text:
User:Rjorjorjo/HW2

EGM6341 HW 2

=Problem 9-2 (2)=

Statement
$$f\left( x \right)=\!\frac{e^x-1}{x}\,\,on\,\,[\!\,0,1\,]\!$$

$$x_0=\!a=\!0,\,x_n=\!b=\,1$$

Consider $$n=\!1,2,4,8,16$$

Constr. $$f_n\left( x \right)=\,\sum_{i=0}^{n}l_{i,n}\left ( x \right)f\left ( x_i \right)$$

Plot $$f.\!\,\, f_n, n=\!1,\,2,\,4,\,8,\,16$$

Comp. $$ I_n=\!\int_{a}^{b}f_n\left ( x \right)dx,\,n=\!1,2,4,8\,$$ and compare to $$ I\!$$

For $$ n=\!4\,$$, plot $$\,l_0,\,l_1,\,l_2 $$ why not $$\,l_3,\,l_4,\,$$ also $$2\,$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * }
 * }

MATLAB
Plot $$f.\!\,\, f_n, n=\!1,\,2,\,4,\,8,\,16:\!$$

For $$ n=\!4\,$$, plot $$\,l_0,\,l_1,\,l_2 $$ why not $$\,l_3,\,l_4,\,$$

For $$ n=\!2\,$$, plot $$\,l_0,\,l_1,\,l_2 \,$$

=Problem 12-2=

Statement
Why :$$E_{n}^{n+1}\left ( x \right )=F^{n+1}\left ( x \right)-0$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg6.pdf|P6-4]], [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * }
 * }

From [[media:Egm6341.s10.mtg6.pdf|P6-4]]

\begin{align} \\E_{n} &= I-I_{n}\; \\ &= \int_{0}^{1}f\left ( x \right )dx-\int_{0}^{1}f_{n}\left ( x \right )dx\,. \end{align} $$

the first derivative of $$E_{n}\,$$ is

 * $$E_n^{1}=\int_{0}^{1}f^{1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{1}\left ( x \right )dx\,$$

the second derivative of $$E_{n}\,$$ is

 * $$E_n^{2}=\int_{0}^{1}f^{2}\left ( x \right )dx-\int_{0}^{1}f_{n}^{2}\left ( x \right )dx\,$$

 * {| style=text-align:center

 * width=2% | $$.\,$$
 * width=2% | $$.\,$$
 * width=2% | $$.\,$$
 * }
 * width=2% | $$.\,$$
 * }
 * }
 * }

 * $$E_n^{n}=\int_{0}^{1}f^{n}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n}\left ( x \right )dx\,$$

 * $$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$

From [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * $$f_{n}\left ( x \right )=\sum _{i=0}^{n}l_{i,n}\left ( x \right ) f\left ( x_i \right)\,$$

 * (i)where $$l_{i,n}\left ( x \right)=l_{i}\left ( x \right)=\prod_{j=0,i \neq j}^{n}\frac{x-x_{j}}{x_{0}-x_{j}}\,$$

 * when $$i=0\,$$
 * {| style=text-align:center

\begin{align} \\n=1, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}} \rightarrow l_{0}=l\left ( x^{1}\right)\; \\n=2, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}} \rightarrow l_{0}=l\left ( x^{2}\right)\; \\.\, \\.\, \\.\, \\n=n, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}...\frac{x-x_{n-1}}{x_{0}-x_{n-1}}\frac{x-x_{n}}{x_{0}-x_{n}} \rightarrow l_{0}=l\left ( x^{n}\right)\, \end{align} $$
 * }
 * }

 * (ii)$$f\left( x_{i} \right)\,$$ is constant, so $$f_n\left( x \right)\,$$ is $$x^{n}\,$$'s function

 * $$f_n^{n+1}\left ( x \right)=0\,$$

 * $$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$

 * $$\rightarrow E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1} 0 dx\,$$

 * $$\rightarrow E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \, $$
 * $$\displaystyle\!$$
 * }
 * }

=Problem 15-1=

Statement
Prove why $$\alpha^{1}\left ( t \right )=F\left ( -t \right)+F\left ( t \right)\,$$

Given
$$\alpha\left ( t \right )=\int_{-t}^{k}f\left ( x\left ( t \right) \right)dt+\int_{k}^{t}f\left ( x\left ( t \right) \right)dt, k\in\![-t,t]\, $$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg14.pdf|P14-2]]
 * }
 * }

(i)$$\int_{k}^{t}F\left ( x\left ( t \right) \right)dt=\mathbb{F}\!\left ( x\left ( t \right) \right)-\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$

 * $$\frac{d}{dt}(\int_{k}^{t}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( t \right) \right))-\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))\, $$

 * since$$k=constant\,\rightarrow\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$ is constant

\begin{align} \\\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( t \right) \right)dt)-\frac{d}{dt}(constant) &=F\left ( t \right)-0\; \\ &= F\left ( t \right)\, \end{align} $$

(ii)$$\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt=\mathbb{F}\!\left ( x\left ( k \right) \right)-\mathbb{F}\!\left ( x\left ( -t \right) \right)\, $$

 * $$\frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))+\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right))\, $$
 * P.S. Chain rule :$$\frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))-(\frac{d(-t)}{dt})\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right))\, $$

 * since$$k=constant\,\rightarrow\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$ is constant

\begin{align} \\\frac{d}{dt}(constant)+\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right)dt) &=0+F\left ( t \right)+0\; \\ &=F\left ( -t \right)\, \end{align} $$

From (i) and (ii)

\begin{align} \\\alpha^{1}\left ( t \right ) &= \frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)+\frac{d}{dt}(\int_{k}^{t}F\left ( x\left ( t \right) \right)dt)\; \\ &= F\left ( -t \right)+F\left ( t \right)\, \end{align} $$
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\ alpha^{1}\left ( t \right )=\! F\left ( -t \right)+F\left ( t \right) \, $$
 * $$\displaystyle\!$$
 * }
 * }

=Problem 15-2=

Statement
Prove why $$G^{(2)}\left ( 0 \right )=0\,$$

Given
$$G^{(1)}\left ( 0 \right )=0\,$$ and $$G^{(1)}\left ( \xi_1 \right )=0\,$$ Apply Rolle's theorem
 * $$ \exists\!\,\,\,\,\xi_2\,\,\,\,\in\!\,\,\,\,[\,0,\xi_1\,]:\,\,\,\,\!G^{(2)}\left ( \xi_2 \right)=0\,$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|P15-1]],[[media:Egm6341.s10.mtg14.pdf|P14-2]]
 * }
 * }

From [[media:Egm6341.s10.mtg14.pdf|P14-2]]

$$G\left( t \right):=\!\,e\left( t \right)-\!t^{5}G\left( 1 \right)\,$$

Do [[media:Egm6341.s10.mtg15.pdf|P15-1]] again to find out $$G^{(2)}\left ( 0 \right)=\!0$$

$$G^{(1)}\left( t \right)=\!\,e^{(1)}\left( t \right)-\!5t^{4}e\,$$

$$G^{(2)}\left( t \right)=\!\,e^{(2)}\left( t \right)-\!20t^{3}e\,$$

We need $$ e^{(2)}\left ( t \right)$$

 * $$e\left( t \right):=\!\alpha\left ( t \right)-\!\alpha_2\left ( t \right)\,$$

 * $$\alpha\left ( t \right)=\!\int_{-t}^{k}F\left ( -t \right)+\!\int_{k}^{t}F\left ( t \right)\,\,\,\,k\in\!\,\,[\,-t,t\,]\,$$

 * $$\alpha_2\left ( t \right)=\!\frac{t}{3}[\!\,F\left ( -t \right)+\!4F\left ( 0 \right)+\!F\left ( t \right)]\!\,$$

$$\rightarrow e^{(2)}\left( t \right):=\!\alpha^{(2)}\left ( t \right)-\!\alpha_2^{(2)}\left ( t \right)\,$$

We need $$\alpha^{(2)}\left ( t \right)\,\,\,$$ and $$\,\,\, \alpha_2^{(2)}\left ( t \right)\,$$

 * $$\alpha^{(1)}\left ( t \right)=\!F\left ( -t \right)+\!F\left ( t \right)\,$$

 * $$\alpha_2^{(1)}\left ( t \right)=\!\frac{1}{3}[\!\,F\left ( -t \right)+\!4F\left ( 0 \right)+\!F\left ( t \right)]\!+\!\frac{t}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!\,$$

$$\alpha^{(2)}\left ( t \right)=\!-\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)\,$$

$$\alpha_2^{(2)}\left ( t \right)=\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( -t \right)+\!\,F^{(1)}\left ( t \right)]\!+\!\,\frac{1}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!+\!\frac{t}{3}[\!\,+\!F^{(2)}\left ( -t \right)+\! F^{(2)}\left ( t \right)]\!\,$$

Finally

\begin{align} \\G^{(2)}\left( t \right) &= e^{(2)}\left( t \right)-\!20t^{3}e\; \\ &= \alpha^{(2)}\left ( t \right)-\!\alpha_2^{(2)}\left ( t \right)-\!20t^{3}e\; \\ &= -\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)-\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)]\!-\!\frac{1}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!-\!\frac{t}{3}[\!\,+\!F^{(2)}\left ( -t \right)+\! F^{(2)}\left ( t \right)]\!-\!20t^{3}e\,. \end{align} $$

When $$t=\!0$$

\begin{align} \\G^{(2)}\left( 0 \right) &= e^{(2)}\left( 0 \right)-\!0\; \\ &= \alpha^{(2)}\left ( 0 \right)-\!\alpha_2^{(2)}\left ( 0 \right)-\!0\; \\ &= -\!F^{(1)}\left ( 0 \right)+\!F^{(1)}\left ( 0 \right)-\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( 0 \right)+\!F^{(1)}\left ( 0 \right)]\!-\!\frac{1}{3}[\!\,-F^{(1)}\left ( 0 \right)+\! F^{(1)}\left ( 0 \right)]\!-\!0-\!0\; \\ &= 0\, \end{align} $$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

G^{(2)}\left ( 0 \right)\,=\!\,0 \, $$
 * $$\displaystyle\!$$
 * }
 * }

title: User:Rjorjorjo/HW2


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

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Facebook improvements

This article deals around improvements that could possibly be integrated into Facebook.

Some time ago, I became aware of Instituut voor Veteranen-NIOOO's "Live and Remember" project (http://www.liveandremember.be/ ), which publishes the personal information of veterans on facebook. Although this is a good project to put ex-combatants back into the spotlights, the project clearly exposes a major defect in Facebook. Facebook is in fact originally designed as a comprehensive "white pages", or a database of (living) individuals to list their contact information However, given the long list of people who are deceased, and the fact that the list of deceased people (which were still alive at the time of their addition) is increasingly longer, the creation of separate categories (for living persons and deceased persons) seems inevitable.

Facebook is the most visited website in social mediia. It is popular in over the world that you can do everything to express something. It's continuosly improving because billion of people use this site.you can comment,post,react,share,upload, and many more. It is recommendable site because ii is very updated of this generation.

At the moment the people in Facebook have not been categorized as such. A categorization, and a slightly different lay-out of the contact information in this category (eg "Location" becomes "Past location", the searching of the name of a deceased person, and their location is split-up into the two categories (to avoid having living people/locations (and vice versa) found in the search), ... "Nickname" should also be added (eg to facilitate the searching on well-known deceased persons). An example of this is eg Avicenna, Dioscorides, Pliny ... A directive might be prepared for the "Personal Data", eg in the example of Avicenna, Dioscorides, ... no extensive information should be listed; inventions and personal achievements can be listed in online encyclopedias such as wikipedia ( a link to the page on wikipedia can be placed above). Group Feed. In spite that there are millions of groups and the average user belongs to several dozens of groups, there is yet no effective way to monitor the groups. A group feed of the groups you belong to would take care part of the problem. Add friends to your different lists while looking at their profiles.Custom feed when opening Home. Before the change in which Home opens with Top News with the ability to wobble to Most Recent, I could have the feed of selected friends show when opening Facebook.Download the emails of your friends. Possible to write more than 20 friends at the time.Send a friend request to fans of pages you manage. Send friend requests to members of groups that you manage. Upload sound files that would play on your Page.See Friends Status in a list, which also include the hidden friends.Able to edit lists of friends in Home as well as in Profile.

Facebook is the most visited website.it is popular in over the world that you can do everything to express something. It's continuosly improving because billion of peope use this site.you can comment,post,react,share,uplpaf and etc. It is recommendable site because it is very updated of this generation.

Facebook could also discard the divisioning in the various countries (eg using seperate websites as http://nl-nl.facebook.com/, http://fr-fr.facebook.com/, ...), so that the aspiration to attain a global “white pages” could be better fulfilled (ie people are not bound to a particular country). Also, in the future, facebook could probably be linked with the personal information the national authorities possess (and with electronic identity cards, ...). This information could then be used to take out non-existent persons, and false personal information from the database. It would also keep the personal information more up-to-date.

Facebook is useful if it is used properly. It is efficient and effective in disseminate information, just one click the message sent and received by all the concerned in a matter seconds. In additional, it is an instrument or way to communicate to your love one no matter they are across the miles. Likewise, it keeps the family members relationship enhances and sustains though they are away from one another. It also plays an important role in education like it serves as medium of communication and distribution of information between teacher and student and it keep us updated anytime.

Finally, there is also the problem that companies are more and more being integrated into Facebook. As Facebook is primarily focused on being a “white pages” and not a “yellow pages”, this information is best also taken out. Instead, perhaps a new database can be made (eg called “Yellow pages Online” or “Companybook”) to place the companies in. This would keep the database simpler and more efficient (especially as people from more and more countries being are added.

Facebook, Inc. held its initial public offering (IPO) in February 2012, and began selling stock to the public three months later, reaching an original peak market capitalization of $104 billion. On July 13, 2015, Facebook became the fastest company in the Standard & Poor's 500 Index to reach a market cap of $250 billion.[11] Facebook has more than 1.86 billion monthly active users as of December 31, 2016.[12] As of April 2016, Facebook was the most popular social networking site in the world, based on the number of active user accounts.[13] Facebook classifies users from the ages of 13 to 18 as minors and therefore sets their profiles to share content with friends only. Facebook is useful if it is used properly. It is efficient and effective in disseminate information, just one click the message sent and received by all the concerned in a matter seconds. In additional, it is an instrument or way to communicate to your love one no matter they are across the miles. Likewise, it keeps the family members relationship enhances and sustains though they are away from one another. It also plays an important role in education like it serves as medium of communication and distribution of information between teacher and student and it keep us updated anytime.

title: Facebook improvements


text:
User:Egm6341.s10.Team4/HW3

= Problem 1: Proof of Tighter Error Bound for Simpson's rule [Simple]=

Given
Refer Lecture slide [[media:Egm6341.s10.mtg17.djvu|17-1]] for problem statement.

From Lecture slide [[media:Egm6341.s10.mtg14.pdf|14-1]], Error for Simple Simpson's rule is given by,

 * {| style="width:90%" border="0" align="center"

\begin{align} E_{2} & = -\frac{(b-a)^5}{2880}f^{(4)}(\xi)\ \, \xi \in [a,b] \\ & = -\frac{h^5}{90}f^{(4)}(\xi)\ \, h := \frac{b-a}{2} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)

 * }.
 * }.

In order to prove the above Error for Simple Simpson's rule [ [[media:Egm6341.s10.mtg15.pdf|proof]] ], we defined

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^5 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (2)

 * }.
 * }.

where,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

e(t) := \int_{-t}^{+t} F(t)dt - \frac{t}{3} \left[F(-t)+4F(0)+F(t)\right] \ \, F(t) := f(x(t))

$$ $$
 * $$\displaystyle \longrightarrow (3)

 * }.
 * }.

Find
 Part I.  Try to replicate the proof, by using the below equations instead of Equation (2) and point out where the proof breaks down.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^4 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (4)

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^6 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (5)

 * }.
 * }.

''' Part II. ''' Using Equation(2) find $$\displaystyle G^{(3)}(0) $$ and follow the same steps in the proof. Observer what happens.

Solution
 Part I. Using Equation (4):

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G(t) := e(t) - t^4 \underbrace{e(1)}_{Constant}

$$

 * }.
 * }.

On differentiating,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(1)}(t) = e^{(1)}(t) - 4t^3 e(1)

$$

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(2)}(t) = e^{(2)}(t) - 12t^2 e(1)

$$

 * }.
 * }.

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(3)}(t) = e^{(3)}(t) - 24t e(1)

$$ $$
 * $$\displaystyle \longrightarrow (6)

 * }.
 * }.

From, HW2-Problem-14-Solution and from the lecture slide [[media:Egm6341.s10.mtg15.pdf|15-2]] we know,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

e^{(3)}(t) = -\frac{t}{3} \left[ F^{(3)}(t) - F^{(3)}(-t)\right]

$$ $$
 * $$\displaystyle \longrightarrow (7)

 * }.
 * }.

Applying Rolle's theorem thrice 15-2], we have shown that

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\exists \ \zeta_3 \in \ (0,\zeta_2) \ such\ that\ G^{(3)}(\zeta_3) = 0,

$$ $$
 * $$\displaystyle \longrightarrow (8)

 * }.
 * }.

Substituting Equations (7) and (8) in (6),

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(3)}(\zeta_3) &= e^{(3)}(\zeta_3) - 24\zeta_3 e(1) \\ &= -\frac{\zeta_3}{3} \left[ F^{(3)}(\zeta_3) - F^{(3)}(-\zeta_3)\right] - 24\zeta_3 e(1) \\ &= 0 \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (9)

 * }.
 * }.

Using DMVT [[media:Egm6341.s10.mtg12.pdf|Derivative Mean Value Theorem, 12-1]], we get

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow \left[ F^{(3)}(\zeta_3) - F^{(3)}(-\zeta_3) \right] &= \left[ \zeta_3 - (-\zeta_3) \right] F^{(4)}(\zeta_4) \\ &= 2\zeta_3 F^{(4)}(\zeta_4) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (10)

 * }.
 * }.

Using Equation (10) in (9),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 24\zeta_3 e(1)

$$ $$
 * $$\displaystyle \longrightarrow (11)

 * }.
 * }.

Solving for $$\displaystyle e(1) $$,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{\zeta_{3}}{36} F^{(4)}(\zeta_4)

$$ $$
 * $$\displaystyle \longrightarrow (12)
 * }.
 * }.

We know from Lecture slide [[media:Egm6341.s10.mtg14.pdf|14-2]],

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1)

$$ $$
 * $$\displaystyle \longrightarrow (13)
 * }.
 * }.

On substituting Equation (12) in (13) gives us,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - h \frac{\zeta_{3}}{36} F^{(4)}(\zeta_4)

$$ $$
 * $$\displaystyle \longrightarrow (14)
 * }.
 * }.

From the relationship between $\displaystyle \xi$ and $\displaystyle \zeta_4$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow \displaystyle F^4(\zeta_4)=\frac{(b-a)}{16} ^4\cdot f^4(\xi)

$$ $$
 * $$\displaystyle \longrightarrow (15)
 * }.
 * }.

Equation (14) then becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - \frac{\zeta_{3}(b-a)^5}{1152} f^{(4)}(\xi)

$$ $$
 * $$\displaystyle \longrightarrow (16)
 * }.
 * }.

This error equation has an extra term $$\displaystyle \zeta_{3} $$ on the numerator as opposed to Equation (1). The proof breaks down in Equation (11), where we expect $$\displaystyle \zeta_{3}^2 $$ to cancel out, only if we had $$\displaystyle -24\zeta_{3}^2 e(1) $$ instead of $$\displaystyle -24\zeta_{3} e(1) $$.

 Part I. Using Equation (5):

On differentiating Equation (5),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow G^{(3)}(t) = e^{(3)}(t) - 120t^3 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (17)

 * }.
 * }.

In this case, Equation (11) becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 120\zeta_3^3 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (18)

 * }.
 * }.

Solving for $$\displaystyle e(1) $$,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{180\zeta_{3}} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (19)
 * }.
 * }.

Substituting in Equation (13),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = - h \frac{1}{180\zeta_{3}} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (20)
 * }.
 * }.

Using Equation (15),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = -  \frac{(b-a)^5}{5760\zeta_{3}} f^{(4)}(\xi)

$$ $$
 *
$$\displaystyle \longrightarrow (21)
 * }.
 * }.

Here also an extra term $$\displaystyle \zeta_{3} $$ is on the denominator as opposed to Equation (1). The proof breaks down in Equation (18), where we expect $$\displaystyle \zeta_{3}^2 $$ to cancel out, only if we had $$\displaystyle -120\zeta_{3}^2 e(1) $$ instead of $$\displaystyle -120\zeta_{3}^3 e(1) $$.

So in order to have a tighter bound on the Error Equation, we choose $$\displaystyle G(t) $$ to be $$\displaystyle e(t)-t^5 e(1) $$, in which case Equation (11) becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow 0 = -\frac{2\zeta_{3}^{2}}{3} F^{(4)}(\zeta_4) - 60\zeta_3^2 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (22)

 * }.
 * }.

Here $$\displaystyle \zeta_{3} $$ term cancels out to give,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{90} F^{(4)}(\zeta_4)

$$ $$
 *
$$\displaystyle \longrightarrow (23)
 * }.
 * }.

and the Error Equation becomes,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow E_{2} = -  \frac{(b-a)^5}{2880} f^{(4)}(\xi)

$$ $$
 *
$$\displaystyle \longrightarrow (24)
 * }.
 * }.

''' Part II. '''

Using Equation (7) in the third derivative of $$\displaystyle G(t) = e(t)-t^5 e(1) $$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G^{(3)}(t) = -\frac{t}{3} \left[ F^{(3)}(t) - F^{(3)}(-t)\right] - 60t^2 e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (25)

 * }.
 * }.

At $$\displaystyle t = 0$$,

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(3)}(0) &= -\frac{0}{3} \left[ F^{(3)}(0) - F^{(3)}(-0)\right] - 60(0)^2 e(1) \\ &= 0
 * $$\displaystyle

\end{align} $$ $$
 *
$$\displaystyle \longrightarrow (26)

 * }.
 * }.

Since $$\displaystyle G^{(3)}(t)$$ is zero at $$\displaystyle t = 0$$ and $$\displaystyle t = \zeta_3$$ [from Equation (8)],

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\exists \ \zeta_4 \in \ (0,\zeta_3) \ such\ that\ G^{(4)}(\zeta_4) = 0,

$$ $$
 *
$$\displaystyle \longrightarrow (27)

 * }.
 * }.

Please note that, $$\displaystyle (0,\zeta_3) $$ is an open interval and $$\displaystyle \zeta_4 \ne 0 $$ or $$\displaystyle \zeta_4 \ne \zeta_3 $$.

Now on differentiating Equation (25),

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

G^{(4)}(t) = -\frac{1}{3} \left[ F^{(3)}(t) - F^{(3)}(-t)\right] -\frac{t}{3} \left[ F^{(4)}(t) + F^{(4)}(-t)\right] - 120t e(1)

$$ $$
 *
$$\displaystyle \longrightarrow (28)

 * }.
 * }.

At $$\displaystyle \zeta_4 $$,

 * {| style="width:90%" border="0" align="center"

\begin{align} \Rightarrow G^{(4)}(\zeta_4) &= -\frac{1}{3} \underbrace{\left[ F^{(3)}(\zeta_4) - F^{(3)}(-\zeta_4)\right]}_{2 \zeta_4 F^{(4)}(\zeta_5)} -\frac{\zeta_4}{3} \left[ F^{(4)}(\zeta_4) + F^{(4)}(-\zeta_4)\right] - 120\zeta_4 e(1) \\ &= 0 \end{align} $$ $$
 * $$\displaystyle
 *
$$\displaystyle \longrightarrow (29)

 * }.
 * }.

where $$\displaystyle \zeta_5 \in [-\zeta_4,\zeta_4]$$.

Solving for $$\displaystyle e(1)$$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{1}{180}  F^{(4)}(\zeta_5) - \frac{1}{360} \left[ F^{(4)}(\zeta_4) + F^{(4)}(-\zeta_4)\right]

$$ $$
 *
$$\displaystyle \longrightarrow (30)

 * }.
 * }.

Using the relationship between $\displaystyle \xi$ and $\displaystyle \zeta_4$ we have,

 * {| style="width:90%" border="0" align="center"

 * $$\displaystyle

\Rightarrow e(1) = -\frac{(b-a)^4}{2880}  f^{(4)}(\xi_2) - \frac{(b-a)^4}{5760} \left[ f^{(4)}(\xi_1) + f^{(4)}(-\xi_1)\right]

$$ $$
 *
$$\displaystyle \longrightarrow (31)

 * }.
 * }.

where $$\displaystyle \xi_1 = x_1 + h \zeta_4 $$ and $$\displaystyle \xi_2 = x_1 + h\zeta_5 $$, as we know $$\displaystyle x = x_1 + ht $$ from [[media:Egm6341.s10.mtg14.pdf|14-2]].

So, as long as $$\displaystyle f(x) \in \mathcal{P}_{3}$$ we know $$\displaystyle f^{(4)}(x) = 0 $$ from [[media:Egm6341.s10.mtg14.pdf|14-1]].

And for which, $$\displaystyle e(1) = 0 \Rightarrow E_{2} = 0 $$.

Author
Solved and Typed by - Egm6341.s10.Team4.andy 12:41, 17 February 2010 (UTC) .

= Problem 2: Error for composite Simpson's rule =

Given
Use composite Simpson's equation (3) on slide [[media:Egm6341.s10.mtg7.pdf|7-2]], which is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]

 * }
 * }

Find
To demonstrate error of composite Simpson's rule is equal to the following:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle 	\left | {E_n}^2\right |\leqslant\dfrac{(b-a)^5}{2880n^4}M_4=\dfrac{(b-a)h^4}{2880}M_4

M_4:=max\left |f^{(4)}(\xi)\right | $$

\xi\in[a,b] $$
 * style= |
 * }.
 * }.

Solution
The error definition can be done as:

{E_n}^2=I-I_n=\int\limits_{a}^{b} f(x)dx-I_n $$

From the composite form of Simpson's rule, we have:

I_n=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)] $$

\Rightarrow {E_n}^2=\int\limits_{a}^{b} f(x)dx-\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]= $$

=\sum_{i=1}^{n/2} [\int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx-\dfrac{h}{3}[f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i})]] $$

where,

h=\frac{b-a}{n} = x_i-x_{i-1} $$

According to Lagrange interpolation error for simple Simpson's rule (refer to [14-1]) for $$\displaystyle [x_{i-1},x_{i+1}] $$ ;

E_2=\dfrac{x_{i+1}-x_{i-1}}{2} f^{(4)}(\xi) $$

\xi\in [x_{i-1},x_{i+1}] $$

\Rightarrow \left |{E_n}^2 \right | \le \sum_{i=1}^{n/2} max \left | \dfrac {(x_{i+1}-x_{i-1})^5}{90\times 2^5} f^{(4)}(\xi)\right | =\dfrac{h^5}{90}\times \sum_{i=1}^{n/2} max \left | f^{(4)}(\xi)\right | $$

h=\dfrac {x_{i+1}-x_{i-1}}{2}=\dfrac {b-a}{n} $$

M_4:=max\left | f^{(4)}(\xi) \right | $$

\xi\in [x_{i-1},x_{i+1}] $$

\overline{M_4}:=\sum_{i=1}^{n/2} max\left |f^{(4)}(\xi) \right | $$

\overline{M_4}\le max\left | \dfrac {n}{2} .M_4 \right | $$

\Rightarrow\left |{E_n}^2\right |\le\dfrac{h^5}{90}\times \dfrac {n}{2}.M_4=\dfrac{h^5}{90}\times \dfrac {\dfrac{b-a}{h}}{2}.M_4 $$

 * {| style="width:20%" border="0"

$$\displaystyle \Rightarrow\left |{E_n}^2\right | \le \dfrac{(b-a)h^4}{180} M_4 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

 * style= |

This answer is different from the asked question, However, this answer is confirmed in the Atkinson's text book "An Intro, Numerical Analysis" 2nd edition (page 257~258)
 * }
 * }

.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&m 15:31, 17 February 2010 (UTC) Reviewed by - Egm6341.s10.Team4.andy 16:23, 17 February 2010 (UTC) .

= Problem 3: Visualization: Error Estimates' Plots=

Given

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

$$I=\int\limits_{0}^{1} f(x)\, dx$$ $$where f(x)=\frac{(e^x-1)}{x}.$$
 *
$$\displaystyle $$
 * }.
 * }.

Find

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

 * 1) Find n such that En=I-In is of the order of 10-6 using the error estimates for the Taylor's series, composite Trapezoidal rule, composite Simpson's rule and compare to numerical results.
 * 2) Numerically find the power of h in the error. Plot error versus h on log graph and measure slope.
 * 1) Find n such that En=I-In is of the order of 10-6 using the error estimates for the Taylor's series, composite Trapezoidal rule, composite Simpson's rule and compare to numerical results.
 * 2) Numerically find the power of h in the error. Plot error versus h on log graph and measure slope.

 *
$$\displaystyle $$
 * }.
 * }.

Solution
The error for Taylor series is analytically given by $$\displaystyle \frac{1}{(n+1)!(n+1)} \leq E_n \leq \frac{e}{(n+1)!(n+1)}$$ The maximum value of En is equal to $$\displaystyle \frac{e}{(n+1)!(n+1)} $$.

The value of the error given in the problem is equal to 10-6.

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

 * $$\Rightarrow \displaystyle \frac{e}{(n+1)!(n+1)}= 10^{-6} $$
 * }.
 * }.

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

 * $$\Rightarrow \displaystyle (n+1)!(n+1)= e \cdot10^6 $$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=8(approximately). $$
 * }.
 * }.

The error for composite Trapezoidal rule is analytically given by $$\displaystyle |E^1_n| \leq \frac{(b-a)^3}{12n^2}M_2$$

where M2 is the maximum of $$\displaystyle f^{''}(x).$$ in the interval [0,1]. $$\Rightarrow a=0,b=1.$$

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

 * Given $$\displaystyle f(x)=\frac{(e^x-1)}{x}.$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow f^'(x)=\frac {xe^x-e^x+1}{x^2}.$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

Plot: .
 * $$\Rightarrow f^{''}(x)=\frac {x^2e^x-2xe^x+2e^x-2}{x^3}$$
 * }.
 * }.

The above function $$\displaystyle f^{''}(x)$$ is plotted with respect to x in the interval [0,1] and a maximum of 0.72 is obtained at x=1.
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle |E^1_n|=\frac{(b-a)^3}{12n^2}M_2=\frac{M_2}{12n^2}$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle \frac{M_2}{12n^2}=10^{-6}$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n^2=\frac{0.72*10^6}{12}$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=245$$(approximately).
 * }.
 * }.

The error for composite Simpson's rule is analytically given by $$\displaystyle |E^2_n| \leq \frac{(b-a)^5}{180n^4}M_4.$$

where M4 is the maximum of $$\displaystyle f^4(x).$$ in the interval [0,1].

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

 * Given $$\displaystyle f(x)=\frac{(e^x-1)}{x}.$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow f^4(x)=\frac {x^4e^x-4x^3e^x+12x^2e^x-24xe^x+24e^x-24}{x^5}.$$
 * }.
 * }.

Plot: .

The above function $$\displaystyle f^4(x)$$ is plotted with respect to x in the interval [0,1] and a maximum of 0.465 is obtained at x=1.

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

 * $$\Rightarrow \displaystyle |E^2_n|=\frac{(b-a)^5}{180n^4}M_4=\frac{M_4}{180n^4}.$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle \frac{M_4}{180n^4}=10^{-6}$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n^4=\frac{0.465*10^6}{180}$$
 * }.
 * {| style="width:70%" border="0" align="center"
 * {| style="width:70%" border="0" align="center"

 * $$\Rightarrow \displaystyle n=8$$(approximately).
 * }.
 * }.

The above analytical values of n are compared to the numerical values of n for an error of 10-6 which in turn are obtained from the matlab code.

 Matlab Code: for $$ E_{Taylors series} $$

 Matlab Code: for $$ E_{n,Trapezoidal} $$

 Matlab Code: for $$ E_{n,simpson} $$

The values of n obtained numerically for an error of 10-6 are 8,200,8 for Taylors series,trapezoidal rule and Simpson's rule which are almost equal to the values 8,245,8 found analytically.

In the next part we are required to find the convergence of the error and power of h in the error numerically. Hence we plot the values of the error and h on a log plot or alternatively plot the values of log(Error) and log(h) in excel plot using the numerical values.

Semilog Plot: .

Semilog Plot: .

Logarithmic Plot: .

The slope of the line in the above graph is obtained to be equal to 2. This is the required power of h in the error formula for trapezoidal rule.

Semilog Plot: . Logarithmic Plot: .

The slope of the line in the above graph is obtained to be equal to 4. This is the required power of h in the error formula for simpsons rule.

Author
Solved and Typed by - Egm6341.s10.team4.anandankala 13:37, 17 February 2010 (UTC) .

= Problem 4: The Error & Ratio of Trapezoidal & Simpson's rule =

Given
Refer Lecture slide [[media:Egm6341.s10.mtg17.djvu|17-3]] [[media:Egm6341.s10.mtg18.djvu|18-1]]for problem statement.

1)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0} e^xcos(x)\; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

2)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0} e^xcos(x)\; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

3)
 * {| style="width:100%" border="0" align="left"

I = \int^{1}_{0} x^3\sqrt{x} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

x\to 0; f^{(4)}(x)\to \infty $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

4)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0}\frac{1}{1+(x-\pi^2)} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

5)
 * {| style="width:100%" border="0" align="left"

I = \int^{1}_{0} \sqrt{x} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

6)
 * {| style="width:100%" border="0" align="left"

I = \int^{2\pi}_{0} e^{cos(x)} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Find
1) $$\displaystyle I = \int^{\pi}_{0} e^xcos(x) \; dx $$

Find the below and make a Table

(i). Compsite Trapozoidal rule $$\displaystyle (I_n) $$ and Error of Trapezoidal$$\displaystyle (E_1) $$ for $$\displaystyle n=2, 4, 8, 16, 32, 64, 128, 256, 512 $$

(ii). The Ratio of Error when $$\displaystyle n $$ is doubled,

(iii). Asymptotic Estimates Ẽ

2) $$\displaystyle I = \int^{\pi}_{0} e^xcos(x) \; dx $$

Find the below and make a Table

(i). $$\displaystyle I_n $$ and Error of Simpson's rule$$\displaystyle (E_2) $$ for $$\displaystyle n=2, 4, 8, 16, 32, 64, 128 $$

(ii). The Ratio of Error when $$\displaystyle n $$ is doubled,

(iii). Asymptotic Estimates Ẽ

3) $$\displaystyle I = \int^{1}_{0} x^3\sqrt x \; dx $$,

4) $$\displaystyle I = \int^{\pi}_{0}\frac{1}{1+(x-\pi^2)} \; dx $$,

5)$$\displaystyle I = \int^{1}_{0} \sqrt{x} \; dx $$,

6)$$\displaystyle I = \int^{2\pi}_{0} e^{cos(x)} \; dx $$

Find the below and make a Table

(i). Error of Trapezoidal$$\displaystyle (E_1) $$ and Simpson's Rule $$\displaystyle (E_2) $$ for $$\displaystyle n=2, 4, 8, 16, 32, 64, 128 $$

(ii). The Ratio of Error when $$\displaystyle n $$ is doubled,

Solution
1)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0} e^xcos(x) \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table are bleow.

 Matlab Code:

 subfunction f(x)

 subfunction dff_f(x)  .

2)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0} e^xcos(x) \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table are bleow.

 Matlab Code:

3)
 * {| style="width:100%" border="0" align="left"

I = \int^{1}_{0} x^3\sqrt x \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table are bleow.

 Matlab Codes for Trapezoidal rule:

 Matlab Codes for Simpson's rule:  .

4)
 * {| style="width:100%" border="0" align="left"

I = \int^{\pi}_{0}\frac{1}{1+(x-\pi^2)} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table is bleow.

 Matlab Code:

 subfunction f4(x)

5)
 * {| style="width:100%" border="0" align="left"

I = \int^{1}_{0} \sqrt{x} \; dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table is bleow.

 Matlab Code:

 subfunction f5(x)

6)
 * {| style="width:100%" border="0" align="left"

I = \int^{2\pi}_{0} e^{cos(x)}dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Matlab code and the Result table is bleow.

 Matlab Code:

 subfunction f6(x)  .

Author
Solved [Sections:1,2,3] by - Egm6341.s10.Team4.roni 14:42, 17 February 2010 (UTC) Solved [Sections:4,5,6] by - Egm6341.s10.Team4.yunseok 15:30, 17 February 2010 (UTC) Typed and reviewed [Sections:2,3] by - Egm6341.s10.Team4.nimaa&m 15:33, 17 February 2010 (UTC) 08:16, 17 February 2010 (UTC) Typed and reviewed [Sections:1,4,5,6] by - Egm6341.s10.Team4.yunseok 15:30, 17 February 2010 (UTC)

.

= Problem 5: Evaluation of a Function Using Rhomberg Interpolation=

Given
Rhomberg interpolation for the composite trapezoidal rule is defined as $$ \displaystyle\ T_k(n)=\frac{2^{2k}T_{k-1}(2n)-T_{k-1}(n)}{2^{2k}-1}$$

Find
We are asked to develop an efficient code that calculates and determines the efficiency of Rhomberg interpolation and compares it to the results to the Taylor Series integration method, using the function $$ \displaystyle\ f(x)=\frac{e^x-1}{x}$$.

Solution
The method for applying the trapezoidal rule has been well defined in previous work, thus it will not be shown again here.

Additionally, each $$ \displaystyle\ f(x_i)$$ was only calculated one time for the entire process, increasing the efficiency of the computation. This was done by calculating $$ \displaystyle\ f(x_i)$$ for a very fine computational domain, and then cherry picking certain point appropriate for the number of points being integrated over.

The Rhomberg interpolation is described above, thus we will move right to out results.

In performing the Rhomberg interpolation, we found that with each additional interpolation of the composite trapezoidal series, an additional level of accuracy was obtained, with a linear return on the level of interpolation. However, even with this increased convergence to the exact solution, Rhomberg interpolation still does not converge nearly as quickly as the Taylor Series approximation of this integral does.

Thus, while the Rhomberg interpolation technique is helpful, the Taylor series method still converges much more quickly as a method of numerical integration.

Figure - Error of Various Integration Methods as a Function of the Number of Sub-Intervals Integrated Over

Code to run calculations -

.

Author
Solved and Typed by -  Egm6341.s10.Team4.riherd 14:38, 17 February 2010 (UTC)

.

= Contributing Members = Egm6341.s10.Team4.riherd 03:11, 17 February 2010 (UTC) - Author of Problems:5 ; Proof Read: 1 Egm6341.s10.Team4.andy 13:01, 17 February 2010 (UTC) - Author of Problems:1 ; Proof Read: 2 Egm6341.s10.Team4.nimaa&m 08:19, 17 February 2010 (UTC) - Author of Problems:2 ; Proof Read: 4 Egm6341.s10.Team4.roni 14:42, 17 February 2010 (UTC) - Author of Problems:2 ; Proof Read: 4 Egm6341.s10.Team4.yunseok 15:31, 17 February 2010 (UTC) - Author of Problems: 4 ; Proof Read: 2,3 Egm6341.s10.team4.anandankala 13:37, 17 February 2010 (UTC)-Author of Problems:3 ; Proof Read: 5

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User:Egm6341.s2010.Team1/HW2

=Problem 9-1(1)=

Find
Use Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]] and Eq. (1) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]] to find expression for $$C_0,C_1,C_2 \,$$ in terms of $$(x_i,f(x_i)), i=0,1,2 \,$$

Given
Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)$$

Eq. (1) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$f_{2}\left ( x \right )=p_{2}\left ( x \right )=C_2 x^2+C_1 x^1+C_0, C_0,C_1,C_2= 3 unknowns \,$$

Solution
From Eq. (2) [[media:Egm6341.s10.mtg7.pdf|Lec. 7-3]]
 * $$l_{i,n}\left ( x \right )=l_{i}\left ( x \right )=\prod_{j=0,j\neq i}^n\frac{x-x_j}{x_i-x_j}$$

Then
 * $$l_{0}\left ( x \right )=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}$$
 * $$l_{1}\left ( x \right )=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}$$
 * $$l_{2}\left ( x \right )=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$$

Also From Eq. (3) [[media:Egm6341.s10.mtg8.pdf|Lec. 8-3]]
 * $$P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)$$

 * $$\Rightarrow P_2(x)=\sum _{i=0}^{2}l_i(x)f(x_i)=l_{0}\left ( x \right )f(x_0)+l_{1}\left ( x \right )f(x_1)+l_{2}\left ( x \right )f(x_2)=C_2 x^2+C_1 x^1+C_0 $$

 * $$\Rightarrow C_2=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}$$
 * $$\Rightarrow C_1=\frac{-f(x_0)(x_1+x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{-f(x_1)(x_0+x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{-f(x_2)(x_0+x_1)}{(x_2-x_0)(x_2-x_1)}$$
 * $$\Rightarrow C_0=\frac{f(x_0)(x_1x_2)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)(x_0x_2)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)(x_0x_1)}{(x_2-x_0)(x_2-x_1)}$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \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)}$$
 * $$ \displaystyle

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

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

$$
 * $$\displaystyle
 * }
 * }

Solution for problem 9-1(1): Egm6341.s10.team1.lei 19:50, 1 March 2010 (UTC)

=Problem 9-2(2)=

Statement
$$f\left( x \right)=\!\frac{e^x-1}{x}\,\,on\,\,[\!\,0,1\,]\!$$

$$x_0=\!a=\!0,\,x_n=\!b=\,1$$

Consider $$n=\!1,2,4,8,16$$

Constr. $$f_n\left( x \right)=\,\sum_{i=0}^{n}l_{i,n}\left ( x \right)f\left ( x_i \right)$$

Plot $$f.\!\,\, f_n, n=\!1,\,2,\,4,\,8,\,16$$

Comp. $$ I_n=\!\int_{a}^{b}f_n\left ( x \right)dx,\,n=\!1,2,4,8\,$$ and compare to $$ I\!$$

For $$ n=\!4\,$$, plot $$\,l_0,\,l_1,\,l_2 $$ why not $$\,l_3,\,l_4,\,$$ also $$2\,$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * }
 * }

MATLAB
Plot $$f.\!\,\, f_n, n=\!1,\,2,\,4,\,8,\,16:\!$$

For $$ n=\!4\,$$, plot $$\,l_0,\,l_1,\,l_2 $$ why not $$\,l_3,\,l_4,\,$$

For $$ n=\!2\,$$, plot $$\,l_0,\,l_1,\,l_2 \,$$

Solution for problem 9-2(2): EGM6341.s10.Team1.Ya-Chiao Chang19:24, 10 February 2010 (UTC)

Proofread problem 9-2(2):

= Problem 9-3(1): Simple to Composite Trapezoidal and Simpson Rules =

Given
Given the simple Trapezoidal:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_1=\dfrac{b-a}{2}[f(a)+f(b)]

 * }
 * }

Find
Find the Composite Trapezoidal rule

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{b-a}{n}[\dfrac{1}{2}f_1+f_2+f_3+...+f_{n-1}+\dfrac{1}{2}f_n]

 * }
 * }

Where:

f_1=f(a), $$

f_n=f(b) $$

Solution
Simple trapezoidal rule uses only one trapezoid to calculate area underneath a function extending this principle to more than one trapezoid leads to the composite function.

The area of a trapezoid is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Area=(b-a)\dfrac{h_1+h_2}{2}

 * }
 * }

Where $$h_1$$ and $$h_2$$ would be $$f(a)$$ and $$f(b)$$ respectively

Assuming five equal width trapezoids the base of each would be $$\dfrac{b-a}{5}$$:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \Rightarrow I_5=\dfrac{b-a}{5}(\dfrac{f_1+f_2}{2}+\dfrac{f_2+f_3}{2}+\dfrac{f_3+f_4}{2}+\dfrac{f_4+f_5}{2})

 * }
 * }

Collecting like terms

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_5=\dfrac{b-a}{5}(\dfrac{1}{2}f_1+f_2+f_3+f_4+\dfrac{1}{2}f_5)

 * }
 * }

Which displays a recognizable patter

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{b-a}{n}(\dfrac{1}{2}f_1+f_2+...+f_{n-1}+\dfrac{1}{2}f_n)

--Egm6341.s10.team1.toddmock 19:57, 1 March 2010 (UTC)
 * }.
 * }.

= Problem 9-3(2): Simple to Composite Trapezoidal and Simpson Rules =

Given
The simple Simpson's rule:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_2=\dfrac{h}{3}[f_0+4f_1+f_2]

 * }.
 * }.

where $$h=\dfrac{b-a}{n}$$ in this case $$n=2$$

Find
Find the composite Simpson's rule known to be

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_2=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...2f_{n-2}+4f_{n-1}+f_n]

 * }.
 * }.

where $$h=\dfrac{b-a}{n}$$

Solution
The simple Simpson's rule comes from calculation of the area underneath a polynomial using two sections

Repeating the pattern of the simple Simpson's rule will yield the composite rule. So if the same function from above is instead split into 5 equal sections then we have the following.

calculation for 2 sets of the sections $$[x_0,x_2]$$ and $$[x_2,x_4]$$ we will obtain

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_0,x_2])=\dfrac{h}{3}[f_0+4f_1+f_2]

 * }
 * }

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_2,x_4])=\dfrac{h}{3}[f_2+4f_3+f_4]
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I([x_0,x_4])=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+f_4]
 * }
 * }

Thus producing the obvious pattern

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

$$ $$ --Egm6341.s10.team1.toddmock 19:57, 1 March 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle I_n=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+...+2f_{n-2}+4f_{n-1}+f_n]
 * }
 * }

=Problem 11-1=

--Egm6341.s10.team1.andrewdugan 20:26, 1 March 2010 (UTC)

=Problem 12-2=

Statement
Why :$$E_{n}^{n+1}\left ( x \right )=F^{n+1}\left ( x \right)-0$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg6.pdf|P6-4]], [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * }
 * }

From [[media:Egm6341.s10.mtg6.pdf|P6-4]]

\begin{align} \\E_{n} &= I-I_{n}\; \\ &= \int_{0}^{1}f\left ( x \right )dx-\int_{0}^{1}f_{n}\left ( x \right )dx\,. \end{align} $$

the first derivative of $$E_{n}\,$$ is

 * $$E_n^{1}=\int_{0}^{1}f^{1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{1}\left ( x \right )dx\,$$

the second derivative of $$E_{n}\,$$ is

 * $$E_n^{2}=\int_{0}^{1}f^{2}\left ( x \right )dx-\int_{0}^{1}f_{n}^{2}\left ( x \right )dx\,$$

 * {| style=text-align:center

 * width=2% | $$.\,$$
 * width=2% | $$.\,$$
 * width=2% | $$.\,$$
 * }
 * width=2% | $$.\,$$
 * }
 * }
 * }

 * $$E_n^{n}=\int_{0}^{1}f^{n}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n}\left ( x \right )dx\,$$

 * $$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$

From [[media:Egm6341.s10.mtg7.pdf|P7-3]]
 * $$f_{n}\left ( x \right )=\sum _{i=0}^{n}l_{i,n}\left ( x \right ) f\left ( x_i \right)\,$$

 * (i)where $$l_{i,n}\left ( x \right)=l_{i}\left ( x \right)=\prod_{j=0,i \neq j}^{n}\frac{x-x_{j}}{x_{0}-x_{j}}\,$$

 * when $$i=0\,$$
 * {| style=text-align:center

\begin{align} \\n=1, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}} \rightarrow l_{0}=l\left ( x^{1}\right)\; \\n=2, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}} \rightarrow l_{0}=l\left ( x^{2}\right)\; \\.\, \\.\, \\.\, \\n=n, \rightarrow & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}...\frac{x-x_{n-1}}{x_{0}-x_{n-1}}\frac{x-x_{n}}{x_{0}-x_{n}} \rightarrow l_{0}=l\left ( x^{n}\right)\, \end{align} $$
 * }
 * }

 * (ii)$$f\left( x_{i} \right)\,$$ is constant, so $$f_n\left( x \right)\,$$ is $$x^{n}\,$$'s function

 * $$f_n^{n+1}\left ( x \right)=0\,$$

 * $$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$

 * $$\rightarrow E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1} 0 dx\,$$

 * $$\rightarrow E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \, $$ $$ Solution for problem 12-2: EGM6341.s10.Team1.Ya-Chiao Chang 19:27, 10 February 2010 (UTC) Proofread problem 12-2:
 * $$\displaystyle
 * }
 * }

=Problem 12-3(1)=

Find
Prove $$P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Given

 * $$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n})$$

Solution

 * $$n=0 \,$$
 * $$q_{1}\left ( x \right ):=(x-x_{0})$$

 * $$q_{1}^{(1)}\left ( x \right )=1$$

 * $$n=1 \,$$
 * $$q_{2}\left ( x \right ):=(x-x_{0})(x-x_{1})$$

 * $$q_{2}^{(2)}\left ( x \right )=2\times 1=2!$$

 * $$n=2 \,$$
 * $$q_{3}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})$$

 * $$q_{3}^{(3)}\left ( x \right )=3\times 2\times 1=3!$$

 * $$n=3 \,$$
 * $$q_{4}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3})$$

 * $$q_{4}^{(4)}\left ( x \right )=4\times 3\times 2\times 1=4!$$

...

 * $$n=n \,$$
 * $$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}...(x-x_{n})$$

 * $$q_{n+1}^{(n+1)}\left ( x \right )=n\times (n-1)\times (n-2)...\times 1=(n+1)!$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\Rightarrow P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$ $$
 * $$\displaystyle
 * }
 * }

Solution for problem 12-3(1): Egm6341.s10.team1.lei19:52, 1 March 2010 (UTC)

= Problem 12-3(2) =

Given
Given the function:

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

$$ $$
 * $$\displaystyle f(x)=log(x)
 * $$\displaystyle f(x)=log(x)
 * $$\displaystyle

 * }
 * }

and

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

$$ $$
 * $$\displaystyle t=2, x_0=3, x_1=4, ..., x_6=9
 * $$\displaystyle t=2, x_0=3, x_1=4, ..., x_6=9
 * $$\displaystyle

 * }
 * }

Find
Plot 3 figures like those on slide [[media:Egm6341.s10.mtg11.pdf|11-2]] for the functions

$$\displaystyle f(x),f_n(x), l_{i,n}(x), q_{n+1}(x)$$

for $$\displaystyle i=3 (x_3=6)$$

and

$$\displaystyle x=5.5$$

Solution
The function f(x) and approximated function are defined as:

 * $$\displaystyle

f(x)=log(x) $$

f_n(x)=\sum_{i=0}^6 l_{i,n}(x)f(x_i)=\sum_{i=0}^6 l_{i,n}(x)log(x_i) $$

Where the Lagrange equation for i=3:

l_{i,n}(x)=\prod_{i=1}^{n} \dfrac{x-x_j}{x_i-x_j} $$

therefore

l_{3,6}(x)=\Big(\frac{x-x_0}{x_3-x_0}\Big)\Big(\dfrac{x-x_1}{x_3-x_1}\Big)\Big(\dfrac{x-x_2}{x_3-x_2}\Big)\Big(\dfrac{x-x_4}{x_3-x_4}\Big)\Big(\dfrac{x-x_5}{x_3-x_5}\Big)\Big(\dfrac{x-x_6}{x_3-x_6}\Big) $$

 * $$\displaystyle

q_{n+1}(t)=(t-x_0)(t-x_1)(t-x_2)...(t-x_n) $$

--Egm6341.s10.team1.toddmock 19:58, 1 March 2010 (UTC)

=Problem 13-2(1)=

Find

 * $$\left | E_{1} \right |\leq \frac{(b-a)^{3}}{12}M_2=\frac{h^{3}}{12}M_2)$$

Given

 * $$n=1, \, q_{2}\left ( x \right )=(x-a)(x-b)$$
 * $$h:=b-a\,$$

Solution
Let:$$M_2:= max f^{(2)} (\xi), \xi \in [a,b]$$
 * $$\left | E_{1} \right |\leq \frac{M_2}{2!}\int_{a}^{b}\left |q_{2}\left ( x \right )\right |dx $$
 * $$=\frac{M_2}{2!}\int_{a}^{b}\underbrace{\left ( x-a \right )}_{\geq 0}\underbrace{\left ( b-x \right )}_{\geq 0}dx $$
 * $$=\frac{M_2}{2!}\int_{a}^{b}[(b+a)x-ab-x^2] dx$$
 * $$=\frac{M_2}{2!}[\frac{1}{2}(b+a)x^2-ab x-\frac{1}{3}x^3]|_{a}^{b}$$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b^3-3ab^2+3a^2b-a^3) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b^2-2ba+a^2)(b-a) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b-a)^2(b-a) $$
 * $$=\frac{M_2}{2!}\times \frac{1}{6}(b-a)^3 $$
 * $$=\frac{M_2}{12}(b-a)^3 $$

and
 * $$h:=b-a\,$$
 * $$=\frac{M_2}{12}h^3 $$
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\Rightarrow \left | E_{1} \right |\leq \frac{(b-a)^{3}}{12}M_2=\frac{h^{3}}{12}M_2$$ $$
 * $$\displaystyle
 * }
 * }

Solution for problem 13-2(1): Egm6341.s10.team1.lei19:54, 1 March 2010 (UTC)

=Problem 13-2(2)=

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

\begin{align} n=2, \\ & q_3(x)=(x-x_o)(x-x_1)(x-x_2)\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

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

\begin{align} \\ & x_o=a, x_2=b, x_1=\frac{a+b}{2} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

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

\left | E_2 \right | = \frac{M_3}{3!}\int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx = \frac{(b-a)^4}{192}M_3 $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

Solution
To capture positive region it is necessary to change the limits and multiply by two

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

\begin{align} \int_{a}^{b}\left|(x-a)(x-\frac{a+b}{2})(x-b) \right |\, dx &= 2\int_{a}^{\frac{a+b}{2}}(x-a)\Big(x-\frac{a+b}{2}\Big)(x-b)\, dx\\ &=2\int_{a}^{\frac{a+b}{2}}x^3-\frac{a^2b}{2}-\frac{ab^2}{2}+\frac{a^2x}{2} + 2abx + \frac{b^2x}{2}-\frac{3ax^2}{2}-\frac{3bx^2}{2}, dx\\ &=\frac{1}{2}x^4-(a+b)x^3+\frac{1}{2}(a^2+4ab+b^2)x^2 - ab(a+b)x\, \Big |^{\frac{a+b}{2}}_a\\
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }.
 * }.

Evaluating at the limits gives

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

\begin{align} &=\frac{1}{2}\Bigg(\Big(\frac{a+b}{2}\Big)^4-a^4\Bigg)-(a+b)\Bigg(\Big(\frac{a+b}{2}\Big)^3-a^3\Bigg)+\frac{1}{2}(a^2+4ab+b^2)\Bigg(\Big(\frac{a+b}{2}\Big)^2-a^2\Bigg) - ab(a+b)\Bigg(\Big(\frac{a+b}{2}\Big)-a\Bigg)\
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }.
 * }.

Simplifying leads to

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

= \frac{1}{32}(b-a)^4 $$
 * $$\displaystyle
 * $$\displaystyle

 * style= |
 * }.
 * }.

Thus the solution becomes

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

$$\displaystyle \left | E_2 \right | = \frac{M_3}{3!}\Bigg(\frac{1}{32}(b-a)^4\Bigg) = \frac{(b-a)^4}{196}M_3 $$ --Egm6341.s10.team1.toddmock 19:56, 1 March 2010 (UTC)
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

=Problem 13-3=

Find
$$I \,$$ and $$I_2 \,$$ on $$[0,1] \,$$

Given

 * $$I=\int_{0}^{1}f_3(x)dx$$
 * $$f_3(x)=3+8x-2x^2+6x^3 \,$$
 * $$I_2=\frac{h}{3}[f(0)+4f(0.5)+f(1)]$$

Solution
Solving for $$I \,$$ is straightforward integration,
 * $$I=\int_{0}^{1}3+8x-2x^2+6x^3dx=\left [ 3x+4x^2-\frac{2}{3}x^3+\frac{3}{2}x^4 \right ]_0^1=\frac{47}{6}$$

Solving for $$I_2 \,$$ by plugging into the formula
 * $$I_2=\frac{1/2}{3}\left [ 3+4(\frac{12}{4}+\frac{16}{4}-\frac{2}{4}+\frac{3}{4})+3+8-2+6 \right ]=\frac{47}{6}$$

Therefore $$I=I_2 \,$$

Solved by EGM6341.s10.Team1.Kumanchik 22:11, 2 March 2010 (UTC)

=Problem 15-1=

Statement
Prove why $$\alpha^{1}\left ( t \right )=F\left ( -t \right)+F\left ( t \right)\,$$

Given
$$\alpha\left ( t \right )=\int_{-t}^{k}f\left ( x\left ( t \right) \right)dt+\int_{k}^{t}f\left ( x\left ( t \right) \right)dt, k\in\![-t,t]\, $$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg14.pdf|P14-2]]
 * }
 * }

(i)$$\int_{k}^{t}F\left ( x\left ( t \right) \right)dt=\mathbb{F}\!\left ( x\left ( t \right) \right)-\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$

 * $$\frac{d}{dt}(\int_{k}^{t}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( t \right) \right))-\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))\, $$

 * since$$k=constant\,\rightarrow\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$ is constant

\begin{align} \\\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( t \right) \right)dt)-\frac{d}{dt}(constant) &=F\left ( t \right)-0\; \\ &= F\left ( t \right)\, \end{align} $$

(ii)$$\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt=\mathbb{F}\!\left ( x\left ( k \right) \right)-\mathbb{F}\!\left ( x\left ( -t \right) \right)\, $$

 * $$\frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))+\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right))\, $$
 * P.S. Chain rule :$$\frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)=\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( k \right) \right))-(\frac{d(-t)}{dt})\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right))\, $$

 * since$$k=constant\,\rightarrow\mathbb{F}\!\left ( x\left ( k \right) \right)\, $$ is constant

\begin{align} \\\frac{d}{dt}(constant)+\frac{d}{dt}(\mathbb{F}\!\left ( x\left ( -t \right) \right)dt) &=0+F\left ( t \right)+0\; \\ &=F\left ( -t \right)\, \end{align} $$

From (i) and (ii)

\begin{align} \\\alpha^{1}\left ( t \right ) &= \frac{d}{dt}(\int_{-t}^{k}F\left ( x\left ( t \right) \right)dt)+\frac{d}{dt}(\int_{k}^{t}F\left ( x\left ( t \right) \right)dt)\; \\ &= F\left ( -t \right)+F\left ( t \right)\, \end{align} $$
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\alpha^{1}\left ( t \right )=\! F\left ( -t \right)+F\left ( t \right) \, $$ $$
 * $$\displaystyle
 * }
 * }

Solution for problem 15-1: EGM6341.s10.Team1.Ya-Chiao Chang 19:28, 10 February 2010 (UTC)

Proofread problem 15-1:

=Problem 15-2(1)=

Statement
Prove why $$G^{(2)}\left ( 0 \right )=0\,$$

Given
$$G^{(1)}\left ( 0 \right )=0\,$$ and $$G^{(1)}\left ( \xi_1 \right )=0\,$$ Apply Rolle's theorem
 * $$ \exists\!\,\,\,\,\xi_2\,\,\,\,\in\!\,\,\,\,[\,0,\xi_1\,]:\,\,\,\,\!G^{(2)}\left ( \xi_2 \right)=0\,$$

Solution

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

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|P15-1]],[[media:Egm6341.s10.mtg14.pdf|P14-2]]
 * }
 * }

From [[media:Egm6341.s10.mtg14.pdf|P14-2]]

$$G\left( t \right):=\!\,e\left( t \right)-\!t^{5}G\left( 1 \right)\,$$

Do [[media:Egm6341.s10.mtg15.pdf|P15-1]] again to find out $$G^{(2)}\left ( 0 \right)=\!0$$

$$G^{(1)}\left( t \right)=\!\,e^{(1)}\left( t \right)-\!5t^{4}e\,$$

$$G^{(2)}\left( t \right)=\!\,e^{(2)}\left( t \right)-\!20t^{3}e\,$$

We need $$ e^{(2)}\left ( t \right)$$

 * $$e\left( t \right):=\!\alpha\left ( t \right)-\!\alpha_2\left ( t \right)\,$$

 * $$\alpha\left ( t \right)=\!\int_{-t}^{k}F\left ( -t \right)+\!\int_{k}^{t}F\left ( t \right)\,\,\,\,k\in\!\,\,[\,-t,t\,]\,$$

 * $$\alpha_2\left ( t \right)=\!\frac{t}{3}[\!\,F\left ( -t \right)+\!4F\left ( 0 \right)+\!F\left ( t \right)]\!\,$$

$$\rightarrow e^{(2)}\left( t \right):=\!\alpha^{(2)}\left ( t \right)-\!\alpha_2^{(2)}\left ( t \right)\,$$

We need $$\alpha^{(2)}\left ( t \right)\,\,\,$$ and $$\,\,\, \alpha_2^{(2)}\left ( t \right)\,$$

 * $$\alpha^{(1)}\left ( t \right)=\!F\left ( -t \right)+\!F\left ( t \right)\,$$

 * $$\alpha_2^{(1)}\left ( t \right)=\!\frac{1}{3}[\!\,F\left ( -t \right)+\!4F\left ( 0 \right)+\!F\left ( t \right)]\!+\!\frac{t}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!\,$$

$$\alpha^{(2)}\left ( t \right)=\!-\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)\,$$

$$\alpha_2^{(2)}\left ( t \right)=\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( -t \right)+\!\,F^{(1)}\left ( t \right)]\!+\!\,\frac{1}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!+\!\frac{t}{3}[\!\,+\!F^{(2)}\left ( -t \right)+\! F^{(2)}\left ( t \right)]\!\,$$

Finally

\begin{align} \\G^{(2)}\left( t \right) &= e^{(2)}\left( t \right)-\!20t^{3}e\; \\ &= \alpha^{(2)}\left ( t \right)-\!\alpha_2^{(2)}\left ( t \right)-\!20t^{3}e\; \\ &= -\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)-\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( -t \right)+\!F^{(1)}\left ( t \right)]\!-\!\frac{1}{3}[\!\,-F^{(1)}\left ( -t \right)+\! F^{(1)}\left ( t \right)]\!-\!\frac{t}{3}[\!\,+\!F^{(2)}\left ( -t \right)+\! F^{(2)}\left ( t \right)]\!-\!20t^{3}e\,. \end{align} $$

When $$t=\!0$$

\begin{align} \\G^{(2)}\left( 0 \right) &= e^{(2)}\left( 0 \right)-\!0\; \\ &= \alpha^{(2)}\left ( 0 \right)-\!\alpha_2^{(2)}\left ( 0 \right)-\!0\; \\ &= -\!F^{(1)}\left ( 0 \right)+\!F^{(1)}\left ( 0 \right)-\!\frac{1}{3}[\!\,-\!F^{(1)}\left ( 0 \right)+\!F^{(1)}\left ( 0 \right)]\!-\!\frac{1}{3}[\!\,-F^{(1)}\left ( 0 \right)+\! F^{(1)}\left ( 0 \right)]\!-\!0-\!0\; \\ &= 0\, \end{align} $$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

G^{(2)}\left ( 0 \right)\,=\!\,0 \, $$ $$
 * $$\displaystyle
 * }
 * }

Solution for problem 15-2(1): EGM6341.s10.Team1.Ya-Chiao Chang 19:29, 10 February 2010 (UTC)

Proofread problem 15-2(1):

=Problem 15-2(2)=

Given
e(t) is given as:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e(t)=\int\limits_{-t}^{t} F(t)\, dt-\dfrac{-t}{3}[F(-t)+fF(t)+F(t)]

 * }
 * }

Find
Determine the third derivative of e(t) which is known to be:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)]

 * }
 * }

Solution
e(t) is the summation of two terms:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e(t)=\alpha(t)-\alpha_2(t)

 * }
 * }

We must use Leibniz integral rule,

For a monovariant function $$g$$:

 * $$ {d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = g(f_2(x)) {f_2'(x)} - g(f_1(x)) {f_1'(x)} $$

Using a point k that lies somewhere in the interval [-t,t]:

\alpha (t)=\int\limits_{-t}^{k} F(t)\, dt+\int\limits_{k}^{t} F(t)\, dt $$

Then evaluation of the first term is:

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \alpha^{(1)}(t)=[F(k)+F(-t)]+[F(t)-F(k)]=F(-t)+F(t)
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \alpha^{(2)}(t)=[-F^{(1)}(-t)+F^{(1)}(t)]=F^{(1)}(-t)
 * }
 * }

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \alpha^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t)
 * }
 * }

Evaluation of the second term:

\alpha_2^{(1)}=\dfrac{-1}{3}[F(-t)+4F(0)+F(t)]+\dfrac{t}{3}F^{(1)}(-t)-\dfrac{t}{3}F^{(1)}(t) $$

\alpha_2^{(2)}(t)=\dfrac{1}{3}F^{(1)}(-t)-\dfrac{1}{3}F^{(1)}(t)+\dfrac{1}{3}F^{(1)}(-t)-\dfrac{t}{3}F^{(2)}(-t)-\dfrac{1}{3}F^{(1)}(t)-\dfrac{t}{3}F^{(2)}(t) $$

=\dfrac{2}{3}F^{(1)}(-t)-\dfrac{2}{3}F^{(1)}(t)-\dfrac{-t}{3}F^{(2)}(-t)-\dfrac{t}{3}F^{(2)}(t) $$

\alpha_2^{(3)}(t)=\dfrac{-2}{3}F^{(2)}(-t)-\dfrac{2}{3}F^{(2)}(t)-\dfrac{1}{3}F^{(2)}(-t)+\dfrac{t}{3}F^{(3)}(-t)-\dfrac{1}{3}F^{(2)}(t)-\dfrac{t}{3}F^{(3)}(t) $$

=-F^{(2)}(-t)-F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)] $$

All that remains is to add the two functions $$\alpha_1^{(3)}$$ and $$\alpha_2^{(3)}$$ :

e{(3)}(t)=\alpha^{(3)}(t)+\alpha_2^{(3)}(t)=F^{(2)}(t)+F^{(2)}(-t)-F^{(2)}(-t)-F^{(2)}(t)+\dfrac{t}{3}[F^{(3)}(-t)-F^{(3)}(t)] $$

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

$$ e^{(3)}(t)=\dfrac{-t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$ --Egm6341.s10.team1.toddmock 19:57, 1 March 2010 (UTC) --Egm6341.s10.team1.andrewdugan 20:27, 1 March 2010 (UTC)
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.

=Problem 16-1=

Find
Plot $$f \,$$ and $$f_n \,$$ for $$n=1,2,3,8,12 \,$$ Plot $$I_n \,$$ vs $$n \,$$ and observe non-convergence Observe the weights, $$w_{i,n} \,$$

Given

 * $$f(x)=\frac{1}{1+x^2}$$
 * $$f_n(x)=\sum_{i}^{n}l_{i,n}f(x_i)$$
 * $$l_{i,n}(x)=\prod_{j=0,j\neq i}^{n}\frac{x-x_j}{x_i-x_j}$$
 * $$x_i=a+\frac{b-a}{n}i$$
 * $$I_n=\sum_{i=0}^{n}\left (\int_{a}^{b}l_i(x)dx \right )f(x_i)$$
 * $$w_{i,n}=\int_{a}^{b}l_i(x)dx$$

Solution
The solution is just plugging into the equations using MATLAB to plot.

For the plot of the estimated function, the oscillations become more wild for higher order polynomials.

For the plot of the integral estimate, the solution does not converge.

And for the plot of the weights, some weights become negative to compensate for the wild oscillations.

MATLAB
Problem solved by EGM6341.s10.Team1.Kumanchik 22:12, 2 March 2010 (UTC)

=Contributing members= EGM6341.s10.Team1.Kumanchik 21:49, 10 February 2010 (UTC)

EGM6341.s10.Team1.Ya-Chiao Chang 19:30, 10 February 2010 (UTC)

Egm6341.s10.team1.andrewdugan 19:41, 10 February 2010 (UTC)

Egm6341.s10.team1.lei 21:12, 10 February 2010 (UTC)

--Egm6341.s10.team1.toddmock 21:03, 10 February 2010 (UTC)

title: User:Egm6341.s2010.Team1/HW2


title: File:11-1pg1.jpg


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text:
18th century European scholarly societies and academies/Académie Delphinale

Summary
Name of Society: Académie Delphinale

Alternate Name(s): Société Littéraire, Société des Sciences et des Arts de la ville de Grenoble,

Country: France

City: Grenoble

Years Active: 1772-Present

The Académie Delphinale was orginally established in 1772. It was formally approved by Louis XVI in 1780 and became known as the Société Littéraire. However in 1789 it once again changed its name to The Académie Delphinale. The existence of this academy was short lived in the 1700's due to the fact that it like many other academies in France was forced to shut down by the Convention in 1793. The Académie Delphinale came back to existence in  1796 but would not regain the title of Académie Delphinale until 1844.

Resources
Scholarly Societies Project Waterloo Library


text:
User:JTR5121819

This is USER:JTR5121819.

Welcome to my user page.

--JTR5121819 19:21, 10 February 2010 (UTC)

title: User:JTR5121819


title: File:Egm6341.s10.Team2.Hw2problem81.jpg


title: File:Egm6341.s10.Team2.Hw2problem82.jpg


title: File:Egm6341.s10.Team2.Hw2problem83.jpg


text: File:F and fn.jpg

title: File:F and fn.jpg


text: File:In.jpg

title: File:In.jpg


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title: File:Wi.jpg


text:
Template:Newpage

When printing, this template starts a new page where it appears. The rest of the current page remains empty. The template uses Page-break and Page-break/end, which are not fully supported by most browsers; however, the particular usage in the template Newpage is supported by several popular browsers.

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text:
18th century European scholarly societies and academies/Societa Italiana

Summary
Name of Society: Societa Italiana

Alternative Name(s): La Societa dei XL, Accademia Nazionale delle Scienze detta dei XL

Country: Italy

City: Verona (within the Veneto region)

Active Dates: 1787- present

The Societa Italiana, or more popularly known as "La Societa dei XL", was founded in 1782 by Antonio Lorgna. Its purpose is to aid in the advancement of scientific knowledge and to support topics involving the history of science.

Reference

 * Accademia Nazionale delle Science detta dei XL


text:
User talk:Miss Alanious

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I see you might work on the same "project" as User:6500jmk4. If this is the case you might want to see this. --Gbaor 14:00, 11 February 2010 (UTC)

title: User talk:Miss Alanious