text:
User talk:Jade Knight/Archives/2010

Dionysios Appreciates The Jade Knight's Helpful Comments
Thank you, Jade Knight for all your good guidance.

This User, still a Neophyte at Wikiversity, never knew for what purpose the "+" button existed. He will henceforth begin to conform to the practice of the real Wikiversity Regulars and have his Text automatically added at the Bottom and Edit his Discussion Page and any other instances which he finds of his deviation from Standard Practice to also conform.

Thank you also for your recognition that there may be some really good ideas among the fevered imaginings of whatever it is that Your Servant thinks he might be doing, with where he wants to go with Wikiversity (Including some perhaps very Creative and Original projects ) -- Creative and Original being great for Wikiversity right now.

He does attempt to keep in Mind that no one else knows what's going inside his head; and it has become apparent to him that his Fellow Wikversitans are often confused by the seemingly erratic pattern of his edits; and he is becoming aware that his commentary often does not help.

Even though the "Third person" and "The Passive Voice" are generally not very effective Rhetorical Devices, they are where Your Servant finds himself.

If he comes to find that it is just a bad habit, he will break the habit. If, on the other hand, "Third Person; Passive Voice" turns out to be Who He Is, he would then have to continue in his (unintentionally) irritating way.

Many Wikiversity Regulars do not use Headers in the Middle of Discussions and the appearance of Headers in the Middle of a Discussion Already in Progress could be very Jarring and Disorienting. And, Your Servant has to add: The magical Arrival of a Heading might tend to recast a Discussion in a Direction that has nothing to do with what was Previously Going On in that Discussion. The Remedy there would be to Edit the Headers until We have some Consensus as to About What We Are Talking. But the Use of Headers tends to make Quickly Visible at least an Idea of About What the Conversation Deals.

Your Servant has filled this Response with subheaders; but certainly he Refrains from Going to Work on Your Discussion Page, Editing It with his Subheadering Interpretations. ¿When you request that the Discussion not be Broken with Added Headers, certainly you are not suggesting that Subheaders not be used within a Comment?

Your Servant has the Goal of communicating what it is he is trying to do. To do so "in simple, easy-to-understand terms" is for him Icing on the Cake. First he desires with desire that he should communicate; then, he desires with desire that he should do so with simplicity.

Fuzzy Thinking, Fuzzy Communications. Your Servant does not mean to be Obscure; it just comes naturally. He strives to to do better.

He does what he can to overcome any and all obstacles to effective communication and thanks you for your most helpful assistance in that regard.

It is a Mutual Admiration Society. Your Servant knows from seeing Your Work that you are an most essential part of This Work and is most sincerely appreciative that you should take the time to be of assistance to a Wikiversity Neophyte. Thank you again for all your good guidance.

(s) Dionysios(talk), a Participant in the Wikiversity Institute for Thought, Date: 2007-07-16 (July 16, 2007) Time: 1631 UTC

 * Strangely, Third Person and Passive Voice are very comfortable for Your Servant. He is aware and reminded of the mild disorientation which can occur and finds First Person and Active Voice equally capable of producing a little disorientation in him.

 * Thank you for your continued guidance on the matter of Subheaders and for removing the unwanted Subheaders. Your Servant will no longer litter our communications with Subheaders.

 * Your Servant cares carefully when making Links, using google: and much of the whole family of Wikimedia projects, including m:, foundation:, b:, n:, w:, q:, wikimedia:, and, of course, wikt:.

 * The primary question, then, with reference to Your Servant's Links is whether they are helpful; and he regrets that you often do not find them so; but, then, often, Your Servant is hoping to do more than provide a definition for a Word.

 * He Links to your User Page because he knows that someone else may come along and want to go there at the point where he made the Link.

 * But, again, thank you for all your good guidance. If you will continue to be patient with Your Servant, you may eventually find reason to rejoice over your efforts bearing fruit.

 * (s) Dionysios(talk), Date: 2007-07-16 (July 16, 2007) Time: 2359 UTC

The Jade Knight (d'viser) 05:06, 23 March 2010 (UTC)

spanish course
Hi, im the teacher you answered in discussion. I won't register if i don't see it's worth it, and now I don't see how can i help teaching Spanish, it's just a mess. I suggest to start not by lessons but by topics: shopping, introducing youself, travelling... In each topic you can put the vocabulary, the grammar related, a little text (better with audio recordings) and some exercises. Then, each 4 or five topics, you can put a test or ask for projects. There should be also chat time (the chat channels of wikipedia or similars can be perfect!) You can order a little bit the courses or answer me in [[w:ca:Usuari:Barcelona]. I will keep looking at the project to see if starts to be a real course. Thanks

The Jade Knight (d'viser) 05:07, 23 March 2010 (UTC)

Could the Proposal to Merge the Advanced General Studies Portal and the School of Advanced General Studies Please be Withdrawn?
Greetings Jade Knight!

The original intention was that the Advanced General Studies portal should eventually become a major portal, directing users to those friendly introductions to learning resources promised on the major portal page as well as providing links to Wikiversity schools and other content development projects in the "Topic:" namespace and that the the school should be a major Wikiversity school.

It is beyond the understanding of Your servant how the proposed collapse of the portal into the school would make anything easier for anyone.

So far, the primary effect of the "decoration" of the pages of the portal and the school with a  template appears to have been to greatly discourage the the founder of the the portal and the school.

Your servant would be very reluctant to continue with work which might very will be destroyed should the proposed Merger actually take place.

Nobody other than the proposer of the merger and the founder of the the portal and the school has posted comments after an urgent Request for Comments was placed on the the portal and school pages nearly one (1) month ago.

Perhaps it is time to withdraw the proposal and let the the founder of the the portal and the school return to development of these pages.

Could the proposal to merge the the portal and the the school please be withdrawn? It could be renewed without prejudice if you were later to find that good progress toward the Advanced General Studies Portal becoming a major portal is not being made.

Best regards,

(s) Dionysios  (talk), a Participant in the Wikiversity School of Advanced General Studies, Date: 2007-08-29 (August 29, 2007) Time:  2156 UTC

The Jade Knight (d'viser) 05:07, 23 March 2010 (UTC)

Welcome Back, My Friend and Adversary!
Welcome Back, My Friend and Adversary! Your Servant has missed you while you have been away. -- Dionysios  (talk), 2007-09-09 (September 9, 2007) Time:  2057 UTC

The Jade Knight (d'viser) 05:07, 23 March 2010 (UTC)

Gastronomy
Thanks for your note. I tried to include as much existing content as possible. I attempted to organise the departmental structure in a hierarchical way to (eventually) encompass as many related/dependent areas as possible. The two major divisions are: (My intention here was that Beverages could be further subdivided to include Alcoholic/non alcoholic. Then into beers, wines and spirits and hot and cold beverages respectively. These will each have production, scientific, historical and service aspects meriting courses.)

The Food division has Departments of:

There is also a major Sub-division of: (This (above) division need to take in commercial and industrial aspects including the restaurant/food service trade with its various aspects.) which has Departments of:

and hosts courses on: My thinking was influenced by the prospectus of the University of Gastronomic Sciences in Italy.

The more practical (as opposed to theoretical) aspects of the course are to be based on conventional cookery and food service courses and examination syllabuses, with additional material to cover a more global perspective.

I hope this helps. Get back to me with further questions. I have been extra busy in the real world lately but I shall be back making contributions before long. --Harkey 10:21, 31 March 2010 (UTC)

William Hamblin congratulates Jade Knight for his fine ground work
Hello Jade Knight, just wanted to say I've seen so very much of you as I have tried to pull together a military history department and link in with the big scheme. I love the wikimedia projects and am fondly looking forward to pulling together the military history and brushing up what I can of tertiary history in general. William Hamblin 12:33, 15 May 2010 (UTC)

Hello
Where have you been? Ottava Rima (talk) 15:00, 29 July 2010 (UTC)

title: User talk:Jade Knight/Archives/2010


text:
User:Egm6341.s11.team3/sub5

= 5.1 =

Given: A number of functions with their finite bounds
I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2}, x \in [0, 6] \ $$

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx, x \in [0, 1] \ $$

Find: Integrate them by different integration methods
1. Find the integral by using Composite Trapezoid, Composite Simpson's rules and plot the functions

a) By using uniform discretization b) By using non-uniform discretization

2. Use Gauss-Lagrange method to approximate the function

Composite Trapezoid rule is given by Vu Quoc's notes
$$ I_n^t = h[\frac{1}{2}f_0 + f_1 +f_2 +.... + f_{n-1} + \frac{1}{2}f_n] $$

Matlab code was written to calculate the generic functions (need to input function and B.C. and interval and the code generates number of elements, approximate integral value, and the error.

For:

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \ $$

Exact versus approximate function plots, blue is exact/analytical and red points are approximate function at 100 points

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

Matab output

Exact versus approximate function plot

Composite Simpson rule is given by Vu Quoc's notes is
$$ I_n^t = \frac{h}{3}[f_0 + 4f_1 +2f_2 +4f_3 +2f_4 +.... + 2f_{n-2} + 4f_{n-1} + f_n] $$

Matlab code was written to calculate the generic functions (need to input function and B.C. and interval and the code generates number of elements, approximate integral value, and the error.

For:

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \ $$

Exact versus approximate function plots, blue is exact/analytical and red points are approximate function at 100 points

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

Matab output

Exact versus approximate function plot

Gauss - Legendre quadrature rule
For this method we had to use the variable transformation since the defined Gauss-Legendre quadrature is in the interval $$ [-1, +1] \ $$

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \in x = [0, 6] \ $$

Therefore we converted to new value "z", that is $$ x \in [0, 6] \rightarrow z \in  [-1, +1] \ $$

Therefore, the conversion became:

$$ z(x) = \frac{x}{3} - 1 \ $$

For check, we checked that z is linear in x as well as boundary conditions.

$$ x = min = 0 \rightarrow z(0) = \frac{0}{3} - 1 = -1 \ $$

$$ x = max = 6 \rightarrow z(6) = \frac{6}{3} - 1 = 1 \ $$

Therefore, it's consistent

Note that 1st thing was done in conversion from "x" to "z(x)" was adjustment of length "z" = 2, while "x" = 6, therefore it was divided by 3 1st and then adjusted to fit the $$ [-1, +1] \ $$ interval.

Finally, since

$$ 3*z(x) = x \ $$

the final integral of Gauss - Legendre quadrature had to be multiplied by factor of "3" to represent the integrated function.

Below is attached matlab code for Gauss - Legendre quadrature for n = 2, 3, 4, 5

and the Gauss - Legendre quadrature integration output on that function

For

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

$$ z = 2x - 1 \ $$

and the Gauss - Legendre quadrature integration output on that function

--Egm6341.s11.team3.JA 03:21, 24 March 2011 (UTC) and Kim

= 5.2 =

Given

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

$$\underline{F} = \underline{I} + \triangle \underline{A} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.1)
 * }

$$\underline{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \underline{A} = \begin{pmatrix} -0.2& 1 \\ -1 & -0.2 \end{pmatrix}, \quad \triangle = 0.02 \quad \underline{x_0}= \dbinom{{{x}_{0}}^{1}}{{{x}_{0}}^{2}} $$

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

$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.2)
 * }

Find
$$ Run \quad LSSM \quad and \quad plot \quad({x}_{j}, j=0,1,...)$$

Solution
$$ By \quad (Eq 5.2.1) \quad $$ $$ \underline{F}=\begin{pmatrix} 0.9960 & 0.200 \\ -0.200 & 0.9960 \end{pmatrix}$$

$$ By \quad (Eq 5.2.2) \quad build \quad code \quad$$

Result is

Given

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

$$\lim_{k \to \infty}\underline{x}_{k}=\lim_{k \to \infty} \underline^{k+1}\underline{{x}_{0}} = \underline{\hat x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.3)
 * }

Find
Find Equilibrium   Point plot $$\underline{\hat x} $$ as big red dot $$\underline{{x}_{0}} $$as big blue dot

Solution
By (Eq5.2.3)

Result is

Given

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

$$ LSSMRN: \quad \underline{{x}_{k+1}} = \underline{F} \quad \underline{x_k} + \underline{G} \quad \underline{w}_{k+1}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 7.0.1)
 * }

$$\underline{G}= \dbinom{1}{1} \alpha\ $$ where $$\alpha\ = 0.5, 1, 2$$

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$

Solution
By Eq(5.2.4)

Result is when $$\alpha\ = 0.5$$

when $$\alpha\ = 1$$

when $$\alpha\ = 2$$

Given
Cauchy Random Noise

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$ in single-slit diffraction

Solution
by (Eq5.2.4) with Cauchy Random Noise

Result is Egm6341.s11.team3.sungsikkim 20:43, 22 March 2011 (UTC)

= 5.3 =

Given:
Given a pdf function $$ f(x) \quad $$ and the defination of Quartile points: $$ Q_1, Q_3 \quad $$
 * {| style="width:100%" border="0"

$$ F(Q_1) = P(x $$ \displaystyle
 * }

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

$$ F(Q_3) = P(x $$ \displaystyle (Eq. 5.3.1)
 * }

The Cauchy pdf $$ C(x_0,r) \quad $$ and Gauss (normal) pdf $$ N(\mu,\sigma) \quad $$ are defined as following:

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

$$ C(x_0,r) := \frac{r}{\pi[r^2+(x-x_0)^2]} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.3.2)
 * }

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

$$ N(\mu,\sigma) := \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.3.3)
 * }

with the half width represented by $$ r^C \quad $$ and $$ r^G \quad $$ respectively.

Find:
1) Find $$\quad Q_1, Q_3 $$ for $$ C(x_0,r) \quad $$.

2) Find $$\quad Q_1, Q_3 $$ for $$ N(\mu,\sigma) \quad $$.

3) Let $$ x_0 = \mu = 0 \quad $$ and $$ r^C = 1 \quad $$. Find $$ \sigma^1 \quad $$ so that $$ r^G =1 \quad $$. Plot $$ C(x_0,r) \quad $$ and $$ N(\mu,\sigma) \quad $$.

4) Find: $$ Q_1^C, Q_3^C \quad $$ for $$ C(x_0,r) \quad $$ and $$ C(0,1) \quad $$, and $$ Q_1^G, Q_3^G \quad $$ for $$ N(\mu,\sigma) \quad $$ and $$ N(0,\sigma^1) \quad $$.

Plot: $$ Q_1^C, Q_3^C \quad $$ for $$ C(0,1) \quad $$,and $$ Q_1^G, Q_3^G \quad $$ for $$ N(0,\sigma^1) \quad $$. Comment on results.

Solution:
1)

Integrate Cauchy pdf function, we get

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

$$ \int C(x_0,r)dx = \int \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} tan^{-1}(\frac{x-x_0}{r}) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.3.4)
 * }

For $$ Q_1 \quad $$, we have

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

$$ F(Q_1) = \int_{-\infty}^{Q_1} \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} [tan^{-1}(\frac{Q_1-x_0}{r}) - tan^{-1}(\frac{-\infty-x_0}{r})] = 0.25 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow tan^{-1}(\frac{Q_1-x_0}{r}) = -\frac{\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow Q_1 = x_0 - r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For $$ Q_3 \quad $$, we have

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

$$ F(Q_3) = \int_{-\infty}^{Q_3} \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} [tan^{-1}(\frac{Q_3-x_0}{r}) - tan^{-1}(\frac{-\infty-x_0}{r})] = 0.75 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow tan^{-1}(\frac{Q_3-x_0}{r}) = \frac{\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow Q_3 = x_0 + r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

2)

Integrate Gauss pdf function with the help from WolframAlpha, we get

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

$$ \int N(\mu,\sigma)dx = \int \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}} dx = \frac{1}{2} erf(\frac{x-\mu}{\sqrt{2}\sigma}) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.3.5)
 * }

For $$ Q_1 \quad $$, we have

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

$$ F(Q_1) = \int_{-\infty}^{Q_1} N(\mu,\sigma)dx = \frac{1}{2} [erf(\frac{Q_1-\mu}{\sqrt{2}\sigma}) - erf(\frac{-\infty-\mu}{\sqrt{2}\sigma})] = 0.25 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow erf(\frac{Q_1-\mu}{\sqrt{2}\sigma}) = -0.5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

solve from WolframAlpha, we get

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

$$ \frac{Q_1-\mu}{\sqrt{2}\sigma} = -0.476936 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow Q_1 = \mu - 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For $$ Q_3 \quad $$, we have

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

$$ F(Q_3) = \int_{-\infty}^{Q_3} N(\mu,\sigma)dx = \frac{1}{2} [erf(\frac{Q_3-\mu}{\sqrt{2}\sigma}) - erf(\frac{-\infty-\mu}{\sqrt{2}\sigma})] = 0.75 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow erf(\frac{Q_3-\mu}{\sqrt{2}\sigma}) = 0.5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

solve from WolframAlpha, we get

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

$$ \frac{Q_3-\mu}{\sqrt{2}\sigma} = 0.476936 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow Q_3 = \mu + 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

3)

From defination of half width:
 * {| style="width:100%" border="0"

$$ \frac{N}{N_{max}} = e^{\frac{-(r^G)^2}{2(\sigma^1)^2}} = 0.5 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

solve from WolframAlpha ,

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

$$ \Rightarrow \frac{-1}{2(\sigma^1)^2} = ln(0.5) = -0.693147 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow \sigma^1 = 0.849322 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.3.6)
 * }

The plot of $$ C(x_0,r) \quad $$ and $$ N(\mu,\sigma) \quad $$ are shown in part 4).

4)

From part 1) and part 2)

For $$ C(x_0,r) \quad $$

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

$$ Q_1^C = x_0 - r $$ and $$ Q_3^C = x_0 + r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For $$ C(0,1) \quad $$

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

$$ Q_1^C = -1 $$ and $$ Q_3^C = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For $$ N(\mu,\sigma) \quad $$

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

$$ Q_1^G = \mu - 0.674489\sigma $$ and $$ Q_3^G = \mu + 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For $$ N(0,\sigma^1) \quad $$, with $$ \sigma^1 \quad $$ given in (Eq. 5.3.6)

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

$$ Q_1^G = -0.572858 $$ and $$ Q_3^G = 0.572858 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

The plot is shown in the following figure:

Comments: By comparing the probability density fuctions of Cauchy and Gauss with the same width, a heavy tail of the Cauchy pdf is observed. Furthermore, because of the heavy tail, the distance between the 1st and 3rd quartile points of Cauchy pdf are larger than that of Gauss pdf.

Egm6341.s11.team3.Xia 2:00, 22 March 2011 (UTC)

Given
Applications and implementation logic for 5 different numerical methods given below ,

1. Taylor Series 2. Composite Trapezoidal rule 3. Composite Simpson's rule 4. Romberg Richardson's rule 5. CTk

Find
The Pros and Cons of the 5 different techniques

Solution
 Taylor Series :

 Pros:

1. Approximates simple or complex functions as a polynomial with remainder representing the error.

2. Higher order T-Series can solve for oscillatory functions as well.

 Cons:

1. One of the main drawback is that the function should be differentiable in the given limits in which it is approximated as a polynomial.

2. Number of iterations can be large depending on the problem. But depending on the problem in-hand we could choose a smaller or larger interval ,which in turn would decrease or increase the processing time / resources respectively.

Composite Trapezoidal rule:

 Pros:

1. Periodic functions can be analyzed with ease and accuracy because the error is dependent on the difference between the odd derivatives of the limits.

2. Better flexibility/operational procedure than simple trapezoidal rule and more accurate.

 Cons:

1. Convergence time lapse is very slow when compared to other techniques.

2. We need to make sure to have sufficiently large number of intervals in the domain.

Composite Simpson's rule:

 Pros:

1. Exactly solves for polynomials of order 3 or less.

2. It is often advantageous to use sub-intervals of different lengths, and concentrate on the places where the integrand is not well-behaved.

 Cons:

1. It is applicable to only even number of intervals in the domains. So the value of 'n' in Composite Simpson's rule is always even. i.e. n=2i; where (i=1,2,3,4....)

2. Not possible to choose odd number of intervals because of the even domain.

 Romberg Richardson's Rule:

 Pros:

1. The successive computation of $$\;I(2n)$$ is less resource consuming and less computational time when $$\;I(n)$$ is already available.

2. Relatively higher accuracy when compared to the other methods and is pretty robust.

3. Applied for energy cascade in turbulence and in weather predictions

4. Applicable to numerical integration of ODEs to estimate the error.

 Cons:

1. Romberg integration is successful only when integrand satisfies the hypotheses of the Euler–Maclaurin.

2. The integrand must have continuous derivatives, though fairly good results may be obtained if only a few derivatives exist

 CTk:

 Pros:

1. Similar to their counter part, it is useful for solving periodic functions exactly

 Cons:

1. The function derivatives have to be solved for first ,at the end points of the domain which is given.

2. Consumes considerably more memory resources compared to Composite rule.

= 5.8  =

Given:

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

$$ E_n^T = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} f(x)dx - \frac{h}{2}(f(x_k) + f(x_{k+1}))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.8.1)
 * }

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

$$ x(t) := \frac{x_k+x_{k+1}}{2} + \frac{h}{2} t \quad $$, $$ t \in [-1,1] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.8.2)
 * }

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

$$ g_k(t) := f(x(t)) \quad $$, $$ x \in [x_k,x_{k+1}] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.8.3)
 * }

Find:
1) Derive the following equation:

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

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

2) Derive the following equation:

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$$ \int_{-1}^{1} (-t)g^{(1)}(t)dt = \int_{-1}^{1} g(t)dt - [g(-1) + g(1)] \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

3) Derive the following equation:

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$$ g_k^{(i)}(t) = (\frac{h}{2})^if^{(i)}(x(t)) \quad $$, $$ x \in [x_k,x_{k+1}] \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Solution:
1) From (Eq. 5.8.1), (Eq. 5.8.2) and (Eq. 5.8.3)

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$$ E_n^T = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} f(x(t))dx(t) - \frac{h}{2}(g_k(-1) + g_k(1))] \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} g_k(t)d(\frac{x_k+x_{k+1}}{2} + \frac{h}{2} t) - \frac{h}{2}(g_k(-1) + g_k(1))] \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

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

2) Integration by parts:

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$$ \int_{-1}^{1} (-t)g^{(1)}(t)dt \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = [(-t) \int g(t)^{(1)}dt]_{-1}^{1} - \int_{-1}^{1} (-t)^{(1)} (\int g(t)^{(1)}dt) dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = [(-t) g(t)]_{-1}^{1} - \int_{-1}^{1} (-t)^{(1)} g(t) dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = [(-1) g(1)-(1) g(-1)] - \int_{-1}^{1} (-1) g(t) dt \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = \int_{-1}^{1} g(t) dt - [g(1)+ g(-1)] \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

3) We can build from i=1:

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$$ g_k^{(1)}(t) = f^{(1)} x^{(1)}(t) \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = f^{(1)} \frac{h}{2} \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

so

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$$ g_k^{(2)}(t) \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = (g_k^{(1)}(t))^{(1)} \frac{h}{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = (f^{(1)} \frac{h}{2})^{(1)} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = f^{(2)} x^{(1)}(t) \frac{h}{2} \quad $$ $$
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 * style="width:95%" |
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 $$ \displaystyle
 * }

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$$ = f^{(2)} (\frac{h}{2})^{2} \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 5.8.4)
 * }

From observation of (Eq. 5.8.4), it is implied:

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$$ g_k^{(i-1)}(t) = f^{(i-1)} (\frac{h}{2})^{i-1} \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

so

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$$ g_k^{(i)}(t) \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = (f^{(i-1)} (\frac{h}{2})^{i-1})^{(1)} \quad $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = f^{(i)} x^{(1)}(t) (\frac{h}{2})^{i-1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = f^{(i)} (\frac{h}{2})^{i} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Egm6341.s11.team3.Xia 20:00, 23 March 2011 (UTC)

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

title: User:Egm6341.s11.team3/sub5


text: File:Nm1 s11 team1 HW5 p4 fig c.png

title: File:Nm1 s11 team1 HW5 p4 fig c.png