, and have productive conversation about our passion for wikis of all stripes.
 * What is Recent Changes Camp, anyway?

At RCC, everybody is welcomed. You don't need to be an expert on anything, and you certainly don't need to consider yourself a geek. Collaboration thrives on diversity! All you need to bring is an open mind, and a willingness to participate, whether by teaching or by taking an active role in discussions. And, don't forget, an unconference is what we make it, so let's make it enlightening and fun.

 * Sessions covering an array of interests

 * Cross-Cultural Collaboration (RoCoCo, 2007 Montréal)
 * Throwing and Catching Swords - Hypersensory Awareness for Better Community (2007 Portland)
 * The Evolutionary Edge of Software Development (2006 Portland)
 * Using wikis for government processes (2008 Palo Alto)
 * Wine tasting (2008 Palo Alto)
 * Nonlinear conversations from the Glass Plate Game (2006 Portland)
 * Who edits Wikipedia? By Ed Chi of PARC's Augmented Social Cognition research group

 * Remember that the schedule will be set during the opening circle.

title: University of Canberra/RCC2012


text:
University of Canberra/RCC

The University of Canberra Recent Changes Camps are part of our outreach to the wiki community, to meet and learn about wiki work, and open, networked practices generally. We plan to host RCC at least once a year from now on. Please see the year links in the above navigation template for documentation of past events.

The last RCC Canberra was held on 20–22 January 2012.

Follow the work people are doing here

 * Click this link for Recent Changes to the University of Canberra RecetChangesCamp pages.
 * Those using RSS to track changes can subscribe here.

The pages with the most recent changes are listed below:

category=University of Canberra/RCC order=descending ordermethod=lastedit mode=unordered shownamespace=true

title: University of Canberra/RCC


text:
Talk:University of Canberra/RCC2012

Who can help organise and run the 2012 RCC at UC?

 * Leighblackall (things are a bit up in the air with me now, so I'll be a bit more unreliable than usual this time, sorry)
 * User:LauraHale
 * Chriswaterguy - happy to be an "unorganiser" in some way, but don't want to take responsibility for logistics or time management. 09:34, 18 October 2011 (UTC)
 * User:Steven Zhang Happy to chip in as much as possible. Steven Zhang 07:45, 16 December 2011 (UTC)
 * Anne Frazer w:en:User:Victorbyron

Nominated dates
20,21,22 January 2012

Food
The yummy chicken place we used last year has closed shop, but is still catering and can deliver. Contact Lucy on 0434056833 Leighblackall 12:09, 3 December 2011 (UTC)

Internet
Wifi was very difficult for most participants in 2011, and many couldn't connect to the service offered. Can this be improved?

Or should participants preferably have a Telstra 3G connection if they want internet? (Note that Vodafone coverage is poor at UC.) --Chriswaterguy 09:43, 18 October 2011 (UTC)

 * We have assurances that the WiFi will be sorted this time.. but I'd bring a 3G connection for back up. Signal should be better also. Leighblackall 12:06, 3 December 2011 (UTC)

Sponsorship and money

 * Need $2000 for catering (Hopefully UC)
 * Need $10000 for travel grants (Likely WMAu)
 * Venue is provided by UC at the http://inspire.edu.au

Getting the word out
Ideas for potential attendees, especially from Australia. Who's missing?

Please add to that page and note if you've made contact already.

Also, please share on FB, G+, Twitter, etc - and ask relevant people if they've heard about it.

All three days

 * Leighblackall 01:35, 31 January 2011 (UTC)
 * LauraHale
 * Chriswaterguy

Keep me informed by email
leighblackall AT gmail DOT com

Sessions Ideas

 * Wikis in government - internal & public. --Chriswaterguy
 * Community building --Chriswaterguy
 * Using wikis in research and publication Leighblackall 00:59, 21 November 2011 (UTC)
 * Leigh, have you heard about openmedicine. They are planning to use wikis for publishing and updating systematic reviews. Conducted a pilot project a few years ago, see http://www.openmedicine.ca/issue/view/12 Juttavd 21:55, 15 December 2011 (UTC)

 * Spam-fighting. (Including: Can we allow anonymous edits and still keep out the latest spambots?) --Chriswaterguy 12:52, 7 December 2011 (UTC)
 * Wikimedia editor retention, addressing issues that drive away editors and ways we can tackle those issues. Steven Zhang 21:32, 15 December 2011 (UTC)

= Discussion =

This is the bit that's like a regular wiki talk page.

Date confirmation?
Will the date be decided soon?

Or, if not, it would be good to know if it's at least narrowed down to definitely Jan 25, 26 or 27 - that would be enough to start arranging my travel plans for Canberra.

Thanks - looking forward to it! --Chriswaterguy 10:28, 18 November 2011 (UTC)

 * Looks like it's confirmed as 20-22 Jan. I'd forgotten it was a three-day event - the time flew. Looking forward to it! --Chriswaterguy 04:50, 26 November 2011 (UTC)
 * Is it really a three day event? I went to a one day event in 2010 and ran out of time.  But is there enough talk for a three day event?  Friday, Saturday Sunday. Graeme Bartlett 12:06, 9 December 2011 (UTC)

 * Hi Graeme, the 2011 one was 3 days and it was easily filled. That said, we don't seem to be getting a big response for the 2012 one so far, (we still haven't promoted it!) so maybe you're right. I am too busy up until early Jan to do much promoting.. so hopefully others can step in there. Leighblackall 04:41, 12 December 2011 (UTC)

Accommodation and travel
So, where are y'all staying? Might be good if we find somewhere close by. If enough of us stay together, we should be able to hire a minivan or something. Ideas? Steven Zhang 21:37, 15 December 2011 (UTC)

 * Last time, I stayed at the Belconnen Premier Inn, but I was the only one. Every morning I walked over to the university via the lake (ca. 20 minutes). It was a nice little exercise considering that we spent the majority of the day inside… This year, I would also prefer to stay where other attendees are staying. Can’t remember where John and Adam stayed last time and if it was nice, but I think it was somewhere in the city. Juttavd 21:38, 15 December 2011 (UTC)

 * Adam and I stayed at Mantra on Northbourne last time. Not sure about accom plans this time, but I'll short have sorted it out tomorrow. John Vandenberg 04:42, 5 January 2012 (UTC)

 * I'll be staying with friends in Belconnen. John Vandenberg 13:34, 6 January 2012 (UTC)

I've found a real gem. I've done some calls and calculations and think I've found a reasonable way we can make this work and be convenient, cheap, and comfortable for all of us.

I called the Canberra Motor village, they have cabins that can accommodate up to 6 people. This particular cabin looks like the best one. At the moment, its approx $170 a night for the cabin, which is just under $30 a head. It's also close to the university, at just over 4kms away. It seems rather comfortable as well, and has the added benefit of included internet and cooking facilities.

It would also allow us to collaborate more after hours (after the conference). I think it's ideal. We should book ASAP. The more of us that are there, the easier it would be to get to and from the city. There are currently 6 Rosella cabins available, (so we can accommodate 36) and 12 of the Wattleglen cabins, also accommodating six, so we can accommodate up to 120 people there. That said I'd be overjoyed if that many attended, but at least we know there is the room for it if needed. Steven Zhang 07:28, 16 December 2011 (UTC)

 * A few of us have been discussing booking accommodation at ^, but due to how funding works have been discussing how to do it. Myself, foks, Sonia and her dad will be staying at the above place, and because it's complex to reimburse individualy (because its one price for 6) I'm happy to pay for accommodation and be reimbursed by WM-AU, but thought it would be appropriate to get the agreement of foks and sonia here as well. Others are welcome to add themselves to the list, the cabin can fit 6 and there's still space for 2 (and 19 other free cabins) Steven Zhang 07:03, 19 December 2011 (UTC)
 * Just to be clear, I didn't ask for funding for my dad: he's just tagging along to get out of Auckland for a bit. A car is also being rented which will take the four of us from Canberra airport to the cabin place late Thursday; if no one has much luggage we can fit another person in there. Sonia 07:10, 19 December 2011 (UTC)
 * Yeah, but the cabin is $169 a night whether one or six stay there. Car hire (which we'd split individually) is $180 for Thurs-Mon, so in theory adding another person would reduce the distributed costs. Steven Zhang 07:14, 19 December 2011 (UTC)

 * I like this idea as I have already pointed out over IRC. I've emailed Craig and Anne to discuss funding options. Jfox 09:34, 19 December 2011 (UTC)

Meetup
There are a few Canberrians who arnt attending, possibly because they dont have a lot of time over this weekend. Maybe we should organise a meetup on Friday or Saturday after the RCC..? We have at least two international attendees, and quite a few interstate attendees, so it would be good to provide an opportunity for the other Canberrians to meetup everyone. John Vandenberg 13:25, 6 January 2012 (UTC)
 * Seems like a good idea, if the people in question would be free on an evening. Sonia 02:41, 12 January 2012 (UTC)

 * Note there is also a Pre-camp dinner, Thursday 19 Jan, in Dickson. --Chriswaterguy 03:34, 16 January 2012 (UTC)

Late Attendees
If people 'turn up' Friday night or Saturday morning, is there any overnight accommodation (ie the 'cabins') still available? - 220.101.22.127 02:11, 20 January 2012 (UTC)

title: Talk:University of Canberra/RCC2012


text: File:HW2.9 2.jpg

title: File:HW2.9 2.jpg


text:
User:Egm6341.s11.team4.fields/HW2.10

=Problem 2.10: Expression for {Ci}=

Refer to lecture slides [[media:nm1.s11.mtg10.djvu|10-2 and 10-4]] for the problem statement.

Given
Equations (2) and (3) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:

 * {| style="width:70%" border="0" align="center"

(10.1)
 * $$\displaystyle P_2 (x) = c_2x^2 + c_1x + c_0 $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(10.2)
 * $$\displaystyle P_2 (x_i) = f(x_i) \, for \, i = 0, 1, 2 $$
 *
 * }
 * }

Objectives
Use equations (10.1) and (10.2) to find an expression for {Ci} in terms of (xi,f(xi)) for i = 0, 1, 2.

Solution
From equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]] we know:

 * {| style="width:70%" border="0" align="center"

(10.3)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{n=2}l_{i,2}(x)f(x_i) $$
 *
 * }
 * }

Where, from equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]

 * {| style="width:70%" border="0" align="center"

(10.4)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 *
 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$

 * {| style="width:70%" border="0" align="center"

(10.5)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(10.6)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(10.7)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 *
 * }
 * }

Using equation (10.3) we get:

 * {| style="width:70%" border="0" align="center"

(10.8)
 * $$\displaystyle P_2 (x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0) + \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1) + \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2) $$
 *
 * }
 * }

Combining terms of $$ x^2 $$  and  $$ x $$ we get:

 * {| style="width:70%" border="0" align="center"

(10.9)
 * $$\displaystyle P_2 (x) = \Big(\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x^2 $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x $$

 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle + \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$

 * }
 * }

After setting equation (10.1) equal to (10.9) we get:

 * {| style="width:70%" border="0" align="center"

(10.10)
 * $$\displaystyle C_2 = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(10.11)
 * $$\displaystyle C_1 = - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big) $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(10.12)
 * $$\displaystyle C_0 = \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 *
 * }
 * }

title: User:Egm6341.s11.team4.fields/HW2.10


text:
User:Egm6341.s11.team4.fields/HW2.11

=Problem 2.11: Simple Simpson's Rule Derivation=

Refer to lecture slides [[media:nm1.s11.mtg10.djvu|10-2 and 10-4]] for the problem statement.

Given
Equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:

 * {| style="width:70%" border="0" align="center"

(11.1)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{2} l_{i,2}(x) f(x_i)$$
 *
 * }
 * }

Simple Simpson's Rule given in equations (2) and (3) on lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

 * {| style="width:70%" border="0" align="center"

(11.2)
 * $$\displaystyle \int_a^bf(x)dx \approx I_2 = \int_a^bP_2(x)dx = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 *
 * }
 * }

Where:
 * {| style="width:70%" border="0" align="center"

(11.3)
 * $$\displaystyle h = \frac{b-a}{2}$$
 *
 * }
 * }

And:
 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle x_0 = a$$

 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle x_1 = \frac{a+b}{2}$$

 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle x_2 = b$$

 * }
 * }

Objectives
Use equation (11.1) to derive the Simple Simpson's Rule (11.2).

Solution
From equation (2) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:

 * {| style="width:70%" border="0" align="center"

(11.4)
 * $$\displaystyle \int_a^bP_2(x)dx = \sum_{i=0}^{2} \int_a^bl_{i,2}(x)dx f(x_i)$$
 *
 * }
 * }

From equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:

 * {| style="width:70%" border="0" align="center"

(11.5)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 *
 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$

 * {| style="width:70%" border="0" align="center"

(11.6)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(11.7)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

(11.8)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 *
 * }
 * }

Using Wolfram Alpha to solve the following integrals we get:

WA

 * {| style="width:70%" border="0" align="center"

(11.9)
 * $$\displaystyle \int_a^b l_{0,2}(x)dx = \int_{a}^{b} \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx = \frac{b-a}{6}$$
 *
 * }
 * }

WA

 * {| style="width:70%" border="0" align="center"

(11.10)
 * $$\displaystyle \int_a^b l_{1,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}dx = -\frac{2}{3}(a-b)$$
 *
 * }
 * }

WA

 * {| style="width:70%" border="0" align="center"

(11.11)
 * $$\displaystyle \int_a^b l_{2,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}dx = \frac{b-a}{6}$$
 *
 * }
 * }

Using Equation (11.4) and the results of equations (11.9), (11.10) and (11.11) we get:

 * {| style="width:70%" border="0" align="center"

(11.12)
 * $$\displaystyle \int_a^bP_2(x)dx = \frac{b-a}{6}f(x_0) + \frac{2}{3}(b-a)f(x_1) + \frac{b-a}{6}f(x_2)$$
 *
 * }
 * }

Simplifying equation (11.12) leaves us with the Simple Simpson's Rule:

 * {| style="width:70%" border="0" align="center"

(11.13)
 * $$\displaystyle I_2 = \int_a^bP_2(x)dx = \frac{b-a}{6}[f(x_0) + 4f(x_1) +f(x_2)] = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 *
 * }
 * }

Where:
 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle h = \frac{b-a}{2}$$

 * }
 * }

title: User:Egm6341.s11.team4.fields/HW2.11


text:
User:Eml5526.s11.team4.rhee/HW2

= Problem 2.2 =

Problem Statement
Consider

Orthonormal basis

Repeated index

 * 1. Find $$ det[b_{jk}] $$

 * 2. Find $$ \mathbf{\Gamma(b_{1},b_{2},b_{3})}=\mathbf{K},\quad det(\mathbf{\Gamma}) $$

 * 3. Find $$ \mathbf{F}=\left\{F_{i}\right\}=\left\{\mathbf{b}_{i}\cdot \mathbf{v}\right\} $$

 * 4. Solve (2-4) for $$ \mathbf{d}=\left\{ v_{j} \right\} $$

 * 5. Use $$ \mathbf{w}_{i}\cdot \mathbf{\mathbb{P}(v)}=0 $$ to find (2-5). What is $$ \vec{\mathbf{K}} $$ and $$ \vec{\mathbf{F}}? \quad \mathbf{d}=\left\{ v_{j} \right\} $$

 * 6. Solve for $$ \mathbf{d} $$ ; Compare to $$ \mathbf{d} $$ in problem 4.

 * 7. Observe symmetry property of $$ \mathbf{K} $$ and $$ \vec{\mathbf{K}}. $$ Discuss pros and cons of 2 methods.(2-4 & 2-5)

Question 1
The determinant of a 3 x 3 matrix using MATLAB ($$\textbf{Ans:} det[b_{jk}]=-8$$)

 Matlab Code

 Result

Question 2
From lecture slide 7-2,

Gram matrix $$ \mathbf{\Gamma(b_{1},b_{2},b_{3})}=\mathbf{K}_{ij}=\mathbf{b_{i} \cdot b_{j}}, \quad i,j=1,2,3 $$

where

 * $$ \mathbf{b_{1}}=\begin{bmatrix}1 & 1 & 1\end{bmatrix}, \mathbf{b_{2}}=\begin{bmatrix}2 & -1 & 3\end{bmatrix}, \mathbf{b_{3}}=\begin{bmatrix}3 & 2 & 6\end{bmatrix} $$
 * }
 * }

Therefore,

And the determinant of $$ \mathbf{\Gamma} $$ is 64.

 Matlab Code

 Result

Question 3
$$ \mathbf{F}=\left\{F_{i}\right\}=\left\{\mathbf{b}_{i}\cdot \mathbf{v}\right\} $$

where
 * $$ \mathbf{v}=5\mathbf{a}_{1}-7\mathbf{a}_{2}-4\mathbf{a}_{3} $$

$$ F_{1}=\mathbf{b}_{1}\cdot \mathbf{v}=(\mathbf{a}_{1}+\mathbf{a}_{2}+\mathbf{a}_{3})\cdot (5\mathbf{a}_{1}-7\mathbf{a}_{2}-4\mathbf{a}_{3}) $$

Likewise, we can obtain a $$ \mathbf{F} $$.

Question 4
Now, we know $$ \mathbf{K} $$ and $$ \mathbf{F} $$ according to problem 2 & 3. So, we can get a $$ \mathbf{d} $$

 Matlab Code

 Result

Now, we can obtain $$ \vec{\mathbf{K}}, \vec{\mathbf{F}} $$ and $$ \mathbf{d} $$ the same as above procedure.

Question 5
From lecture slide 7-2,4,

$$ \vec{\mathbf{K}}=\left[K_{ij}\right]=\mathbf{w}_{i}\cdot \mathbf{b}_{j} = \mathbf{a}_{i}\cdot \mathbf{b}_{j} $$

where
 * $$ \mathbf{w}_{i}=\mathbf{a}_{i}, \qquad \mathbf{b}_{1}=\mathbf{a}_{1}+\mathbf{a}_{2}+\mathbf{a}_{3}, \quad \mathbf{b}_{2}=2\mathbf{a}_{1}-\mathbf{a}_{2}+3\mathbf{a}_{3}, \quad \mathbf{b}_{3}=3\mathbf{a}_{1}+2\mathbf{a}_{2}+6\mathbf{a}_{3} $$

From equ(2-5) and using equ(2-1)

$$ \vec{\mathbf{F}}=\left\{F_{i}\right\}=\left\{\mathbf{a}_{i}\cdot \mathbf{v}\right\} $$ So,

Question 6
$$ \mathbf{d}=\vec{\mathbf{K}}^{-1}\vec{\mathbf{F}} $$

 Matlab Code

 Result

Question 7
We can observe that $$ \mathbf{K} $$ is Symmetric and $$ \vec{\mathbf{K}} $$is Nonsymmetric.

= Problem 2.9 =

Solution
From lecture slide

1. $$ \mathbf{b}=\left\{b_{j}, j=0,1,2\right\}=cos(jx+\phi) $$

 * $$ \mathbf{d}=\left\{d_{j}, j=0,1,2\right\} $$

2. Ess. b.c.:

 * $$ u^{h}(x)|_{\Gamma _{g}}=g \Rightarrow \sum_{j=1}^{n}d_{j}b_{j}(x)|_{\Gamma _{g}}=g $$

Nat. b.c.:

 * $$ \frac{\partial u^{h}(x)}{\partial x}n(x)|_{\Gamma _{h}}=h \Rightarrow \sum_{j=1}^{n}d_{j}{b}'_{j}(x)|_{\Gamma _{h}}=-hn(x) $$

3. From lecture slide

4. $$ \mathbf{K}\mathbf{d}=\mathbf{F} $$ From lecture slide 10-4 & 10-5,

 * $$\mathbf{K}_{ij}=\int_{\Omega }b_{i}(x)\left[\frac{\mathrm{d} }{\mathrm{d} x}\left(a_{2}(x)\frac{\mathrm{d} }{\mathrm{d} x}b_{j}(x)\right)\right]dx $$

 * $$ \mathbf{F}_{i}=-\int_{\Omega}b_{i}(x)f(x)dx$$

where
 * $$ a_{2}=2, \quad f=3 $$

= Cont'd (team wiki) =

Matrix Form
There are now three equations for three unknowns which can be expressed in matrix form.

$$ \begin{align} \begin{bmatrix} cos(\phi) & \cos(\phi+1) & \cos(\phi+2)\\ 0 & \sin(\phi) & 2\sin(\phi)\\ 0 & 2\left\{sin(\phi+1)-sin(\phi)\right\}cos(\phi) & 8(\phi+1)cos(\phi) \end{bmatrix}\begin{bmatrix}d_0\\d_1\\d_2\end{bmatrix}=\begin{bmatrix}0\\4\\3cos(\phi) \end{bmatrix} \end{align} $$

$$ \mathbf{K} $$ is not symmetric.

$$ \phi = \pi/4 $$
Solving the system for $$ \phi = \pi/4 $$

$$ \begin{bmatrix}d_0\\d_1\\d_2\end{bmatrix}=\begin{bmatrix}1.7007\\ 5.6651\\ -0.0041\end{bmatrix} $$

 * Exact solution: $$ u(x) = \frac{-3}{4}x^2-4x+\frac{19}{4} $$

title: User:Eml5526.s11.team4.rhee/HW2


text: File:RCC UC 2011 - Open space discussion on ustream.jpg


text: University of Canberra/RCC2011/Discussion About Open Space


text:
User:Hydriz

Hi, I am Hydriz. Please visit my userpage on Meta for more information about me.

title: User:Hydriz


text:
University of Canberra/Submission to the 2011 review of professional development review processes

This page invites people to collaborate on a submission to the 2011 Review of the Universty of Canberra's Academic Professional Development Review process. The following email was sent to staff on 27 January 2011:

"As noted in an all staff notice late last year it was proposed that we undertake a review and evaluation of the academic PDR process. The object of the review is to assess the effectiveness of the implementation of the current PDR policy, relevant procedures and systems issues and to make recommendations for improvements to ensure that the PDR process supports the University’s strategic objectives. This survey is the first stage of the review process. The survey instrument is on OPUS and is for all ongoing and fixed term (of 6 months or more) academic staff. Your responses will make an invaluable contribution to the PDR process. A second stage will involve an external facilitator who will convene focus groups to discuss emergent issues. If for any reason you are unable to either complete the survey or contribute to the focus group discussions, you are invited to submit written comments, suggestions and ideas addressed to the Director HR. Rest assured your views will be treated in the strictest confidence."

Requests for comment
Q15: The PDR process could be improved by (maximum 1000 characters allowed).
 * 1) In the 2009 PDR there was clear evidence of panels trying to coerce academic staff onto a teaching only track. This was not evident in 2010.
 * 2) There is a lack of specificity in the PDR guidelines about goal-setting (e.g., in 2010 I was told to be more specific; in 2011 I was told to choose fewer goals)
 * 3) The feedback structure for the PDR is top-down rather than 360 degrees
 * 4) There was a lack of consistency in feedback from the supervisory panel between 2010 and 2011 (in my case it was a different panel each time)
 * 5) HR continues to shift the goal posts on performance expectations - such changes should not be made during a 12 month review period when it is already underway
 * 6) HR have never responded to the feedback provided on the 2010 review in the Moodle site
 * 7) Until this review there has been no objective assessment of staff satisfaction with the PDR process
 * 8) The spreadsheet system in 2009 and online system in 2010 both had very significant limitations. Much could be done to improve this aspect.

Q16: Other ways that PDR could be improved in the future are ? (maximum 1000 characters allowed)
 * 1) The mapping of the university's 39 strategic steps directly onto PDR evaluation is somewhat primitive - e.g., how is performance in each of these areas to be weighted with regard to overall satisfactoriness? This is not made explicit anywhere, hence it is difficult to know what to prioritise as a staff member. It is also unclear how achievement of specific performance goals compared to performance on the performance expectations is in determining overall satisfactoriness. There remains much opaque, subjective judgement on behalf on the panel.
 * As a result, performance and goals for each of the performance expectations were not discussed in the 30 min meeting with the panel and only the bold expectations were discussed.


text:
User:Eml5526.s11.team4.yang/hw2

=Homework 2.3=

Refer to lecture slide [[media:fe1.s11.mtg8.djvu|mtg-8]] for the problem statement.

Given
Use { ai } (basis),let $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)$$, then equation
 * {| style="width:100%" border="0"

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

\underline{w}(x)\underline{P}(\underline{v})=0\begin{matrix} {} & \forall  \\ \end{matrix}\underline{w}(x)=\sum\limits_{i=1}^{n}(x)

$$     (Eq 1) becomes
 *
 * }
 * {| style="width:100%" border="0"

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

\underline{w}(x)\underline{P}(\underline{v})=0\begin{matrix} {} & \forall  \\ \end{matrix}\{{{\beta }_{1}},{{\beta }_{i}}..\}\in {{\mathbb{R}}^{n}}\begin{matrix} {} & s.t.\underline{w}(x)=\sum\limits_{i=1}^{n} \\ \end{matrix}

$$     (Eq 2)
 *
 * }

Objectives
Show that equation (2) is equivalent to
 * {| style="width:100%" border="0"

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

\underline(x)\underline{P}(\underline{v})=0\begin{matrix} {} & i=1,2,...n \\ \end{matrix}

$$     (Eq 3)
 *
 * }

Solutions
1). First, we try to derive Eq.(3) using Eq.(2).

Eq. (1) is valid for any function of w (x), which is a linear combination of these linearly independent bases. Thus, the Eq. (1) is valid for some particular sets of $${{\beta }_{i}}$$ value to get $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)$$. For example, we choose $${{\beta }_{1}}=1,{{\beta }_{2}}={{\beta }_{3}}=....{{\beta }_{n}}=0$$, then $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)={{\underline{a}}_{1}}(x)$$. By substituting $$\underline{w}(x)$$ into Eq.(2), we obtain:
 * {| style="width:100%" border="0"

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

\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$ Which is the equation for Eq.(3) of i=1.
 *
 * }

Then, in similar way, we choose $${{\beta }_{2}}=1,{{\beta }_{1}}={{\beta }_{3}}=....{{\beta }_{n}}=0$$, then we obtain:
 * {| style="width:100%" border="0"

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

\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$
 *
 * }

Which is the equation for Eq.(3) of i=2.

Similarly, we continuing to choose $${{\beta }_{i}}=1,{{\beta }_{1}}={{\beta }_{2}}=..={{\beta }_{i-1}}={{\beta }_{i+1}}..={{\beta }_{n}}=0$$ then we obtain:
 * {| style="width:100%" border="0"

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

\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$ Which is the equation for Eq.(3) of any value of i. So, Eq. (3) is derived with Eq.(2).
 *
 * }

2). Secondly, we try to derive Eq.(3) using Eq.(2).

{ ai(x), i=1,2…n } is a family of linearly independent basis functions, and according to Eq.(2) we have:

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

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

\underline(x)\underline{P}(v)=0

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

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

\underline(x)\underline{P}(v)=0

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

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

\underline(x)\underline{P}(v)=0

$$
 *
 * }

Linearly combining the above equations, by choosing any set of $${{\beta }_{i}}$$value, we obtain:
 * {| style="width:100%" border="0"

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

\begin{align} & {{\beta }_{1}}[\underline(x)\underline{P}(v)]+{{\beta }_{2}}[\underline(x)\underline{P}(v)]+..+{{\beta }_{i}}[\underline(x)\underline{P}(v)]+...{{\beta }_{n}}[\underline(x)\underline{P}(v)] \\ & ={{\beta }_{1}}\centerdot 0+{{\beta }_{2}}\centerdot 0+..+{{\beta }_{i}}\centerdot 0+..+{{\beta }_{n}}\centerdot 0 \\ & =0 \\ \end{align}

$$
 *
 * }

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

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

\begin{align} & {{\beta }_{1}}[\underline(x)\underline{P}(v)]+{{\beta }_{2}}[\underline(x)\underline{P}(v)]+..+{{\beta }_{i}}[\underline(x)\underline{P}(v)]+...{{\beta }_{n}}[\underline(x)\underline{P}(v)] \\ & =[\sum\limits_{i=1}^{n}(x)]\underline{P}(v) \\ & =\underline{w}(x)\underline{P}(v) \\ & =0 \\ \end{align}

$$
 *
 * }

Because the value set of $${{\beta }_{i}}$$undefinedis arbitrary, Eq.(2) is valid for any w (x).

So, we obtain Eq.(2) using Eq.(3).

According the above two aspects, we can conclude that Eq.(2) is equivalent to Eq. (3).

title: User:Eml5526.s11.team4.yang/hw2


text:
University of Canberra/RCC2011/Participants

Participants

 * 1) Laura Hale - University of Canberra
 * 2) Leigh Blackall - University of Canberra
 * 3) Jutta von Dincklage - Cancer Council Australia
 * 4) Rob Fitzgerald - University of Canberra
 * 5) John Vandenberg - Wikimedia Australia, University of New England
 * 6) Jani Patokallio - Lonely Planet
 * 7) Rose Grozdanic - Australian Flexible Learning Framework
 * 8) Michael Coghlan - TAFESA, NewLearning
 * 9) Kirsty Sharp - Tasmanian Polytechnic
 * 10) Tom Worthington - Australian National University
 * 11) Diane Phillips - University of Canberra
 * 12) MarkDilley - AboutUs
 * 13) SatuSuro - Wikipedia editor - Western Australia
 * 14) Chriswaterguy - Wikipedian & co-founder of Appropedia
 * 15) Adam Jenkins - University of South Australia
 * 16) Billinghurst - no organisation - Fri - Sat for parts of days, ?Sun.
 * 17) Marghanita da Cruz - Ramin Communications - definitely attending all day Friday
 * 18) Stephan Ridgway - Sydney - Travelling down Friday, definitely attending Sat. and Sun Morning
 * 19) Robyn Jay - Free
 * 20) James Neill - University of Canberra - attending all day Friday
 * 21) Ken -  Lonely Planet
 * 22) Mark Spain


text:
University of Canberra/RCC2011/Wikisource

Wikisource

 * Participants: Billinghurst, John Vandenberg, MarkDilley, Jani Patokallio, Marghanita

 * Topics:
 * No notability requirements
 * mw:Extension:Proofread Page functionality
 * Help:transclusion of 'Page' (physical structure) into logical structure
 * page and line breaks in printed edition; discussion about NLA newspapers project, now part of NLA trove.
 * division of content between language subdomains: English texts go on Main Page, German texts go on Main Page, etc.
 * author: pages
 * s:Wikisource:Translations, inc. Bible (Wikisource) (November 2005-) and Romance of the Three Kingdoms (April 2006-)
 * Jani mentioned Humanity Declaration which would be an interesting translation project


title: File:Fe1.s11.team6.HW2.7.fig1.JPG


title: File:Fe1.s11.team6.HW2.7.fig2.JPG