text:
QuickSort

Quicksort belongs to the group of dynamically dividing algorithms and works with recursions. It is not stable.

The main idea behind Quicksort is to divide a group of data records into two subgroups in such a way that all the elements in the lower subgroup are smaller than those of the upper subgroup. In order to do this division, a pivot element is selected. In the original version of the algorithm this was in a fixed way either the first or the last element. The best choice for a pivot is a value which sorts half of the records to each subgroup. Without having a look at all the key values a good selection is not possible. And if the pivot is selected just by chance, any choice is as good as all the others; and selecting the first or last element makes the implementation of the algorithm a little bit easier.

We assume that we use the last element as the pivot. We start at the beginning of the list, until we find an element which is bigger than the pivot and remember it. Then we start at the end of the list and search for the first element which is smaller than the pivot. These records are then swapped. Then, beginning from the remembered positions, again the first bigger and the first smaller element are looked up and swapped. This is repeated until the two pointer meet in the middle. At the end of a split operation it is checked whether the pivot itself has to be moved.

Then the algorithm takes both subgroups and calls itself for the next division operation. If a subgroup contains only one or no element, there is no further division possible and the process returns to the calling level. Since both subgroups are sorted and all the elements of the lower subgroup were smaller than those of the upper subgroup, all the elements of the returned group are sorted without any more handling.

Quicksort is Bubblesort type algorithm. Have critical in Descendency array sort (9,8,7,6...1). Can changed, when use random pivot position. (still have one critical, but not in desc array)

Timing
On each level the records are divided into two subgroups. With n data records we get
 * L = lg2(n)

levels. For every subgroup on any level all the records have to be compared with the pivot. On any level the number of compare operations is n. With an even distribution the number of swaps will be ½ n. The time for a sort can be estimated with

 * T = C * n * lg2(n) + D * ½ * n *  lg2(n)

 * T = (C + D * ½) * n * lg2(n)

 * T = E * n * lg2(n)

where E is a constant.

If the input shall be sorted ascending, the list is already sorted descending, and the last element is selected as a pivot, all elements but one will go into one of the sublists. The number of levels will then be n and not lg2(n). The timing can then be aproximated as a worst case by

 * T = E * n2

Variations
In order to make sure that the worst case will not happen, there are some strategies. The simplest is to check the last, the first and the element in the middle and to select the one that is nearer to the average of the three. With little effort the chance to make a bad selection can be lowered significantly.

Another possibility for improvement is to select the pivot just by chance. With big numbers of data records this method makes it improbable to run into a worst case.

Another possibility is to calculate the average of the key values and to select the key as a pivot which has the smallest distance to the average. We have to sum up all the values in order to calculate the average which makes n operations. Then all key values are compared with the average and if a key is found with a smaller distance to the average the candidate is changed. For this again n operations are needed. The effort for the hole selection increases in a linear fashion with the number of records. This method does not guarantee that both subgroups will have the same number of records, but with evenly distributed key values the selection will be pretty good.

The search for an optimal pivot, for example by generating histograms, causes a much higher effort.

Example
The pivot-element is shown in red and the sublists in green and yellow. We show the variant where the last element of a domain is selected as pivot.

Step 1:  The pivot is 771.

The first element bigger than the pivot is 901. The last element smaller than the pivot is 522. These elements are swapped.

Step 2: Now we handle the lower subgroup of step 1. The pivot is 522; the first element bigger is 529 and the last element smaller is 020. The two swap. No more swaps possible.

Step 3: Now the lower subgroup of step 2. The pivot is 227. The first bigger value is 387; the last smaller value is 011; the two swap. Now the first bigger value is 310 and the last smaller value is 174; the two swap.

Step 4:  The lower subgroup of step 3 is divided now. The pivot is 20. The first bigger value is 174 but there is no last smaller value between 174 and the pivot: therefor the pivot and 174 swap.

Step 5:  The upper subgroup can not be divided. Now the lower subgroup is divided but there will be no further subdivisions

Result 1: The sorted lists of step 4 and step 5 are joined

Step 6: We continue with the upper subgroup of step 3. The pivot is 227; the first bigger value is 310 but there is no last smaller value between the pivot and 310; therefor they swap.

Step 7: we sort the upper subgroup of step 6

Result 2: The sorted lists of step 6 and step 7 are joined

Step 8: Now we come to the upper subgroup of step 2. There is no smaller last element therefor the two records swap.

Result 3: Result 2 and Step 8 are joined

Step 9: We sort the upper subgroup of step 1. The pivot is 771; since there is no last smaller element, the two records are swapped.

Result: The data of step 9 are added.

title: QuickSort


text: Category:Introduction to Italian

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

title: Category:Introduction to Italian


text: Category:New Zealand Law and Practice Examination


text:
Using Wikiversity/What is Wikiversity

Wikiversity is an active-learning community where everyone can learn from each other through discussion and learning projects. People of all ages from all over the world come to Wikiversity to participate in the learning community.

History
Wikiversity's community initially developed within the Wikibooks project. Some people within the Wikibooks community sought to create focus groups that could be used to organize works and help people collaborate on common goals. Study aids, books for learners, books for professionals, books for children, research, and works useful for classroom use were among the initial targets for these focus groups. "Wikiversity" was initially created as part of the effort to create works useful for classroom use, but was later expanded by a movement within the Wikibooks community that sought to expand Wikibooks' scope beyond textbooks to include hosting of additional types of learning resources. Karl Wick, the founder of Wikibooks, started Some ideas about Wikiversity.

Wikiversity was created as an independent project in August 2006.

Scope
Wikiversity is as much about building a community as it is about learning and research. You are encouraged to develop and pursue your own learning or research goals. You are encouraged to ask and answer questions. You are encouraged to share what you know and what you would like to know. No subject is out of scope.

Wikiversity hosts flexible educational resources for use in multiple contexts. Wikibooks develops textbooks that ultimately are for use off the shelf as-is in a specific context while Wikisource hosts previously published works that are either out of copyright or have been released under a compatible copyleft license. Wikipedia another project is building an encyclopedia of human knowledge. The educational resources that Wikiversity hosts may be useful to teachers and students alike. We believe that as learning resources develop course work will be shared with other communities and projects.

title: Using Wikiversity/What is Wikiversity


text: Category:Vancouver Flower Clock Project

title: Category:Vancouver Flower Clock Project


text:
User:Kilbad

Visit me at my Wikipedia page...

title: User:Kilbad


text:
User talk:Kilbad

Visit me at my Wikipedia talk page...

title: User talk:Kilbad


text:
User:Jacopete/What is Hierarchical Task Analysis

What is a Hierarchical Task Analysis (HTA)?
Annett and Stanton refer to HTA as "exploring tasks through a hierarchy of goals indicating what a person is expected to do, and plans indicating the conditions when subordinate goals should be carried out" (Task Analysis, p. 9).

This sounds quite simple. Right? It actually is simple, but there is a systematic approach to completing an HTA. However, let's first look at the advantages:

 * It allows for easy comparison among multiple task anaylses of the same goal

 * A completed task anaylsis will provide a springboard for a needs assessment

 * Since HTA is conducted through observation, the observer will minimally interfere with the activity and worker.

In the name of fairness, we should also consider a disadvantage:

 * While interference may be minimal during observation, before and after the observation takes time the worker could be using to complete other responsibilities.


text:
User talk:Rasim

.

title: User talk:Rasim


text:
User:Mikefife

Mike Fife.

Lives: Leeds, West Yorkshire, UK. Age: 48

Interests: Music, cycling.

Links and Mentors

 * Forum:User Page Design - Discuss related issues here.

Bren Travis

title: User:Mikefife


text:
User talk:Mikefife

 Hello Mikefife, 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. --Ottava Rima (talk) 17:19, 22 March 2010 (UTC)

title: User talk:Mikefife


text:
User talk:Avraham

 Hello Avraham, 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. --Ottava Rima (talk) 17:31, 22 March 2010 (UTC)

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

Welcome to Wikiversity
I always enjoy working with you on Wikipedia, and nice to see you over here. Jacob Frank says so, too. Hope you decide to stay. This is a place where you have quite a lot of leeway to create interesting and innovative learning resources, so if you have any ideas for projects that wouldn't fit within the scope of other parts of Wikimedia, you can try them here. --AFriedman (talk) 07:34, 23 March 2010 (UTC)

title: User talk:Avraham


text:
University of Florida/Egm6341/s10.Team2/HW5

Statement
P. 26-3

To show that the value of the constants $$\; c_{6}=0$$ and $$c_{5}=\frac{-7}{360}$$ in the polynomials $$P_{4}\left ( t \right )$$ and $$P_{5}\left ( t \right )$$

Solution
From | page 26-2, we know that:

$$P_{5}\left ( t \right )=\frac{-t^{5}}{120}+\frac{t^{3}}{36}+ C_{5}t+C_{6}$$

We select $$\; P_{5}\left ( t \right )$$ such that it is a odd function and $$\; P_{5}\left ( 1 \right )=P_{5}\left ( -1 \right )=P_{5}\left ( 0 \right )=0$$

Therefore, using $$P_{5}\left ( 0 \right )=0$$, we get:

$$P_{5}\left ( 0 \right )=\frac{-0^{5}}{120}+\frac{0^{3}}{36}+ C_{5}*0+C_{6}=0$$

 * {| style="width:60%" border="0"

$$  \displaystyle \boldsymbol \therefore \; C_{6}=0 $$     (1)
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 *
 * }

Similarly, using $$\; P_{5}\left ( 1 \right )=0$$, we get:

$$P_{5}\left ( 1 \right )=\frac{-1^{5}}{120}+\frac{1^{3}}{36}+ C_{5}*1+C_{6}=\frac{-1}{120}+\frac{1}{36}+C_{5}=0 $$

 * {| style="width:60%" border="0"

$$  \displaystyle \boldsymbol \therefore \; C_{5}=\frac{1}{120}-\frac{1}{36}=-\frac{7}{360} $$     (2)
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 *
 * }

Author
Egm6341.s10.team2.lee 08:58, 24 March 2010 (UTC) proofread --Egm6341.s10.team2.niki 14:53, 24 March 2010 (UTC)

Statement
P. 27-1

 Continue the proof of Trapezoidal Rule Error to steps 4a and 4b and determine $$\boldsymbol{ P_6(t)} $$ and $$\boldsymbol{ P_7(t) }$$

Solution
From steps 3a and 3b( P.26-3) we get the expression

 * {| style="width:60%" border="0"

$$  \displaystyle \boldsymbol{E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_D} $$     (P.26-3)
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 *
 * }

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$$  \displaystyle P_4(t)=\frac{-t^4}{24}+\frac{t^2}{12}-\frac{7}{360}=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_5$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$

P_5(t)=\frac{-t^5}{120}+\frac{t^3}{36}-\frac{7t}{360}=c_1(\frac{t^5}{5!})+c_3(\frac{t^3}{3!})+c_5(t) $$     (P.26-3)
 *
 * }

Step 4a

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

$$  \displaystyle \int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_udt=[g^{(5)}(t)P_6(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}[{P_6(t)g^{(6)}(t)}}_E]dt $$ where, $$P_6(t)=\int P_5(t)=c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

$$ P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

Step 4b

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$$  \displaystyle \int_{-1}^{+1}[{\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_u}]dt=E=[g^{(6)}(t)P_7(t)]_{-1}^{+1}-\int_{-1}^{+1}P_7(t)g^{(7)}(t)dt $$ where, $$P_7(t)=\int P_6(t)==c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

$$P_7(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta$$

Selecting $$\displaystyle{P_7(t)}$$ such that

$$\displaystyle{P_7(+1)=P_7(-1)=P_7(0)=0}$$ ,we get

$$\displaystyle{P_7(t=0)=0\Rightarrow \beta=c_8=0}$$

$$P(-1)=0=-\frac{(-1)^7}{7!}+\frac{1}{6}(\frac{(-1)^5}{5!})-\frac{7}{360}(\frac{(-1)^3}{3!})-\alpha\Rightarrow \alpha=c_7=\frac{31}{15120}$$

Summary

 * {| style="width:60%" border="0"

$$  \displaystyle P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$

P_7(t)=\int P_6(t)=\frac{-t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\frac{31}{15120} t $$
 *
 * }

Author
--Egm6341.s10.team2.niki 22:04, 23 March 2010 (UTC)

--Proofread by Egm6341.s10.team2.lee 10:14, 26 March 2010 (UTC)

Statement
find $$\; t_k(x)$$ on P. 27-1

Solution
From Page 27-1, we know:

$$x\left ( t_{k} \right )=\frac{1}{2}\left ( x_{k} + x_{k+1} \right ) + t_{k}\frac{h}{2}$$

Now rearranging the terms on both sides to write $$\; t_{k}$$ in terms of $$\; x$$, we get:

$$t_{k}\left ( x \right )=\frac{2x-\left (x_{k}+x_{k+1} \right )}{h}$$ $$ \; \;$$ where the size of the interval, $$\; h=x_{k+1}-x_{k}$$

It can also be easily verified by substituting $$\; x=x_{k}$$ which yields $$t_{k}\left ( x=x_{k} \right )= \frac{x_{k}-x_{k+1}}{x_{k+1}-x_{k}}=-1$$

Also when $$\; x=x_{k+1}$$ yields $$t_{k}\left ( x=x_{k+1} \right )= \frac{x_{k+1}-x_{k}}{x_{k+1}-x_{k}}=1$$

Author
Srikanth Madala (SM)

Statement
P. 27-2

Find the values of $$d_{1}=\bar{d_{2}}, d_{2}=\bar{d_{4}}$$ and $$d_{3}=\bar{d_{6}} $$

Clue : Use the formula: $$d_{i}=\bar{d_{2i}}= \frac{P_{2i}\left ( 1 \right )}{2^{2i}}$$

Solution
We know from the formula:

$$d_{i}=\bar{d_{2i}}= \frac{P_{2i}\left ( 1 \right )}{2^{2i}}$$

From | Mtg 26,slide 26-1, we know the formula:

$$P_{2}\left ( t \right )= \frac{-t^{2}}{2!}+\frac{1}{6}$$

$$P_{2}\left ( 1 \right )= \frac{-1^{2}}{2!}+\frac{1}{6}$$

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

 * style="width:92%; padding:10px; border:2px solid #8888aa" |
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$$d_{1}=\bar{d_{2}}= \frac{P_{2}\left ( 1 \right )}{2^{2}}=\frac{\frac{-1}{3}}{4}=\frac{-1}{12} $$

(1)
 *
 * }

From | Mtg 26,slide 26-3, we know the formula:

$$P_{4}\left ( t \right )= \frac{-t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}$$

$$P_{4}\left ( 1 \right )= \frac{-1^{4}}{24}+\frac{1^{2}}{12}-\frac{7}{360}= \frac{1}{45}$$

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

 * style="width:92%; padding:10px; border:2px solid #8888aa" |
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$$d_{2}=\bar{d_{4}}= \frac{P_{4}\left ( 1 \right )}{2^{4}}=\frac{\frac{1}{45}}{16}=\frac{1}{720}$$

(2)
 *
 * }

From problem 2 on this HW, where we deduced that:

$$P_6(t)=\frac{-t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{15120} $$

$$P_6(1)=\frac{-1^6}{720}+\frac{1^4}{144}-\frac{7*1^2}{720}+\frac{31}{15120}=\frac{-2}{945}$$

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

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$$d_{3}=\bar{d_{6}}= \frac{P_{6}\left ( 1 \right )}{2^{6}}=\frac{\frac{-2}{945}}{64}=\frac{-1}{30240}$$

(3)
 *
 * }

Author
--Srikanth Madala 01:44, 24 March 2010 (UTC) Srikanth Madala (SM)

Statement
P. 28-2 derive

Derive the Following Equation for the error of the trapezoidal rule:

$$ E_{n}^{1} = I - T_{0}(n) = \sum_{r=1}^{l} h^{2r}\tilde{d}_{2r} \left[ f^{2r-1}(b) - f^{2r-1}(a) \right] - \frac{h^{2l}}{2^{2l}} \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} P_{2}l(t_{k}(x))f^{2l}(x)dx $$

Solution
The Following Equation is to be derived:

$$ E_{n}^{1} = I - T_{0}(n) = \sum_{r=1}^{l} h^{2r}\tilde{d}_{2r} \left[ f^{2r-1}(b) - f^{2r-1}(a) \right]  -  \frac{h^{2l}}{2^{2l}} \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} P_{2}l(t_{k}(x))f^{2l}(x)dx  $$

$$ E= \int_{-1}^{+1} P_{1}(t)g^{1}(t) dt = \underbrace{ \frac{h}{2} \left[ P_{2}g^{1} + P_{4}g^{3} + ... + P_{2l}g^{2l-1} \right]_{-1}^{+1} }_{A} - \underbrace{ \int_{-1}^{+1} P_{2l}(t)g^{2l}(t)dt }_{B} $$

Since $$ P_{2} $$ is an even function then:

$$ \left[ P_{2}(t)g^{1}(t) \right]_{-1}^{+1} = P_{2}(+1) \left[ g^{1}(+1) - g^{1}(-1) \right] $$

With this property then A can be rewritten as follows:

$$ \sum_{i}^{l}P_{2r}(+1)\left[g^{2r-1}(+1) - g^{2r-1}(-1) \right] $$

A transformation of variable is needed from 't' to 'x' and is as follows:

$$ g_{k}^{i}(t) = (\frac{h}{2})^{i}f^{i}(x(t)) \quad \quad x \epsilon [x_{k}, x_{k+1} ] $$

With this transformation A can be rewritten as follows:

$$ \frac{h}{2} \sum_{i=1}^{l} P_{2r}(+1)(\frac{h}{2})^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

The Following is then defined:

$$ \tilde{d}_{2r} = \frac{P_{2r}(+1)}{2^{2r}} $$ as discussed in [28-2] of the lectures.

A is then rewritten as follows:

$$ \frac{h}{2} \sum_{i=1}^{l} \tilde{d}_{2r} 2^{2r} \left(\frac{h}{2} \right) ^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

$$ \frac{h}{2} \sum_{i=1}^{l} \tilde{d}_{2r} 2 h^{2r-1} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

$$ \sum_{i=1}^{l} \tilde{d}_{2r} h^{2r} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] $$

The Next Step is to simplify the term 'B'

$$ B= \int{-1}^{+1} P_{2l}(t)g^{2l}(t) dt $$

Using the previously mentioned transformation of variable the following can be stated:

$$ \int_{-1}^{+1}P_{2l}(t) \left(\frac{h}{2}\right)^{2l}f^{2l}(x(t))dt $$

With the previously performed simplifications the following is written:

$$ E_{n}^{1} = A - B = \sum_{i=1}^{l} \tilde{d}_{2r} h^{2r} \left[f^{2r-1}(b) - f^{2r-1}(a) \right] - \frac{h^{2l}}{2^{2l}}\int_{-1}^{+1}P_{2l}(t) \left(\frac{h}{2}\right)^{2l}f^{2l}(x(t))dt $$

This is recognized as the Error for the trapezoidal rule.

Author
--Egm6341.s10.Team2.GV 22:09, 25 March 2010 (UTC)

--Proofread by Egm6341.s10.team2.lee 10:13, 26 March 2010 (UTC)

Statement
P. 28-2 bernoulli nos

Solution

Author
Egm6341.s10.team2.lee 10:12, 26 March 2010 (UTC)

Statement
P. 28-2

 Redo steps in the proof of the Trapezoidal Rule error by trying to cancel terms with odd order derivatives of "g"

Solution
We begin with equation (5) on P. 21-1 which is the result of transformation of variables on equation (1) P. 21-1

(| Prob 4 HW4) (P. 21-1)

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

$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{1}g_k(t)dt-[g_k(-1)+g_k({+1})]] $$   (5) on P.21-1 From | Prob 5 HW4, we can express the above equation as:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{+1}(-t)g^{(1)}(t)dt] $$   (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Integrating the term withing the square brackets by "Integration by parts" | Prob 7 HW4 we can rewrite (1) as follows
 * {| style="width:100%" border="0"

$$  \displaystyle E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[[P_2(t)g^{(1)}(t)]_{-1}^{+1}-\int_{-1}^{+1}P_2(t)g^{(2)}(t)dt] $$   (2) In order to eliminate terms with even powers of $$\displaystyle{h}$$ we need to remove terms with odd derivatives of $$\displaystyle{g(t)}$$.Therefore, the boundary term in eqn (2) above must be set to zero by selection of $$\displaystyle{P_2(t)}$$.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We have $$\displaystyle{P_2(t)}$$ from eqns (1 and 2)P. 21-3
 * {| style="width:100%" border="0"

$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (1)p21-3
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 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle c_2=0 $$   (2)p21-3 Setting $$\displaystyle{P_2(-1)=0}$$ gives $$\displaystyle{c_3=1/2}$$ and hence we get
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle P_2(t)=c_1(\frac{t^2}{2!})+c_3 $$   (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

So following this method, the next term to be eliminated will have P4(t)
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$$  \displaystyle P_4(t)=c_1(\frac{t^4}{4!})+c_3(\frac{t^2}{2!})+c_4(t)+c_5 $$   (4)
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Setting $$\displaystyle{P_4(-1)=0 and P_4(1)=0 }$$ and solving we get $$\displaystyle{c_4=0 and c_4=-5/24 }$$.Continuing on these lines we get the eqn (2) in the form

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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{[P_2g^{(1)}]_{-1}^{+1}-[P_3g^{(2)}]_{-1}^{+1}}+[P_4g^{(3)}]_{-1}^{+1}-[P_5g^{(4)}]_{-1}^{+1}+[P_6g^{(5)}]_{-1}^{+1}-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt $$   (5) Dropping all the terms with odd order derivatives of $$g(t)$$ we get
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}{-[P_3g^{(2)}]_{-1}^{+1}}+-[P_5g^{(4)}]_{-1}^{+1}+-[P_7g^{(6)}]_{-1}^{+1}-\int_{-1}^{+1}P_7g^{(7)}dt$$ (6) In general on integrating "m" times we get:
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}(\sum_{i=1}^{m}-[P_{(2i+1)}g^{(2i)}]_{-1}^{+1})-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (7) manipulating the terms yields,
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$$  \displaystyle E_n^1=\frac{h}{2}\sum_{k=0}^{n-1}[\sum_{i=1}^{m}-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\int_{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt $$   (8)
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Transforming g(t) back to f(x) we get [see [| prob 6 HW4]

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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}\sum_{k=0}^{n-1}[(P_{(2i+1)}(-1)( f^{(2i)}(x_k))-(P_{(2i+1)}(+1) f^{(2i)}(x_{k+1}))]-(\frac{h}{2})^{2m+1}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9)
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Since the polynomial is odd, it can be extracted out to give the eqn in the form
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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}P_{(2i+1)}(-1)\left [\sum_{k=0}^{n-1}[( f^{(2i)}(x_k)+f^{(2i)}(x_{k+1})]\right ]-(\frac{h}{2})^{2m+2}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9)
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The problem with this formula is evident at first sight, from this expression.The HO derivatives add up in the first term of the expression and since they will not cancel out (9) cannot be reduced to a simpler form. To see the difference in the two approaches more clearly we compare the equations from the two methods. From (1) P. 27-1 we have
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$$  \displaystyle E_n^1=\sum_{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-\frac{h^{(2l)}}{2^{(2l)}}\sum_{k=0}^{n-1}\int_{x_k}^{x_{k+1}}P_{2l}(t_k(x))f^{(2l)}(x)dx $$   (10)
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$$  \displaystyle E_n^1=\sum_{i=1}^{m}(\frac{h}{2})^{2i+1}P_{(2i+1)}(-1)\left [\sum_{k=0}^{n-1}[( f^{(2i)}(x_k)+f^{(2i)}(x_{k+1})]\right ]-(\frac{h}{2})^{2m+2}\sum_{k=0}^{n-1}(\int_{x_k}^{x_{k+1}}P_{(2m+1)}(x) f^{(2m+1)}(x))dx) $$   (9) Comparing the eqns 9 and 10 we see that the first term on 9 has a summation of HO derivatives at different points in the square brackets as opposed to 10 which has the difference of the higher order derivative at only the end points. In this lies the difference of the two methods. 10 is general and easier to apply as we need to evaluate the differential of F(x) at only the end points while 9 is computationally cumbersome since it needs the derivative evaluated at all points.
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Author
--Egm6341.s10.team2.niki 14:52, 24 March 2010 (UTC)

===Problem 8: Calculating the coefficients $$C_{3},C_{5} \; and \; C_{7}$$ and the pairs of polynomials $$\left (P_{2},P_{3} \right ),\; \left (P_{4},P_{5}  \right ),\; \left (P_{6},P_{7}  \right )$$ using the Recurrence formula===

Statement
P. 29-3

Use the recurrence formula on P.29-2 to obtain the coefficients and expressions of the pairs of polynomials $$\left (P_{2},P_{3} \right ),\; \left (P_{4},P_{5}  \right ),\; \left (P_{6},P_{7}  \right )$$

Solution
We know the recurrence formulae:

$$P_{2i}(t)=\sum_{j=0}^{i} C_{2j+1} \; \frac{t^{2(i-j)}}{[2(i-j)]!}$$

$$P_{2i+1}(t)=\int P_{2i}(t)$$

$$P_{2i+1}(t)=\sum_{j=0}^{i} C_{2j+1} \; \frac{t^{2(i-j)+1}}{[2(i-j)+1]!}+C_{2i+2}$$

Since all the $$\; P_{2i+1}(t)$$ polynomials are odd functions and are set to zero at +1 and -1, we have:

$$P_{2i+1}(1)=\sum_{j=0}^{i} C_{2j+1} \; \frac{1}{[2(i-j)+1]!}=0$$

$$\sum_{j=0}^{1} C_{2j+1} \; \frac{1}{[2(1-j)+1]!}=0\Rightarrow \frac{C_{1}}{3!}+\frac{C_{3}}{1!}=0 \;\Rightarrow \frac{C_{3}}{1!} = -\frac{C_{1}}{3!}= -\frac{(-1)}{3!}=\frac{1}{6}$$

similarly,

$$\sum_{j=0}^{2} C_{2j+1} \; \frac{1}{[2(2-j)+1]!}=0\Rightarrow \frac{C_{1}}{5!}+\frac{C_{3}}{3!}+\frac{C_{5}}{1!}=0 \;\Rightarrow \frac{C_{5}}{1!} = -\frac{C_{1}}{5!}-\frac{C_{3}}{3!}= -\frac{(-1)}{5!}-\frac{\frac{1}{6}}{3!}=\frac{-7}{360}$$

$$\sum_{j=0}^{3} C_{2j+1} \; \frac{1}{[2(3-j)+1]!}=0\Rightarrow \frac{C_{1}}{7!}+\frac{C_{3}}{5!}+\frac{C_{5}}{3!}+\frac{C_{7}}{1!}=0 \;\Rightarrow \frac{C_{7}}{1!} = -\frac{C_{1}}{7!}-\frac{C_{3}}{5!}-\frac{C_{5}}{3!}= -\frac{(-1)}{7!}-\frac{\frac{1}{6}}{5!}-\frac{\frac{-7}{360}}{3!}=\frac{31}{15120}$$

Now evaluating the polynomials using the coefficients:

$$P_{2}(t)=\sum_{j=0}^{1}C_{2j+1} \; \frac{t^{2(1-j)}}{[2(1-j)]!}=C_{1}\frac{t^{2}}{2!}+C_{3}=-\frac{t^{2}}{2}+\frac{1}{6}$$

$$P_{3}(t)=\sum_{j=0}^{1}C_{2j+1} \; \frac{t^{[2(1-j)+1]}}{[2(1-j)+1]!}=C_{1}\frac{t^{3}}{3!}+C_{3}\frac{t^{1}}{1!}=-\frac{t^{3}}{6}+\frac{t}{6}$$

Similarly,

$$P_{4}(t)=\sum_{j=0}^{2}C_{2j+1} \; \frac{t^{2(2-j)}}{[2(2-j)]!}=C_{1}\frac{t^{4}}{4!}+C_{3}\frac{t^{2}}{2!}+C_{5}\frac{t^{0}}{0!}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}$$

$$P_{5}(t)=\sum_{j=0}^{2}C_{2j+1} \; \frac{t^{[2(2-j)+1]}}{[2(2-j)+1]!}=C_{1}\frac{t^{5}}{5!}+C_{3}\frac{t^{3}}{3!}+C_{5}\frac{t^{1}}{1!}=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}$$

$$P_{6}(t)=\sum_{j=0}^{3}C_{2j+1} \; \frac{t^{2(3-j)}}{[2(3-j)]!}=C_{1}\frac{t^{6}}{6!}+C_{3}\frac{t^{4}}{4!}+C_{5}\frac{t^{2}}{2!}+C_{7}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}$$

$$P_{7}(t)=\sum_{j=0}^{3}C_{2j+1} \; \frac{t^{[2(3-j)+1]}}{[2(3-j)+1]!}=C_{1}\frac{t^{7}}{7!}+C_{3}\frac{t^{5}}{5!}+C_{5}\frac{t^{3}}{3!}+C_{7}\frac{t^{1}}{1!}=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}$$

Author
Srikanth Madala (SM)

Statement
To write a explanatory commentary on kessler's Matlab code for the evaluation of polynomials and their coefficients in the Trapezoidal error P. 30-2

Solution
Matlab code

Author
Srikanth Madala (SM)

Statement
P. 30-4

An Ellipse as given:

$$ r (\theta) = \frac{a(1-e^{2})}{1-e \cos( \theta)} $$

where:

$$ eccentricity=e= \sin(\frac{\pi}{12}) $$

$$ a=1 $$

Calculate the Circumference of the elipse as follows:

Method 1:

$$ C= \int_{0}^{ 2 \pi } \left[r^{2} + (\frac{dr}{d\theta})^{2} \right] ^{\frac{1}{2}} $$

Method 2:

$$ C = 4aE(e^{2}) $$

$$ E(e^{2}) = \int_{0}^{\frac{\pi}{2}} \left[1 - e^{2}\sin(\theta) \right] ^{\frac{1}{2}}d\theta $$

Method 1
The Circumference will be calculated by using the following:

$$ C= \int_{0}^{ 2 \pi } \left[r^{2} + (\frac{dr}{d\theta})^{2} \right] ^{\frac{1}{2}} $$

And then integrated numerically using 3 methods:

- Composite Trapezoidal - Clencurt - Romberg Table

The approximation for the exact integral was calculated using the "quad" Matlab command. It was approximated to be 6.176517851674653

Composite Trapezoidal
Matlab code for composite Trapezoidal calculation

Matlab code for calculation of the integrand

Clencurt
Matlab code

Romberg Table
Romberg Table for n=16

6.176517900218236  6.176517900385847   6.176517900341150   6.176517900343991   6.176517900343945   6.176517900343941   6.176517900343945   6.176517900343943   6.176517900343947   6.176517900343945   6.176517900343941                   0   6.176517900343944   6.176517900343946   6.176517900343945   6.176517900343941                   0                   0   6.176517900343946   6.176517900343946   6.176517900343941                   0                   0                   0   6.176517900343946   6.176517900343942                   0                   0                   0                   0   6.176517900343943                   0                   0                   0                   0                   0

Error of Romberg Table

 * 1.0e-007 *

-0.485435833752490 -0.487111941893659  -0.486664966103945   -0.486692917078813  -0.486692899315244  -0.486692934842381   -0.486692908197028  -0.486692925960597  -0.486692917078813   -0.486692925960597  -0.486692925960597  -0.486692881551676   -0.486692925960597  -0.486692890433460                   0   -0.486692899315244                   0                   0

Columns 4 through 6

-0.486693378931591 -0.486692917078813  -0.486692881551676   -0.486692917078813  -0.486692881551676                   0   -0.486692881551676                   0                   0                    0                   0                   0                    0                   0                   0                    0                   0                   0

Matlab code

Method 2
$$ \Complex = 4aE(e^{2}) $$

$$ E(e^{2}) = \int_{0}^{\frac{\pi}{2}} \left[1 - e^{2}\sin(\theta) \right] ^{\frac{1}{2}}d\theta $$

The integral is calculated using 3 methods:

- Composite Trapezoidal - "Clencurt" Matlab Command - Romberg Table

It will then by used to find the circumference as previously stated.

Composite Trapezoidal
$$ \Complex = 4 \times I = 6.147423421522488 $$

Romberg Table
1.536876035753892  1.536849128589532   1.536849129666574   1.536849129666558   1.536849129666558   1.536849129666558   1.536855855380622   1.536849129599258   1.536849129666558   1.536849129666558   1.536849129666558                   0   1.536850811044599   1.536849129662352   1.536849129666558   1.536849129666558                   0                   0   1.536849550007914   1.536849129666295   1.536849129666558                   0                   0                   0   1.536849234751700   1.536849129666542                   0                   0                   0                   0   1.536849155937831                   0                   0                   0                   0                   0

$$ I = 1.536849129666558 $$

$$ \Complex = 4 \times I = 4 \times 1.536849129666558 =  6.147396518666232 $$

Both Methods yield similar but not equal results. Also it is noted that the error associated with method 1 is less than that of method 2

Author
--Egm6341.s10.Team2.GV 22:07, 25 March 2010 (UTC)

Statement
Derive the infinitesimal arc length using the Law of cosine

Author
Egm6341.s10.team2.lee 08:57, 24 March 2010 (UTC)

Signatures
Solved problems 4 and 9. Proofread 5,10--Srikanth Madala (SM)

Solved problem 2 and 7. Proofread 1,3,8,10--Egm6341.s10.team2.niki| Niki Nachappa (NN) 17:17, 24 March 2010 (UTC)

Proofread problems 3, 8, 11--Egm6341.s10.team2.patodon 20:57, 24 March 2010 (UTC)

Solved problems 5 and 10. Proofread 1, 4, 11 --Egm6341.s10.Team2.GV 22:14, 25 March 2010 (UTC)

Solved problems 1,6, 11. Proofread 2, 7, 9 --Egm6341.s10.team2.lee 16:28, 26 March 2010 (UTC)


text:
User:Rjorjorjo/HW5

=Find out the constants of integration function $$P_{n}(t)$$=

Statement
Find $$C_{5}\,$$ and $$C_{6}\,$$ Given $$P_{5}(t)= \!\,\int P_{4}(t)dt \!\,= \!\,-\!\,\frac{t^{5}}{120}+ \!\,\frac{t^{3}}{12}+\!\,C_{5}t+\!\,C_{6}$$

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|P26-3]]
 * }
 * }

Solution
(i)Select $$P_{5} :\!\,P_{5}(\pm \!\,1)=\!\,0$$ and $$P_{5} (0)=\!\,0$$

(ii)
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P_{5}(0)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
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\Rightarrow-\frac{0^{5}}{120}+\frac{0^{3}}{36}+C_{5}0+C_{6} = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
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\Rightarrow\,C_{6} = 0 $$ (iii) $$P_{5}(+ \!\,1)=\!\,0$$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
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\Rightarrow-\frac{1^{5}}{120}+\frac{1^{3}}{36}+C_{5}1 = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
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\Rightarrow-\frac{3}{360}+\frac{10}{360}+C_{5} = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
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\Rightarrow C_{5} = -\frac{7}{360} $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

=Redo steps in proof of Trapezoidal error=

Statement
Redo steps in proof of Trapezoidal error by try to cancel terms with odd order of derivative of g

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|P28-4]]
 * }
 * }

Solution
From

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|P26-1]] [[media:Egm6341.s10.mtg21.djvu|P21-3]]
 * }
 * }

\begin{align} \\E &= \int_{+1}^{-1}\underbrace{(-t)}_{P_{1}(t)}g^{1}(t)dt\; \\ &= [\!\,P_{2}(t)g^{1}(t)]\!\,_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_{2}(t)g^{2}(t)dt}_{A}\,. \end{align} $$

In order to cancel odd order of derivative of g, select $$P_{2}(t) = 0 \,\,at\,\, t=\pm\!\,1$$

\begin{align} \\P_{2}(t) &= \int P_{1}(t)dt\; \\ &= -C_{1}\frac{t^{2}}{2}+C_{3}\,where C_{2} = 0\; \\ &= -\frac{t^{2}}{2}+\alpha\!\,\; \end{align} $$ $$P_{2}(t) = 0 \,\,at\,\, t=1$$

$$\rightarrow \alpha\!\,=\frac{1}{2}$$

$$\rightarrow C_{1} = -1\,\,C_{3} = \frac{1}{2}$$

$$ A = [P_{3}(t)g^{2}(t)]_{-1}^{+1} + \underbrace{\int_{-1}^{+1}\underbrace{P_{3}(t)}_{U^{'}}\underbrace{g^{3}(t)}_{V}dt}_{B}$$

$$ B = [UV]_{-1}^{+1} - \int UV^{'} = [P_{4}(t)g^{3}(t)]_{-1}^{+1} - \underbrace{\int_{-1}^{+1}P_{4}(t)g^{4}(t)dt}_{C}$$

In order to cancel odd order of derivative of g, select $$P_{2}(t) = 0 \,\,at\,\, t=\pm\!\,1$$

$$ t = 1 $$

\begin{align} \\P_{4}(1) &= \int P_{3}(t)dt\; \\ &= -\frac{1}{4!\!\,}+\frac{1}{4}+C_{4}+C_{5}\; \\ &= 0\; \end{align} $$

$$\rightarrow C_{4}+C_{5} = \frac{1}{24}-\frac{1}{4} = -\frac{5}{24}$$

$$ t = -1 $$

\begin{align} \\P_{4}(-1) &= \int P_{3}(t)dt\; \\ &= -\frac{1}{4!\!\,}+\frac{1}{4}-C_{4}+C_{5}\; \\ &= 0\; \end{align} $$

$$\rightarrow -C_{4}+C_{5} = \frac{1}{24}-\frac{1}{4} = -\frac{5}{24}$$

$$ C_{5} = -\frac{5}{24}\,\,C_{4} = 0 $$

$$ C = [P_{5}(t)g^{4}(t)]_{-1}^{+1} - \underbrace{\int_{-1}^{+1} P_{5}(t)g^{5}(t)dt}_{D}$$

$$ D = \underbrace{[P_{6}(t)g^{5}(t)]_{-1}^{+1}}_{= 0} - \int_{-1}^{+1} P_{6}(t)g^{6}(t)dt$$

$$ E = -[P_{3}(t)g^{2}(t)+P_{5}(t)g^{4}(t)+...+P_{2l+1}(t)g^{2l}(t)]_{t = -1}^{t = +1} - \int_{-1}^{+1}P_{2l+2}(t)g^{2l+2}(t)dt$$

$$ P_{3}(t) $$ is odd function $$ \rightarrow P_{3}(-1) = -P_{3}(1)$$

$$ [\,P_{3}(t)g^{2}(t)]_{-1}^{1} = P_{3}(1)g^{2}(1) - P_{3}(-1)g^{2}(-1) = P_{3}(1)[g^{2}(1) + g^{2}(-1)]$$

$$ E = -\sum_{r = 1}^{l}P_{2r+1}(1)[g^{2r}(1) + g^{2r}(-1)] - \int_{-1}^{+1}P_{2l+2}(t)g^{2l+2}(t)dt$$

From [[media:Egm6341.s10.mtg21.djvu|p.21-2]] $$ g_{k}^{i} = (\frac{h}{2})^{i}f^{i}(x(t))\,\,x\in\!\,[x_{k},\,x_{k+1}] $$

$$ E = -\sum_{r = 1}^{l}P_{2r+1}(1)[(\frac{h}{2})^{2r}f^{2r}(x(1)) + (\frac{h}{2})^{2r}f^{2r}(x(-1))] - \int_{-1}^{+1}P_{2l+2}(t)(\frac{h}{2})^{2l+2}f^{2l+2}(x)\frac{2}{h}dx$$

$$ \overline{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$

$$ E = -\sum_{r = 1}^{l}2\overline{d_{2r+1}}h^{2r} [f^{2r}(b) + f^{2r}(a)] - (\frac{h}{2})^{2l+2}\frac{2}{h}\sum_{k = 0}^{n - 1}\int_{x_{k}}^{x_{k+1}}P_{2l+2}(t_{k}(x))f^{2l+2}(x)dx$$

\begin{align} \\E_{n}^{1} &= \frac{h}{2}E\; \\ &= -\sum_{r = 1}^{l}\overline{d_{2r+1}}h^{2r-1} [f^{2r}(b) + f^{2r}(a)] - (\frac{h}{2})^{2l+2}\sum_{k = 0}^{n - 1}\int_{x_{k}}^{x_{k+1}}P_{2l+2}(t_{k}(x))f^{2l+2}(x)dx\; \end{align} $$

=Compute the curve length of ellipse=

Statement
Compute C by using two methods:

$$ \displaystyle a)\, C = \int_{\theta = 0}^{\theta = 2\pi}dl \,\,\rightarrow From(4)$$ [[media:Egm6341.s10.mtg30.djvu|P30-3]]

$$ dl = d\theta[r^{2} + (\frac{dr}{d\theta})^{2}]^{\frac{1}{2}}\,\,\rightarrow From(1) $$ [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle b)\, C = 4aE(e^{2}) \,\,\rightarrow From(5)$$ [[media:Egm6341.s10.mtg30.djvu|p.30-4]]

$$ E(e^{2}) = \int_{0}^{\frac{\pi}{2}}[1 - e^{2}sin^{2}\theta]^{\frac{1}{2}}d\theta \rightarrow From(6)$$ [[media:Egm6341.s10.mtg30.djvu|p.30-4]]

Solution
$$ \begin{align} \displaystyle a)\, C = \int_{\theta = 0}^{\theta = 2\pi}dl \\ &=\int_{\theta = 0}^{\theta = 2\pi}d\theta[r^{2} + (\frac{dr}{d\theta})^{2}]^{\frac{1}{2}} \\\end{align}\ $$

$$ r(\theta) = \frac{1 - e^{2}}{1 - ecos\theta} $$ $$ e = sin(\frac{\pi}{12}) $$

$$ \begin{align} C = \int_{\theta = 0}^{\theta = 2\pi}dl \\ &=\int_{\theta = 0}^{\theta = 2\pi}[(\frac{1 - e^{2}}{1-ecos\theta})^{2} + (\frac{(e^{3} - e)sin\theta}{e^{2}cos^{2}\theta - 2ecos\theta + 1})^{2}]^{\frac{1}{2}}d\theta\\ &=\int_{\theta = 0}^{\theta = 2\pi}[1 + (\frac{e\cancelto{1}{(e^{2} - 1)}sin\theta}{(ecos\theta - 1)^{\cancelto{1}{2}}})^{2}(\frac{\cancelto{1}{1-ecos\theta}}{\cancelto{1}{1 - e^{2}}})^{2}]^{\frac{1}{2}}d\theta\\ &=[1 + (\frac{esin\theta}{ecos\theta - 1})^{2}]^{\frac{1}{2}}d\theta \\\end{align}\ $$

Use Matlab to find the value of C

Get I1 = 1.6041  Elapsed time is 0.049997 seconds.

$$ \displaystyle b)\, C = 4a\int_{0}^{\frac{\pi}{2}}[1-e^{2}sin^{2}\theta]^{\frac{1}{2}}d\theta \ $$

Here $$ a\, =\, 1\, $$

Use Matlab to find the value of C

Get I2 = 6.1766  Elapsed time is 0.057046 seconds.

title: User:Rjorjorjo/HW5


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CheckUser

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Sandbox

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Sandbox Server

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Deletion

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School of...

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FAQ

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Custodianship

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Wikiversity Day

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Welcome

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Instructional design/Task analysis/Anatomy of an HTA

Parts of a whole
Below is an image of a completed HTA (minus the objectives). The most basic HTA has three components:

 * The Task = the main goal being observed.
 * Subtasks = actions that must be completed before accomplishing the main goal/task.
 * Elements = actions that must be completed before accomplishing a subtask.

In rare occasions, an element may be broken down into steps, in which case those steps must be completed in order to complete the element.

This image first appeared in the HCI textbook

An HTA begins with the task to be analyzed. It's a novel idea, I know. In this case, the task is making a cup of tea and is labeled with a 0. The task is then broken down into subtasks labeled 1-6. There may be times that a task to be analyzed will be fairly simple and not require subtasks. Shown here, the first subtask has been broken down into elements, which are labeled 1.1-1.4. These elements must be completed before moving to the next subtask of emptying the pot, which will have a new set of elements.

Click next to continue.