text:
User:Egm4313.s12.team3.droll

=Problem 4: Find a General Solution=

Problem Statement
For the following two problems find the general solution for the equations and then check the answer by substitution.
 * {| style="width:100%" border="0"

$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ $$\displaystyle 10y''-32y'+25.6y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part A
This problem is very similar to [| EGM4313 Team 3's Report 1 Problem 5, Part B]
 * {| style="width:100%" border="0"

$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ (4.1) From the characteristic equation (found on K 2011 p.54 (3)):
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.2) Using (4.2) and substituting the corresponding values for a and b you get:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ (4.3) Next we will check for the possibility of a Real Double Root (K 2011 p.55 "Case II") using:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a^2-4b=0 $$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (2\pi)^2-4(\pi^2)=0 $$ $$  \displaystyle 4\pi^2 - 4\pi^2=0 $$ $$  \displaystyle 0=0 $$ Since the Equation is a Real Double Root you now use:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.5) Solving for $$\displaystyle -a/2 $$ equals:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{-a}{2}=\frac{-2\pi}{2}=-\pi $$ (4.6) Substituting (4.6) into (4.5) leaves:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{-\pi x} $$ (4.7) Taking the first and second derivative of (4.7) gives:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y'=-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x} $$ (4.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y''=\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x} $$ (4.9) Finally inserting (4.7),(4.8), and (4.9) into (4.1) allows a check of the answer:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x})+2\pi (-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x})+\pi^2 ((C_1+C_2x)e^{-\pi x})=0 $$ (4.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Through simple canceling of like terms from (4.10) gives:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part B
Going over many of the same steps in Part A we have the following equation:
 * {| style="width:100%" border="0"

$$\displaystyle 10y''-32y'+25.6=0 $$ (4.11) Which when divided by 10 equals:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y''-3.2y'+2.56=0 $$ (4.12) From the characteristic equation (again found on K 2011 p.54 (3)):
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.13) Using (4.12) and substituting the corresponding values for a and b you get:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2-3.2 \lambda+2.56=0 $$ (4.14) We will again check for the possibility of a Real Double Root (also on K 2011 p.55 "Case II") using:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a^2-4b=0 $$ (4.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (-3.2)^2-4(2.56)=0 $$ $$  \displaystyle 10.24 - 10.24=0 $$ $$  \displaystyle 0=0 $$ Since this equation is also a Real Double Root you use:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.16) Solving for $$\displaystyle -a/2 $$ equals:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{-a}{2}=\frac{-(-3.2)}{2}=1.6 $$ (4.17) Substituting (4.17) into (4.16) gives the equation:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{1.6x} $$ (4.18) Taking the first and second derivative of (4.18) gives:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y'=((1.6)(C_1+C_2x)+C_2)e^{1.6x} $$ (4.19)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y''=2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})) $$ (4.20) BY inserting (4.18),(4.19), and (4.20) into (4.11) the answer can be verified:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

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

$$  \displaystyle 10(2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})))-32(((1.6)(C_1+C_2x)+C_2)e^{1.6x})+25.6((C_1+C_2x)e^{1.6x})=0 $$ (4.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As in Part A simple canceling of similar terms gives:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

=Problem 3: Spring-Dashpot-Mass System in Series=

Problem Statement
For the following system (as can be seen in the figure below) comprised of a spring, a dashpot, and a mass in series solve for the equation of motion and draw the free body diagrams.

Problem Solution
For information regarding the order and linearity of ordinary differential equations, see Loc Vu-Quoc class EGM 4313 notes Sec 1. Also, from Sec 1 of Loc Vu-Quoc's class EGM4313 (p.1-4).

Kinematics

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

$$\displaystyle {y}=y_k+y_c $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

 *

 *

 * }

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

$$\displaystyle {y_k}=Spring\,Displacement $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$\displaystyle {y_c}=Dashpot\,Displacement $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Kinetics

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

$$\displaystyle {a}=d^2y/dt^2={y}'' $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From th previous equation for acceleration and the mass (represented by the variable m) the sum of the forces can represented as follows:
 * {| style="width:100%" border="0"

$$\displaystyle m{y}''=f(t)-f_I $$     (2) Where:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle f_I=The\,Internal\,Forces\,on\,the\,Mass $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Taking the second derivative of (1):
 * {| style="width:100%" border="0"

$$\displaystyle {(y=y_k+y_c)}'' $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {y}={y_k}+{y_c}'' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From the free body diagrams above through geometry :
 * {| style="width:100%" border="0"

$$\displaystyle f_I=f_k $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle f_k=f_c $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which finally means that:
 * {| style="width:100%" border="0"

$$\displaystyle f_I=f_c=f_k $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle f_k=k{y_k} $$     (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle f_c=C{y_c}' $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From (7) we can solve for $$\displaystyle {y_c}'$$
 * {| style="width:100%" border="0"

$$\displaystyle k{y_k}=C{y_c}' $$     (10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {y_c}'=(k/C){y_k} $$     (11) Then solving for $$\displaystyle {y_c}''$$ equals:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {(y_c')}'=((k/C)y_k)'\,\,==>{y_c}''=(k/C){y_k}' $$     (12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then plugging (12) into (4):
 * {| style="width:100%" border="0"

$$\displaystyle {y}=y_k+(k/C){y_k}' $$     (13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting (13) for $$\displaystyle {y}''$$
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})=f(t)-f_I $$     (14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And solving for $$\displaystyle f(t)$$ with $$\displaystyle f_I=$$ (5)
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})+f_k=f(t) $$     (15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally with (8) substituted into $$\displaystyle f_k$$ in (15) gives the final answer of:
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})+k{y_k}=f(t) $$     (16)
 * style="width:95%" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

title: User:Egm4313.s12.team3.droll


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

title: File:Pea2.s12.sec43.d.djvu


text:
User:Egm4313.s12.Team15.j.shugart

Hey

title: User:Egm4313.s12.Team15.j.shugart


text:
User:Taoyao

I don't have anything particular I want to write here, but would like people to know I created this account in good faith!

title: User:Taoyao


text:
User:Egm6322.s12.team1.zheng.zx/R8

Problem 4: Plot the following pairs in separate figures
 Solved without assistance

Given
From [[media:pea1.f11.mtg42.djvu|Mtg 42]] Pg 42-2

Legendre polynomials

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ & {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1) \\ & {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

Legendre functions

$$\begin{align} & {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{Q}_{1}}(x)=\frac{1}{2}x\log \left( \frac{1+x}{1-x} \right)-1 \\ & {{Q}_{2}}(x)=\frac{1}{4}(3{{x}^{2}}-1)\log \left( \frac{1+x}{1-x} \right)-\frac{3}{2}x \\ & {{Q}_{3}}(x)=\frac{1}{4}(5{{x}^{3}}-3x)\log \left( \frac{1+x}{1-x} \right)-\frac{5}{2}{{x}^{2}}+\frac{2}{3} \\ \end{align}$$ And the fourth pair: $$ \begin{align} P_4(x)&=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \\ Q_4(x)&= ((105*atan(x*i)*i)*x^4 + 105*x^3 + (-90*atan(x*i)*i)*x^2 - 55*x + 9*atan(x*i)*i)/24 \end{align} $$

(Q4 is the result calculated by Matlab)

Find
1.Plot the above 5 pairs in separate figures

2.Observe even-ness and odd-ness of {Pi,Qi},i=1,...,4 and guess the value of the scalar products: $$ =\int^{\mu=+1}_{\mu=-1}P_i(\mu)Q_i(\mu)d\mu $$

Solution

So, for Pn: when n is odd, Pn is odd; when n is even, Pn is even;

for Qn: when n is odd, Qn is even; when n is even, Qn is odd;

Thus, PnQn is always odd

So, $$ =\int^{\mu=+1}_{\mu=-1}P_i(\mu)Q_i(\mu)d\mu=0 $$ for i=0,...,n

title: User:Egm6322.s12.team1.zheng.zx/R8


text:
Category:User nds

Users in this category indicate they have knowledge of language Low German.

title: Category:User nds


text:
Category:User nds-N

Users in this category indicate they have skill level N for language Low German.

title: Category:User nds-N


text:
Category:User simple

Users in this category indicate they have knowledge of language Simple English.

title: Category:User simple


text:
Category:User simple-N

Users in this category indicate they have skill level N for language Simple English.

title: Category:User simple-N


text:
Category:User nl-4

Users in this category indicate they have skill level 4 for language Dutch.

title: Category:User nl-4


text:
User:Abdars

Dear visitors
Hi! My name is Abdullah Danish Shervani... Thanks for paying a visit to my page. Please visit my talk page to leave a message.

Appreciated!

--  abdars

title: User:Abdars


text:
Category:User nds-4

Users in this category indicate they have skill level 4 for language Low German.

title: Category:User nds-4


text:
User:Egm4313.s12.team19.chiamchittrong

My contribution to Report 1

For Report 1, I did Problem R1.4

title: User:Egm4313.s12.team19.chiamchittrong


text:
User talk:Jade~enwikiversity

 Hello Jade, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

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

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Mathonius 23:28, 31 January 2012 (UTC)

Cars 3
Hello Jade, welcome to the English Wikiversity. A page you recently created, Cars 3, was outside of Wikiversity's scope and has been deleted; you might be looking for Wikipedia instead. Regards, Mathonius 23:28, 31 January 2012 (UTC)
 * Please stop adding content that's not covered by the project scope. Your articles, List of episodes of Deadly 360 and Coming Soon to Sky Movies Premiere, are not suitable for Wikiversity and have therefore been deleted. Mathonius 10:12, 7 February 2012 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Jade. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Jade~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:39, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Jade~enwikiversity


text:
User:Egm6322.s12.team2

Welcome to the public Wikiversity Web site for Team 2 of the University of Florida's Spring 2012 Principles of Engineering Analysis course (EGM 6322)

Please add your name on Team2 members =Homework Assignments= Report #8

Report #9

Report #10

Report #11

Report #12

Report #13

Report #14

=Team2 Members= [http://en.wikiversity.org/wiki/User:Egm6322.s12.team2. Manuel Steele]

Matt Shields

[http://en.wikiversity.org/wiki/User:Egm6322.s12.team2. Sungsik Kim] Lang Xia

=Useful Links= Class wiki

Course Report table

Course Lecture plan

title: User:Egm6322.s12.team2


text:
User:Egm6322.s12.team2/HW8

= R 8.1 - Gram matrix for Fourier basis functions =

Given
Fourier basis functions:
 * {| style="width:100%" border="0"

$$  \displaystyle \textbf{B} = \left\{1,cos(n\omega\theta),sin(n\omega\theta)\right\}, n = 1,2,....     $$ (8.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \mathbf{\Gamma} = [\mathbf{\Gamma_{ij}}] = [(\mathbf b_i \cdot \mathbf b_j)] \in \mathbb{R} $$     (8.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Definition of inner product for functions:
 * {| style="width:100%" border="0"

$$  \displaystyle \langle g,h \rangle = \int_a^b g(x)h(x)dx $$     (8.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
Part 1: Construct the Gram matrix for the Fourier basis functions (3)p.40-1 using the scalar product (2)p.42-8. What is the property of the Gram matrix? Part 2: Show that $$\left\{cos\theta,sin\theta\right\}$$ are linearly independent using the Wronskian; see (2)p.35-1

Solution
Part 1: To form the Gram matrix for the Fourier basis functions given in Eq. 8.1.1, it is necessary to compute the inner product for all combinations of basis functions; ie, $$g(\theta)=1,h(\theta)=cos(n\omega\theta); g(\theta)=1,h(\theta)=sin(n\omega\theta); g(\theta)=cos(n\omega\theta),h(\theta)=cos(n\omega\theta);$$ etc. Using Eq. 3 for all permutations, the components of the Gram matrix are computed below. The diagonal terms are given by:

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

$$  \displaystyle \begin{align} \langle 1,1 \rangle &= 1 \\ &=\Gamma_{11} \end{align} $$     (8.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \begin{align} \langle \text{cos}(n\omega\theta),\text{cos}(n\omega\theta) \rangle& = \int_0^T \text{cos}^2(n\omega\theta)d\theta \\ &= \int_0^T \frac{1}{2} + \frac{1}{2}\text{cos}(2n\omega\theta) d\theta \\ &=\left[\frac{\theta}{2} + \frac{\text{sin}(2n\omega\theta)}{4n\omega}\right]_0^T \\ &= \frac{T}{2} + \frac{\text{sin}(2n\omega T)}{4n\omega} \\ & =\Gamma_{22} \end{align} $$     (8.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \begin{align} \langle \text{sin}(n\omega\theta),\text{sin}(n\omega\theta) \rangle& = \int_0^T \text{sin}^2(n\omega\theta)d\theta \\ &= \int_0^T \frac{1}{2} - \frac{1}{2}\text{cos}(2n\omega\theta) d\theta \\ &=\left[\frac{\theta}{2} - \frac{\text{sin}(2n\omega\theta)}{4n\omega}\right]_0^T \\ &= \frac{T}{2} - \frac{\text{sin}(2n\omega T)}{4n\omega} \\ & =\Gamma_{33} \end{align} $$     (8.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

And the off-diagonal terms, noting that due to symmetry of the integral, $$\Gamma_{ij} = \Gamma_{ji}$$ for $$i\neq j$$ :

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

$$  \displaystyle \begin{align} \langle 1,\text{cos}(n\omega\theta) \rangle& = \int_0^T (1)(\text{cos}(n\omega\theta))d\theta \\ &= \frac{1}{n\omega}[\text{sin}(n\omega\theta)]_0^T \\ &= \frac{1}{n\omega}\text{sin}(n\omega T)\\ & =\Gamma_{12} = \Gamma_{21} \end{align} $$     (8.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \begin{align} \langle 1,\text{sin}(n\omega\theta) \rangle& = \int_0^T (1)(\text{sin}(n\omega\theta))d\theta \\ &= -\frac{1}{n\omega}[\text{cos}(n\omega\theta)]_0^T \\ &= -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) \\ & =\Gamma_{13} = \Gamma_{31} \end{align} $$     (8.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \begin{align} \langle \text{cos}(n\omega\theta),\text{sin}(n\omega\theta) \rangle& = \int_0^T (\text{cos}(n\omega\theta))(\text{sin}(n\omega\theta))d\theta \\ &= \int_0^T \frac{1}{2}\text{sin}(2n\omega\theta) d\theta \\ &= -\frac{1}{4n\omega}[\text{cos}(2n\omega\theta)]_0^T \\ &= -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right) \\ & =\Gamma_{23} = \Gamma_{32} \end{align} $$     (8.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substituting the results of Eq. 8.1.4-8.1.9 into Eq. 8.1.2 yields the Gram matrix for the Fourier basis functions:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma} = \begin{bmatrix} 1 & \frac{1}{n\omega}\text{sin}(n\omega T) & -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) \\ \frac{1}{n\omega}\text{sin}(n\omega T) & \frac{T}{2} + \frac{\text{sin}(2n\omega T)}{4n\omega} & -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right)\\ -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) & -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right) & \frac{T}{2} - \frac{\text{sin}(2n\omega T)}{4n\omega} \end{bmatrix} $$     (8.1.10)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

By inspection, it is clear that the columns of $$\mathbf{\Gamma}$$ are linearly independent as one cannot form one vector from a linear combination of the other two; thus, the Gram matrix for the Fourier basis functions is invertible for all values of n.

Part 2: Wronskian:
 * {| style="width:100%" border="0"

$$  \displaystyle W = \begin{vmatrix} u_1 & u_2\\ \dot{u_1} & \dot{u_2} \end{vmatrix} $$     (8.1.11) in which $$\dot{u}$$ represents the derivative of u. Thus, for the Fourier basis functions in Eq. 8.1.1,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} W &= \begin{vmatrix} \text{cos}\theta & \text{sin}\theta\\ \dot{\text{cos}\theta } & \dot{\text{sin}\theta} \end{vmatrix}\\ &= \begin{vmatrix} \text{cos}\theta & \text{sin}\theta\\ -\text{sin}\theta & \text{cos}\theta \end{vmatrix} \\ &= \text{cos}^2\theta + \text{sin}^2\theta = 1 \neq 0 \end{align} $$     (8.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

As the determinant is nonzero the basis functions are linearly independent.

= R 8.2 Fourier Series, Euler's Identity, and Fourier Coefficients=

Given
Fourier Series:

$$f(\theta) = a_0 + \sum_{n=1}^\infty a_n \frac{1}{2}(e^{in\omega\theta} + e^{-in\omega\theta}) + \sum_{n=1}^\infty b_n \frac{1}{2i}(e^{in\omega\theta} - e^{-in\omega\theta}) $$

Find
Verify (4). $$\int_0^T e^{i(n-m) \omega \theta} d\theta = $$ $$ \begin{cases} 0, & \text{if} n \ne m \\ T, & \text{if} n=m \end{cases}$$ Recover from (1) on p. 42-12 the Fourier coefficients found with the trigonometric identities above.

From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
Apply Euler's identity. $$e^{ikx} = \cos kx + i\sin kx$$

$$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ The first part of the solution requires the derivation of (4) from p. 42-11. First, let n=m and substitute into the derivation with the Euler identity. The solution requires expansion of the terms in the integral into terms of sine and cosine. The key is to recognize that the expansion of the Euler identity for the positive and negative exponents as shown below. $$e^{-in\omega \theta} = \cos{n\omega \theta} - i\sin{n\omega \theta}$$ $$e^{in\omega \theta} = \cos{n\omega \theta} + i\sin{n\omega \theta}$$ $$e^{in\omega \theta} + e^{-in\omega \theta} = \cos{n\omega \theta} + \cos{n\omega \theta} - i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the imaginary terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2\cos{n\omega \theta}$$

Divide the exponential summation by 2, and we have. $$(e^{in\omega \theta} + e^{-in\omega \theta})/2 = \cos{n\omega \theta}$$ Similarly, the pattern can be repeated for the sine term but leaves the imaginary component.

$$e^{in\omega \theta} - e^{-in\omega \theta} = \cos{n\omega \theta} - \cos{n\omega \theta} + i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the real terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2i\sin{n\omega \theta}$$

This final term can be divided by 2i to get the equivalent value below. $$(e^{in\omega \theta} - e^{-in\omega \theta})/2 = \sin{n\omega \theta}$$ The Euler identity applies to the integral below. $$\int_0^T e^{i(n-m) \omega \theta} d\theta = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}}$$ Note that if n=m then the following values occur. $$\cos{(n-m) \omega \theta} = \cos{0} = 1$$ $$\sin{(n-m) \omega \theta} = \sin{0} = 0$$ By substituting the values back into the integral, a simple integration is achieved for m=n. $$ \begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \int_0^T{(1 + 0)} \\ & = T \end{align}$$ Next, we concentrate on the condition where n does not equal m below. In this case, apply the substitution $$\text{T} = 2\pi / \omega$$. $$\begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^T \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^2{\pi / \omega} \\ & = \frac{1}{(n-m)\omega}[(\sin{(n-m)\omega 2{\pi / \omega}} - i\cos{(n-m)\omega 2{\pi / \omega}} - (\sin{0} - i\cos{0})] \\ & = \frac{1}{(n-m)\omega}[(\sin{2\pi (n-m)} - i\cos{2\pi (n-m)}) + i] \\ & = \frac{1}{(n-m)\omega}[0 - i + i] \\ & = 0 \end{align}$$ The 2nd part of problem R8.2 requires the recovery of Fourier coefficients. First, the defintion below must be restated as given for the problem.  $$ \bar a_n := \begin{cases} 2 a_0, & \text{for } n=0 \\ a_n - ib_n, & \text{for } n>0 \\ a_n + ib_n, & \text{for } n<0 \end{cases}$$ Next, we revisit the equation of interest below. $$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ In the equation above, if n=m, then we can re-examine the integral above.  First, in the summation, all terms for n go to 0 except where n=m, so the summation term no longer appears. Subsequently, the integral derivation of (4) on p. 42-11 is applied. So this gives the following equation. $$\int_0^T f(\theta) e^{-in \omega \theta}d\theta = \tilde{a_n} T = \tilde{a_m} T$$ Note that the constant can be expanded into the equation from which we recover the Fourier coefficients. $$\begin{align} \tilde{a_n} & = \frac{1}{2}\bar{a_n} \\ & = \frac{1}{T} \int_0^T f(\theta) e^{-in\omega\theta} d\theta \end{align}$$

The equation can be solved in 3 parts: n=0, n>0, and n<0. Let's start with n=0 for which the Fourier coefficient is defined below. Through team collaboration, the derivation below must include to proper definition of Fourier coefficients as noted by Wolfram at url http://mathworld.wolfram.com/FourierSeries.html. The key is to use the relationship for the boundaries of the integral where T=2L. $$\begin{align} \tilde{a_0} & = \frac{1}{T}\int_0^T f(\theta) e^{-in\omega \theta}d\theta & = \frac{1}{2L}\int_0^{2L} f(\theta)d\theta \end{align}$$ Next, the Fourier coefficient for n>0 can be derived with the given definitions and the Euler identity. The identity of the even function for cosine is used. $$\begin{align} {a_n - ib_n} & = \frac{2}{T}\int_0^T f(\theta) e^{-in\omega \theta}d\theta \\ & = \frac{2}{T}\int_0^T f(\theta)(\cos(-in\omega\theta) + i\sin(-n\omega\theta) d\theta \end{align}$$ In the deriviation above for n>0, the Fourier coefficient is the real part of the complex number in the integration. $$\begin{align} a_n & = \frac{2}{T} \mathbf{Re}\int_0^T f(\theta)(\cos{(-n\omega\theta)} + i\sin{(-n\omega\theta)}) \\ & = \frac{1}{L}\int_0^{2L} f(\theta) \cos{(-n\omega\theta)} \\ & = \frac{1}{L}\int_0^{2L} f(\theta) \cos{(n\pi\theta / L)} \end{align}$$ Next, we derive the Fourier coefficient for n<0 by simply taking the imaginary component of the main integral. It must be noted that sine is an odd function to solve the problem. $$\begin{align} b_n & = \frac{2}{T}\mathbf{Im} \int_0^T f(\theta)(\cos{(-n\omega\theta)} + i\sin{(-n\omega\theta)}) \\ & = \frac{2}{T}\int_0^T f(\theta) \sin{(-n\omega\theta)} \\ & = -\frac{1}{L}\int_0^{2L} f(\theta) \sin{(n\pi\theta /L)} \end{align}$$

=R8.3 Fourier Series and Real Components=

Given
We are given (2) and (3) on p. 42-10 below. $$f(\theta) = \frac{1}{2} \sum_{n=-\infty}^{+\infty} \bar a_n e^{in\omega\theta} = \sum_{n=-\infty}^{+\infty}\tilde a_n e^{in\omega\theta}$$ $$\bar a_n := \begin{cases} 2 a_0, &\text{for }n=0 \\ a_n-ib_n, & \text{for }n>0 \\ a_n + ib_n, &\text{for }n<0 \end{cases}$$

Find
Show that, equivalent to (2)-(3) p. 42-10, the Fourier series can be written as $$f(\theta) = \text{Re} \left(\sum_{n={\color{red} 0}}^{+\infty} \tilde a_n e^{in \omega \theta} \right)$$

From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
The first step is to write the original equation for the Fourier Series using exponential terms. $$ f(\theta) = \frac{1}{2}\sum_{n=-\infty}^{+\infty} \bar{a_n} e^{in\omega\theta} = \frac{1}{2}\sum_{n=-1}^{-\infty}(a_n + ib_n)e^{-in\omega\theta} + 2a_0 e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta}$$ The equation can be simplified by recognizing the conjugate values. $$\bar{a_n}^* = \bar{a_{-n}}$$ The conjugate method can be applied to the summation below. $$\sum_{n=-1}^{-\infty} (a_n + ib_n) e^{in\omega\theta} = \sum_{n=1}^{+\infty} (a_n - ib_n)e^{-in\omega\theta}$$ Substitute the right hand term above back into the main equation for $$f(\theta)$$. The derivation steps below show the real component for the solution. $$ \begin{align} f(\theta) & = \frac{1}{2}(\sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta} + 2{a_0}e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2}(2{a_0}e^{in\omega\theta}(e^{-in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2} (2a_0 e^{in\omega\theta} + \sum_{n=0}^{\infty}(a_n - ib_n)\cos{n\omega\theta} \\ & = \frac{1}{2} \sum_{n=0}^{\infty} \bar{a_n} \cos{n\omega\theta} \\ & = \sum_{n=0}^{\infty} \tilde{a_n} \cos{n\omega\theta} \\ & = \text{Re} \sum_{n=0}^{\infty} \tilde{a_n}(cos{n\omega\theta} + i\sin{n\omega\theta}) \\ & = \text{Re} \sum_{n=0}^{+\infty} \tilde{a_n} e^{in\omega\theta} \end{align} $$

=R*8.4 -Legendre Function =

Given
The following pairs: $$\displaystyle \{P_{0},\, Q_{0}\},\,\{P_{1},\! Q_{1}\},\cdots\,,\{P_{4},\, Q_{4}\}$$ (8.4.1)

where $$\displaystyle \{P_{i},\, Q_{i}\} $$ are Legendre polynomials.

Find
1.Plot the 8.4.1 in separate figures. 2.Observe even-ness and odd-ness of 8.4.1 and guess the value of the scalar products:

$$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du=?$$

Solution
1.The Legendre polynomial is as follows: $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 $$

$$\displaystyle {P}_{1}(x) = x $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1) $$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x) $$

$$\displaystyle {P}_{4}(x) = \frac{35\,{x}^{4}}{8}-\frac{15\,{x}^{2}}{4}+\frac{3}{8} $$ and $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) $$

$$\displaystyle {Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 $$

$$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x $$

$$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

$$\displaystyle {Q}_{4}(x) = \frac{35\,{x}^{4}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{15\,{x}^{2}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{8}+\frac{3\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{35\,{x}^{3}}{8}+\frac{55\,x}{24} $$ Using Maxima with GNUplot we get the following figures: 2. It is easy to find that $$\displaystyle {P}_{0}(x) = 1={P}_{0}(-x) $$

$$\displaystyle {P}_{1}(x) = x= -{P}_{1}(-x) $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1)= {P}_{2}(-x)$$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x)= -{P}_{2}(-x) $$ Thus, we are able to guess that $$\displaystyle {P}_{2n}(x)={P}_{2n}(-x) $$ are even-ness while $$\displaystyle {P}_{2n+1}(x)=-{P}_{2n+1}(-x) $$ are odd-ness. Using the same method, we will find that $$\displaystyle {Q}_{2n}(x)=-{Q}_{2n}(-x) $$ are odd-ness while $$\displaystyle {Q}_{2n+1}(x)={Q}_{2n+1}(-x) $$ are even-ness. From the above, we can easily find that $$\displaystyle {P}_{i}(x){Q}_{i}(x) $$ are odd-ness, Since the integral of smooth odd functions from -a to a are zero. That's to say $$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du= 0$$

Given
Consider the functions $$\displaystyle g_{i}:\,\mathbb{R}\rightarrow\mathbb{R}$$ and the function $$\displaystyle f=\sum_{i}g_{i}\,:\,\mathbb{R}\rightarrow\mathbb{R}$$

(8.5.1)

Find
Show that: 1. If $$\displaystyle g_{i}$$ is odd, then $$\displaystyle f $$ is odd. 2. If $$\displaystyle g_{i}$$ is even, then $$\displaystyle f$$  is even.

Solution
1. If $$\displaystyle g_{i} $$ is odd, we have $$\displaystyle g_{i}(-x)=-g_{i}(x) $$ then $$\displaystyle {\color{blue}f(-x)}=\sum_{i}g_{i}(-x)=g_{1}(-x)+g_{2}(-x)+\cdots+g_{i}(-x)$$ $$\displaystyle =-g_{1}(x)-g_{2}(x)-\cdots-g_{i}(x)=-\sum_{i}g_{i}(x)={\color{blue}-f(x)}$$ thus $$\displaystyle f$$ is odd. 2. If $$\displaystyle g_{i}$$ is even, we have $$\displaystyle g_{i}(-x)=g_{i}(x)$$ then $$\displaystyle {\color{blue}f(-x)}=\sum_{i}g_{i}(-x)=g_{1}(-x)+g_{2}(-x)+\cdots+g_{i}(-x)$$ $$\displaystyle =g_{1}(x)+g_{2}(x)+\cdots+g_{i}(x)=\sum_{i}g_{i}(x)={\color{blue}f(x)}$$ thus $$\displaystyle f$$ is even.

= R 8.6 - Legendre Polynomials =

Given

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

$$ P_n(x) = \sum^{[n/2]}_{i=0}(-1)^i\frac{(2n-2i)!x^(n-2i)}{2^ni!(n-i)!(n-2i)!} $$     (8.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find

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

$$ \displaystyle P_{2k}(x) is even$$ (8.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle P_{2k+1}(x) is odd$$ (8.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
Even Function $$\displaystyle F(x) = -F(x) $$

Odd Function $$ \displaystyle F(x) = F(-x) $$

When n=2k

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

$$ \displaystyle P_{2k}(x) = \sum^{[2k/2]}_{i=0}(-1)^i\frac{(4k-2i)!x^(2k-2i)}{2^2ki!(2k-i)!(2k-2i)!}$$ (8.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

above equation, x term is the factor to deciede even or odd function.

Therefore, extract x term and expand it

Let assume that k = 5

Then, equation is reduced below

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

$$ \displaystyle \sum_{i=0}^{[5]}x^{10-2i}= x^{10} + x^8 + x^6 + x^4 + x^2 + x^0 $$ (8.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

since 6 terms are even functions, therefore, $$ P_{2k} $$ is even function

when n=2k+1

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

$$ \displaystyle P_{2k+1}(x) = \sum^{[2k+1/2]}_{i=0}(-1)^i\frac{(2(2k+1)-2i)!x^{((2k+1)-2i)}}{2^(2k+1)i!(2k+1)-i)!((2k+1)-2i)!} $$     (8.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

extract x term and expand it

Assuming k=5

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

$$ \displaystyle \sum_{i=0}^{[5]}x^{11-2i}= x^{11} + x^9 + x^7 + x^5 + x^3 + x^1$$ (8.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since those terms are odd function, therefore, $$P_{2k+1} $$ is odd function

Egm6322.s12.sungsik 18:02, 1 February 2012 (UTC)

= R 8.7 - Change of basis to Legendre polynomials =

Given
Consider the following polynomial:

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

$$  \displaystyle \begin{align} q &\in \mathcal{P}_n \Rightarrow q(x) = \sum_{i=0}^n c_ix^i \\ n &= 5; c_0 = -4,c_1=-7,c_2=3,c_3=2,c_4=-5,c_5=1 \end{align} $$     (8.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Find $$d_i$$ such that
 * {| style="width:100%" border="0"

$$  \displaystyle q(x) = \sum_{i=0}^n d_iP_i(x) \in \mathcal{P}_n $$     (8.1.14) and plot Eq. 8.1.12 and 8.1.13 on separate figures
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
The goal of this problem is essentially to complete change of basis such that the polynomial q(x) given in Eq. 8.1.13 is mapped onto the orthogonal basis of Legendre polynomials $$P_i$$. This requires the coefficients to the polynomials which can be obtained using Eq. (3)p.42-16 in the class notes, adapted here in Eq. 8.1.15:
 * {| style="width:100%" border="0"

$$  \displaystyle d_i = \frac{2i+1}{2} \int_{x=-1}^{x=1} P_i(x) q(x)dx $$     (8.1.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The Legendre polynomials are well known and can be found online; see for example Wolfram's website (http://mathworld.wolfram.com/LegendrePolynomial.html). Thus, it is only necessary to multiply the polynomials and integrate from -1 to 1; this yields the required coefficients $$d_0$$ through $$d_5$$. For brevity, the algebra for the first coefficient will be included and only the values of the remaining coefficients will be presented. All calculations were verified using Mathematica.

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

$$  \displaystyle \begin{align} d_0& = \frac{2(0)+1}{2} \int_{x=-1}^{x=1} (1) (x^5-5x^4+2x^3+3x^2-7x-4)dx \\ &= \frac{1}{2}\left[\frac{1}{6}x^6 - x^5 + \frac{1}{2} x^4 +x^3 -\frac{7}{2}x^2 -4x\right]_{-1}^1 \\ &= -4 \end{align} $$     (8.1.16)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle d_1=-5.370; d_2= -0.857; d_3=1.244;d_4=-1.143;d_5=0.127 $$     (8.1.17)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

To verify that the mapping of q(x) onto the Legendre polynomial basis functions, the original polynomial and the summation of $$\sum_{i=0}^{n=5}d_i P_i$$ are plotted separately from $$-10<x<10$$. As expected, the figures are identical.

= References for Report 8 = Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 42) [[media:pea1.f11.mtg42.djvu|Mtg 42]] 23 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) [[media:pea2.s12.sec43.b.djvu|Mtg 43]] 24 Jan 2012.

= Contributing Members =

Egm6322.s12.team2.Xia I solved R*8.4 and R8.5, proofread R*8.1 and R*8.2

title: User:Egm6322.s12.team2/HW8


text:
User:Egm6322.s12.team2/HW9

= R 9.1 - Coefficients for heat conduction on a sphere =

Given

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

$$  \displaystyle f(\theta) = {T}_{0}cos(\theta) $$     (9.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(\theta) = {T}_{0}cos^{6}(\theta) $$     (9.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(\theta) = {T}_{0}exp[-\frac{2\theta}{\pi}^2] $$     (9.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(\mu) = {T}_{0}sinh(1-\mu^2) $$     (9.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
1. Find $$ A_{i}$$ 2. Compute the first 5 non-zero coefficients 3. Plot Exact Function, Fourier-Legendre series with 3 Terms and 5 Terms

Solution
Problem 1.
 * {| style="width:100%" border="0"

$$  \displaystyle \int_{-1}^{1}({p}_{2k}\times Even) \ne 0, \int_{-1}^{1}({p}_{2k}\times Odd) = 0, \int_{-1}^{1}({p}_{2k+1}\times Even) = 0, \int_{-1}^{1}({p}_{2k+1}\times Odd) \ne 0
 * style="width:95%" |
 * style="width:95%" |

$$     (9.1.5)
 *
 * }

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

$$  \displaystyle \mu = cos(\theta) $$     (9.1.6) Converting to $$\theta$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle f(\mu) = {T}_{0}cos(\mu) $$     (9.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(\mu) = {T}_{0}(\mu)^{6} $$     (9.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(\mu) = {T}_{0}exp[-(\frac{2{cos}^{-1}(\mu)}{\pi})^2] $$     (9.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle f(\mu) = \sum_{j=0}^\infin {A}_{j}{P}_{j}(\mu) $$     (9.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)F(\mu)d(\mu) $$     (9.1.11) Therefore,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)T_{0}\mu d(\mu) $$     (9.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)T_{0}(\mu)^{6}d(\mu) $$     (9.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu){T}_{0}exp[-\frac{2{cos}^{-1}(\mu)}{\pi}^2]d(\mu) $$     (9.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu){T}_{0}sinh(1-\mu^2)d(\mu) $$     (9.1.15) Problem 2. Substitute j into Eq 9.1.11 to Eq 9.1.15 $$ A_{1}= T_{0},A_{3}=0,A_{5}=0,A_{7}=0, A_{9}=0$$ $$ A_{0}= T_{0}, A_{2}=\frac{1}{7}T_{0}, A_{4}=\frac{10}{21}T_{0}, A_{6}=\frac{24}{77}T_{0}, A_{8}=0$$ $$ A_{0}= 0.417T_{0}, A_{1}=0.478T_{0}, A_{2}=0.097T_{0}, A_{3}=0.0076T_{0}, A_{4}=-0.00157T_{0}$$ $$ A_{0}= 0.7459T_{0}, A_{2}=-0.7997T_{0}, A_{4}=0.06671T_{0}, A_{6}=-0.01341T_{0}, A_{8}=0.00046T_{0}$$ Problem 3. Use Equation(9.1.10),Expand the fourier-legendre function Egm6322.s12.sungsik 01:54, 7 February 2012 (UTC) Egm6322.s12.sungsik 02:05, 10 February 2012 (UTC)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

= R 9.2 - Coefficients for general solution of heat conduction on a sphere =

Given
Boundary condition on the surface of the sphere:
 * {| style="width:100%" border="0"

$$  \displaystyle f(\mu) = T_0(1-\mu ^2)^2 $$     (9.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Fourier-Legendre series for the boundary condition in 9.2.1:
 * {| style="width:100%" border="0"

$$  \displaystyle f(\mu) = T_0(1-\mu ^2)^2 = \sum_{j=0,2,4}A_jP_j(\mu) $$     (9.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

General solution for heat conduction on a sphere:
 * {| style="width:100%" border="0"

$$  \displaystyle u(r,\theta) = \sum_{j=0}^\infty \bar{A_j}r^jP_j(\mu) $$     (9.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \sum_{j=0}^\infty \bar{A_j}P_j(\mu) = \sum_{j=0}^\infty A_jP_j(\mu) \Rightarrow \bar{A_j} = A_j, \forall j $$ (9.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
The purpose of this problem is to show that the coefficients of the general formulation of the heat conduction problem on a sphere (obtained through solving the Laplace equation via separation of variables) are equal to the coefficients obtained from expanding the Fourier-Legendre series subject to the boundary condition of Eq. 9.2.1. As this condition is valid when r=1, Eq. 9.2.3 becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle u(r=1,\theta) = \sum_{j=0}^\infty \bar{A_j}P_j(\mu) $$     (9.2.5) Applying the boundary conditions and comparing to the Fourier-Legendre expansion of 9.2.2:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \begin{align} \sum_{j=0}^\infty \bar{A_j}P_j(\mu) &= f(\mu) = T_0(1-\mu^2)^2 \\ &= \sum_{j=0}^\infty A_jP_j(\mu) \end{align} $$     (9.2.6) Recall from [[media:Pea2.s12.sec43.a.djvu|class notes]] that the only nonzero indices of $$A_j$$ are j=0,2,4. This is due to the orthogonality and oddness/eveness of the Legendre polynomials. This is covered in the lecture notes but will be reviewed here for clarity.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We know that the Legendre polynomials, by definition, are a subset of all polynomials:
 * {| style="width:100%" border="0"

$$  \displaystyle P_j\in\mathcal P_j $$     (9.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Any set of polynomials of degree n are spanned by the same number of orthogonal polynomials (for our purposes, Legendre polynomials will suffice). This also indicates that a polynomial of degree m > n is orthogonal to the subspace $$\mathcal{P_n}$$ as no linear combination of degree n polynomials can produce a new polynomial of a higher order. This is written as:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} P_m &\bot \{P_j,j=0,1,2,\cdots,n\}\forall m \ge n+1\\ P_m &\bot \mathcal P_n \forall m \ge n+1 \end{align} $$     (9.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The boundary condition of the problem, given in Eq. 9.2.1, is a 4th order polynomial; thus, when computing the coefficients of the Fourier-Legendre expansion in Eq. 9.2.2 according to the formula
 * {| style="width:100%" border="0"

$$  \displaystyle A_j = \frac{\langle P_j,f(\mu) \rangle}{\langle P_j,P_j \rangle} $$     (9.2.9) the numerator on right hand side will be zero for j>4 due to the orthogonality of the polynomials (the inner product of orthogonal functions is zero). Thus,
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \langle P_j,f(\mu) \rangle = 0, \forall j \ge 5 $$     (9.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

It should also be noted that the boundary condition $$f(\mu)$$ is an even function whereas Legendre polynomials of order 2k+1 for k = 0,1,2,... are odd functions; thus, since the inner product of odd and even functions are zero,
 * {| style="width:100%" border="0"

$$  \displaystyle A_{2k+1} \sim \langle P_{2k+1},f(\mu) \rangle = 0 \text{ for } k=0,1,2,\cdots $$     (9.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Eq. 9.2.10 and 9.2.11, the only surviving coefficients are $$A_0, A_2 \text{ and } A_4$$. Thus, after evaluating these coefficients using Eq. 9.2.9, Eq. 9.2.6 becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} \sum_{j=0}^\infty \bar{A_j}P_j(\mu) &= \sum_{j=0,2,4} A_jP_j(\mu) \\ &= \frac{8T_0}{15}P_0(\mu) - \frac{16T_0}{21}P_2(\mu) + \frac{8T_0}{35}P_4(\mu) \end{align} $$     (9.2.12) As the expansion of the left hand side of Eq. 9.2.12 is in terms of the same Legendre polynomials as the right hand side, it is clear that the coefficients must be identical. For $$j\neq 0,2,4$$, the values of $$\bar{A_j}$$ are zero and for $$j=0,2,4$$ the coefficients must be the same as $$A_0, A_2, \text{and} A_4$$ given in the [[media:Pea2.s12.sec43.a.djvu|class notes]]. Thus,
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \bar{A_j}=A_j,\forall j $$ (9.2.13)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=R*9.3 Identities of Legendre Equation=

Given
We are given the 1st and 2nd homogeneous solutions of the Legendre Equation from p.41-4, 41-5, 7-1, and 37-4. $$\{P_j, j=0,1,2,...\}$$ $$\{Q_j, j=0,1,2,...\}$$

Find
Derive the following identities for the 2nd homogeneous solutions of the Legendre Equation. $$Q_0(x) = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right) = \tanh ^{-1} x$$ $$Q_1(x) = \frac{x}{2} \log\left(\frac{1+x}{1-x} \right) - 1 = x \tanh^{-1}x - 1$$ Also find the expression for the $$Q_2(x)$$ in terms of the inverse hyperbolic tangent of x $$Q_2(x) = \frac{1}{4}(3x^2 - 1) \log \left(\frac{1+x}{1-x} \right) - \frac{3}{2}x $$

From lecture notes [[media:Pea2.s12.mtg43.b.djvu|Mtg 43b]]

Solution
The Legendre differential equation was defined in lecture page 7-1 for EGM 6321. $$L_2(y) := (1-x^2)y'' - 2xy' + n(n+1)y = 0$$ King (2003) very clearly derives the solutions for these equations in pages 5-6 and 31-33. First assert the method of Frobenius. Choose a power series solution of the form below. $$y = \sum_{i=0}^{\infty}a_i x^{i+c}$$ Substitute the term above into Legendre's equation to get a series representation. $$(1-x^2)\sum_{i=0}^{\infty}a_i(i+c)(i+c-1)x^{i+c-2} - 2x \sum_{i=0}^{\infty} a_i(i+c)x^{i+c-1} + n(n+1) \sum_{i=0}^{\infty}a_ix^{i+c} = 0$$ Rearrange the equation to simplify the grouping of terms below. $$\sum_{i=0}^{\infty}a_i(i+c)(i+c-1)x^{i+c-2} +\sum_{i=0}^{\infty} a_i(n(n+1) - (i + c)(i + c + 1))x^{i+c}$$ Further simplification gives a form that allows derivation of homogeneous solutions. $$a_0c(c-1)x^{c-2} + a_1c(c+1)x^{c-1} + \sum_{i=2}^{\infty}(a_i(i+c)(i+c-1) + a_{i-2}(n(n+1) - (i+c-2)(i+c-1))) x^{i+c-2} = 0$$ The series definitions gives an indicial equation of the form c(c-1) = 0. Apply Frobenius General Rule 1 on p. 14 of King (2003) - "If the indicial quation has two distinct roots, $$c = \alpha, \beta (\alpha < \beta) , $$whose difference is an integer, and one of the coefficients of $$x^k$$ becomes indeterminate on putting $$c = \alpha$$, both solutions can be generated by putting $$c = \alpha$$ in the recurrence relation." Therefore, the solution of Legendre's equation is given in expanded form below. $$y = a_0[1 - \frac{n(n+1)}{2!}x^2 - {3*2 - n(n+1))(n(n+1)}{4!}x^4 + ...] + a_1[x + \frac{(2*1 - n(n+1))}{3!}x^3$$ $$+ \frac{(4*3 - n(n+1))(2*1 - n(n+1))}{5!}x^5 + ...]$$ The simplified form of the soltuion is given below. This equation gives rise to the Legendre polynomials. $$y = AP_n(x) + BQ_n(x)$$ In the solution, the Legendre polynomials are given by the following summation. $$P_n(x) = \sum_{r=0}^m(-1)^r \frac{(2n-r)!x^{n-24}}{x^nr!(n-r)!(n-24)!}$$ The first three terms are listed as $$P_0, P_1, and P_2$$. $$P_0(x) = 1, P_1(x) = x, P_2(x) = \frac{1}{2} (3x^2 - 1)$$ These terms provide the basis for finding the Legendre functions: $$Q_0(x), Q_1(x), and Q_2(x)$$. The key is to utilize the reduction of order formula. For a homogeneous differential equation, we have the definition below. $$y'' + a_1(x)y' + a_0(x)y = 0$$ If a solution $$u_1(x)$$ exists, then a 2nd solution is given by this integral. $$u_2(x) = u_1(x)\int^x \frac{1}{u_1^2(t)} \exp{-\int^t a_1(s)ds} dt$$ If we apply this equation to Legendre polynomials, then the following form is given for the differential equation. $$y'' - \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0$$ The astute mathematician must note that the integral for the inner integral must be derived first in the reduction of order formula. $$\int^t a_1(s)ds = \int^t \frac{2s}{1-s^2} ds = \log(1 - t^2)$$ So the Legendre function can be derived by using substitution variables of t. In general, the equation below gives the nth Legendre function for the nth Legendre polynomial. $$Q_n(x) = P_n(x)\int^x{\frac{1}{{P_n(t)}^2}} \exp{-\log{(1-t^2)}} dt$$ For n=0, the following derivation occurs for $$Q_0(x)$$. $$ \begin{align} Q_0(x) & = P_0(x)\int^x {\frac{1}{{P_0(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = \int^x{\frac{dt}{(1-t^2)}} \\ & = \int^x{(\frac{1}{2(1+t)} + \frac{1}{2(1-t)})}dt \\ & = [\frac{1}{2}\log{\frac{1+t}{1-t}}]^x \\ & = \frac{1}{2} \log{\frac{(1+x)}{(1-x)}} \end{align} $$ Similarly, for n=1, $$Q_1(x)$$ can be derived. $$ \begin{align} Q_1(x) & = P_1(x)\int^x {\frac{1}{{P_1(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = x\int^x{\frac{dt}{t^2(1-t^2)}} \\ & = x\int^x{(\frac{1}{t^2} + \frac{1}{2(1+t)} + \frac{1}{2(1-t)})}dt \\ & = x[-\frac{1}{t} + \frac{1}{2}\log{\frac{1+t}{1-t}}]^x \\ & = x\frac{1}{2} \log{\frac{(1+x)}{(1-x)}} - 1 \end{align} $$ Lastly, for n=2, the integration gives $$Q_2(x)$$. $$ \begin{align} Q_2(x) & = P_2(x)\int^x {\frac{1}{{P_2(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = \frac{1}{4}(3x^2-1) \log{\frac{1+x}{1-x}} - \frac{3}{2}x \end{align} $$ Note that in each Legendre function there exists a term that can be simplified into the inverse hyperbolic tangent. Apply the definition from Wolfram for inverse hyperbolic tangent at http://mathworld.wolfram.com/InverseHyperbolicTangent.html. $$ \begin{align} \frac{1}{2} \log{(1+x)}{(1-x)} & = \frac{1}{2} \log{(1+x)} - \log{(1-x)} \\ & = \frac{1}{2} \ln{(1+x)} - \ln{(1-x)} \\ & = \tanh^{-1}{\frac{(1+x)}{(1-x)}} \end{align} $$ Therefore, the following definitions are derived for 9.3. $$Q_0(x) = \frac{1}{2} \log{\frac{(1+x)}{(1-x)}} = \tanh^{-1}{\frac{(1+x)}{(1-x)}}$$ $$Q_1(x) = x\frac{1}{2} \log{\frac{(1+x)}{(1-x)}} - 1 = x\tanh^{-1}{\frac{(1+x)}{(1-x)}}-1$$ $$Q_2(x) = \frac{1}{4}(3x^2-1) \log{\frac{1+x}{1-x}} - \frac{3}{2}x = \frac{1}{2}(3x^2-1) \tanh^{-1}{\frac{1+x}{1-x}} - \frac{3}{2}x $$ Note 4 tilde command by Manuel Steele below. Egm6322.s12.team2.steele.m2 08:24, 10 February 2012 (UTC)

=R 9.4 -Verify the second homogeneous solutions of Legendre equation =

Given
Given the following equations $$\displaystyle Q_{n}(x)=P_{n}(x)\tanh^{-1}x-2\sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)$$
(9.4.1) $$\displaystyle J\,:=1+2\left[\frac{n-1}{2}\right]=1+2\mathrm{Int}(\frac{n-1}{2})$$
(9.4.2)

Find
Verify the expressions for $$\displaystyle Q_{0},\, Q_{1},\, Q_{2}$$ as shown on the following equations $$\displaystyle Q_{0}(x)=\frac{1}{2}\log(\frac{1+x}{1-x})=\tanh^{-1}x$$
(9.4.3) $$\displaystyle Q_{1}(x)=\frac{x}{2}\log(\frac{1+x}{1-x})-1=x\tanh^{-1}x-1$$
(9.4.4) $$\displaystyle Q_{2}(x)=\frac{1}{2}(3x^{2}-1)\log(\frac{1+x}{1-x})-\frac{3}{2}x= ?$$
(9.4.5)

Solution
The Legendre polynomial is as follows:

$$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$
(9.4.6) where $$\displaystyle m = \frac{n}{2}$$.

Thus $$\displaystyle {P}_{-1}(x) = 0$$

$$\displaystyle {P}_{0}(x) = 1 $$

$$\displaystyle {P}_{1}(x) = x $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1) $$

Then from 9.4.1 we have $$\displaystyle Q_{0}(x)=\underbrace{P_{0}(x)}_{=1}\tanh^{-1}x-2\left(\frac{-2+1}{-1+1}\right)^{\dagger}\underbrace{P_{0-1}(x)}_{=0}=\tanh^{-1}x $$ $$\displaystyle Q_{1}(x)=\underbrace{P_{1}(x)}_{=x}\tanh^{-1}x-2\frac{2-2+1}{2-1+1}\underbrace{P_{1-1}(x)}_{=1}=x\tanh^{-1}x-1$$ $$\displaystyle Q_{2}(x)=\underbrace{P_{2}(x)}_{(3x^{2}-1)/2}\tanh^{-1}x-2\sum_{j=1,3}^{3}\frac{4-2j+1}{(4-j+1)j}P_{2-j}(x)$$ $$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-2\left(\frac{3}{4}\underbrace{P_{1}(x)}_{=x}+\frac{4-6+1}{(4-3+1)\cdot3}\underbrace{P_{-1}(x)}_{=0}\right)$$ $$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-\frac{3}{2}x$$ $$\displaystyle \dagger $$ Although $$\displaystyle P_{-1}(0)$$, the fact that 1-1=0  makes this term meaningless. We'd better define $$\displaystyle \sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}=0 $$, if $$\displaystyle n=0 $$

= R 9.5 - Legendre Function =

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

$$  \displaystyle Q_n(x) = P_{n}(x)\tanh^{-1}x-2\sum_{j=1,3,5,...}^J\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x) $$     (9.5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle J := 1 + 2\left[\frac{n-1}{2}\right] = 1+2\text{Int}\left(\frac{n-1}{2}\right) $$     (9.5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
$$Q_{n}$$ is odd or even depending on the index n

Solution

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

$$  \displaystyle \tan^{-1}(x) = \text{Odd}$$ (9.5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle P_{2k} = \text{Even}$$ (9.5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle P_{2k+1} = \text{Odd}$$ (9.5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When n=even From Eq(9.5.1) $$\text{Even}\times\text{Odd}-\text{Odd}= \text{Odd}-\text{Odd}$$ when n=Odd $$\text{Odd}\times\text{Odd} = \text{Even}-\text{Even}$$ Hence, $$ Q_{0},Q_{2},...,Q_{2k} = \text{Odd Function}, k=1,2,3,...$$ $$ Q_{1},Q_{3},...,Q_{2k+1} = \text{Odd Function}, k=1,2,3,...$$ Egm6322.s12.sungsik 02:22, 7 February 2012 (UTC)

= R 9.6 - Equalities in spherical coordinates =

Given
Definition of $$\rho$$: Boundary condition on the surface of the sphere:
 * {| style="width:100%" border="0"

$$  \displaystyle \rho = \frac{r_P}{r_Q} $$     (9.6.1) Cosine law:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2r_Pr_Q\text{cos}\gamma $$     (9.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Show the following equalities:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} (r_{PQ})^2 &= \left[\left(\frac{r_P}{r_Q}\right)^2 + 1 - 2\frac{r_P}{r_Q}\text{cos}\gamma\right]\\ &= (r_Q)^2[\rho^2 + 1 -2\rho\text{cos}\gamma] \end{align} $$     (9.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \text{cos}\gamma = \text{cos}\theta_Q\text{cos}\theta_P\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_Q\text{sin}\theta_P $$     (9.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
The first inequality in Eq. 9.6.3 can be shown trivially simply by substituting the definition of $$\rho$$ from Eq. 9.6.1 into the equation. This immediately yields the solution.
 * {| style="width:100%" border="0"

$$  \displaystyle (r_{PQ})^2 = [(\rho)^2 + 1 - 2(\rho)\text{cos}\gamma] $$     (9.6.5) Showing the equality in Eq. 9.6.4 requires somewhat more effort. It is first necessary to define the coordinates of points P and Q in spherical coordinates, as shown in the [[Media:Pea2.s12.sec43.c.djvu|class notes]]:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

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

$$  \displaystyle \begin{align} \vec{x_p} &= (r_P\text{cos}\theta_P\text{cos}\phi_P,r_P\text{cos}\theta_P\text{sin}\phi_P,r_P\text{sin}\theta_P)\\ \vec{x_q} &= (r_Q\text{cos}\theta_P\text{cos}\phi_Q,r_Q\text{cos}\theta_Q\text{sin}\phi_Q,r_Q\text{sin}\theta_Q) \end{align} $$     (9.6.6) The distance d between two points $$\vec{x_1}, \vec{x_2}$$ in Cartesian space is given by the three-dimensional Pythagorean theorem:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2} $$     (9.6.7) Substituting the spherical coordinates in Eq. 9.6.6 into Eq. 9.6.7 yields an expression for $$r_{PQ}$$:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle r_{PQ} = \sqrt{A + B + C} $$ (9.6.8) where
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle \begin{align} A &= (r_P\text{cos}\theta_P\text{cos}\phi_P - r_Q\text{cos}\theta_P\text{cos}\phi_Q)^2 = (r_P\text{cos}\theta_P\text{cos}\phi_P)^2 - 2r_Pr_Q\text{cos}\theta_P\text{cos}\phi_P\text{cos}\theta_Q\text{cos}\phi_Q + (r_Q\text{cos}\theta_Q\text{cos}\phi_Q)^2\\ B &= (r_P\text{cos}\theta_P\text{sin}\phi_P - r_Q\text{cos}\theta_P\text{sin}\phi_Q)^2 = (r_P\text{cos}\theta_P\text{sin}\phi_P)^2 - 2r_Pr_Q\text{cos}\theta_P\text{sin}\phi_P\text{cos}\theta_Q\text{sin}\phi_Q + (r_Q\text{cos}\theta_Q\text{sin}\phi_Q)^2\\ C &= (r_P\text{sin}\theta_P - r_Q\text{sin}\theta_Q)^2 = (r_P\text{sin}\theta_P)^2 - 2r_Pr_Q\text{sin}\theta_P\text{sin}\theta_Q + (r_Q\text{sin}\theta_Q)^2 \end{align} $$     (9.6.9) Expanding and collecting like terms yields:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle r_{PQ}^2 = r_P^2(\underbrace{\text{cos}^2\theta_P[\underbrace{\text{cos}^2\phi_P+\text{sin}^2\phi_P}_{=1}] + \text{sin}^2\theta_P}_{=1}) + r_Q^2(\underbrace{\text{cos}^2\theta_Q[\underbrace{\text{cos}^2\phi_Q+\text{sin}^2\phi_Q}_{=1}] + \text{sin}^2\theta_Q}_{=1}) - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\phi_P\text{cos}\theta_Q\text{cos}\phi_Q + \text{cos}\theta_P\text{sin}\phi_P\text{cos}\theta_Q\text{sin}\phi_Q + \text{sin}\theta_P\text{sin}\theta_Q] $$     (9.6.10) Again collecting like terms in the term on the far right yields:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle r_{PQ}^2 = r_P^2 + r_Q^2 - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\theta_Q(\text{cos}\phi_P\text{cos}\phi_Q + \text{sin}\phi_P\text{sin}\phi_Q) +     \text{sin}\theta_P\text{sin}\theta_Q] $$     (9.6.11) Recognizing the term inside the parenthesis as the sum of cosines identity shown in Eq. 9.6.12:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle \text{cos}(u\pm v) = \text{cos}u\text{cos}v\mp\text{sin}u\text{sin}v $$     (9.6.12) Eq. 9.6.11 can now be written as:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle r_{PQ}^2 = r_P^2 + r_Q^2 - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\theta_Q\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_P\text{sin}\theta_Q] $$     (9.6.13) Comparing this expression for $$r_{PQ}$$ with the one given by the law of cosines in Eq. 9.6.2 indicates that the term in the brackets must be equal to cos$$\gamma$$ for the equality to hold true:
 * style="width:92%|
 * style="width:92%|
 *

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

$$  \displaystyle \text{cos}\gamma = \text{cos}\theta_P\text{cos}\theta_Q\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_P\text{sin}\theta_Q $$     (9.6.14)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=R9.7 Identities of Binomial Theorem=

Given
We are given the following Binomial Equation from 43-21. $$(1+x)^r = \sum^{\infty}_{k=0} \binom{r}{k} \, x^k, \text{for} |x| <1$$

Find
Derive the following identities for the Binomial Theorem. $$(1-x)^{-\frac{1}{2}} = \sum^{\infty}_{i=0} \alpha_i x^i$$ $$\alpha_i = \frac{1 \cdot 3 \cdots(2i-1)}{2 \cdot 4 \cdots(2i)}$$

From lecture notes [[media:Pea2.s12.mtg43.c.djvu|Mtg 43c]]

Solution
The first step is to write the general equation of the Binomial Theorem. $$(x+y)^n = \sum^n_{k=0} \binom{n}{k} x^{n-k}y^k$$ Note that the Binomial Theorem can be extended to infinity when $$n \equiv r \in \mathbb{R}$$. $$(x+y)^r = \sum^{\infty}_{k=0} \binom{r}{k} x^{r-k} \, y^k$$ The biniomial coefficent can be expanded as follows. $$\binom{r}{k} = \frac{r}{(r-1) \cdots(r-k+1)}{k!}$$ Note that if we set y=1 then the equation can be simplified. $$(1+x)^r = \sum{\infty}_{k=0} \binom{r}{k} \, x^k, \text{for} |x| < 1$$ The Binomial Equation was adapted by Isaac Newton for real numbers beyond non-negative integers. The key is to factor (n-k)! from the binomial coefficient. Then if r=-s, we get the following form of the Binomial Equation. $$\frac{1}{{(1-x)}^s} = \sum^{\infty}_{k=0} \binom{s+k-1}{k} x^k$$ Note that if s = 1/2, then the equation can be further simplified. $$\frac{1}{(1-x)^{\frac{1}{2}}} = \sum^{\infty}_{k=0} \binom{k-{\frac{1}{2}}}{k} x^k$$ Realize that the binomial coefficient is a fraction, so the numerator and denominator can be multiplied by 2. This will give us the form for the solution. $$\frac{1}{(1-x)^{\frac{1}{2}}} = \sum^{\infty}_{k=0} \binom{2k-1}{2k} x^k$$ Therefore, the given steps derive the desired form below where we let i=k. $$(1-x)^{-\frac{1}{2}} = \sum^{\infty}_{i=0} \alpha_i x^i$$ $$\alpha_i = \frac{1 \cdot 3 \cdots(2i-1)}{2 \cdot 4 \cdots(2i)} = \binom{2k-1}{2k}$$

Manuel Steele 4 tilde command signature Egm6322.s12.team2.steele.m2 08:26, 10 February 2012 (UTC)

=R* 9.8 -Generating Legendre polynomials =

Given
Part 1 Recurrence relation RR2: $$\displaystyle (n+1)P_{n+1}-(2n+1) \mu P_n+nP_{n-1}=0 $$
(9.8.1) with $$\displaystyle P_0(\mu)=1 $$ and $$ \displaystyle P_1(\mu)= \mu$$ Part 2 Define: $$ \displaystyle A(\mu ,\rho):=1-2\mu\rho+\rho^2=:1-x$$ $$\displaystyle -x:=-2\mu\rho+\rho^2$$
(9.8.2) and the binomial theorem: $$\displaystyle (1-x)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\alpha_{i}x^{i}$$
(9.8.3) $$\displaystyle \alpha_{i}=\frac{1\cdot3\cdots(2i-1)}{2\cdot4\cdots(2i)}$$
(9.8.4)

Find
Finding $$ \displaystyle \{P_{2},\cdots,P_{6}\}$$ in terms of condition 1, and 2 respectively. and compare the result

Solution
Part 1 From 9.8.1, we have $$\displaystyle P_{n+1}=\frac{2n+1}{n+1}\mu P_{n}-\frac{n}{n+1}P_{n-1}$$
(9.8.5) with $$\displaystyle P_0(\mu)=1 $$ and $$ \displaystyle P_1(\mu)= \mu$$ then $$\displaystyle \begin{array}{l} \\ P_{2}=\frac{3}{2}\mu P_{1}-\frac{1}{2}P_{0}=\frac{1}{2}(3\mu^{2}-1)\\ \\ P_{3}=\frac{5}{3}\mu P_{2}-\frac{2}{3}P_{1}=\frac{1}{2}(5\mu^{3}-3\mu)\\ \\ P_{4}=\frac{7}{4}\mu P_{3}-\frac{3}{4}P_{2}=\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\\ \\ P_{5}=\frac{9}{5}\mu P_{4}-\frac{4}{5}P_{3}=\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\\ \\ P_{6}=\frac{11}{6}\mu P_{5}-\frac{5}{6}P_{4}=\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5) \end{array}$$

Part 2 Since $$\displaystyle x:=2\mu\rho-\rho^{2}=\rho(2\mu-\rho)$$ From the 9.8.3 and 9.8.4 we have $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{\infty}\alpha_{i}x^{i}=\sum_{i=0}^{\infty}\alpha_{i}\rho^{i}(2\mu-\rho)^{i}$$ when computing for $$\displaystyle P_{6}$$, we can see that the highest order is $$\displaystyle i=6$$，thus $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{6}\alpha_{6}x^{6}=\sum_{i=0}^{6}\alpha_{6}\rho^{6}(2\mu-\rho)^{6}$$ $$\displaystyle =\frac{231{\left(2\mu\rho-{\rho}^{2}\right)}^{6}}{1024}+\frac{63{\left(2\mu\rho-{\rho}^{2}\right)}^{5}}{256}+\frac{35{\left(2\mu\rho-{\rho}^{2}\right)}^{4}}{128}+\frac{5{\left(2\mu\rho-{\rho}^{2}\right)}^{3}}{16}+\frac{3{\left(2\mu\rho-{\rho}^{2}\right)}^{2}}{8}+\frac{2\mu\rho-{\rho}^{2}}{2}+1$$ $$\displaystyle =\frac{231\,{\rho}^{12}}{1024}-\frac{693\,\mu\,{\rho}^{11}}{256}+\frac{3465\,{\mu}^{2}\,{\rho}^{10}}{256}-\frac{63\,{\rho}^{10}}{256}-\frac{1155\,{\mu}^{3}\,{\rho}^{9}}{32}+\frac{315\,\mu\,{\rho}^{9}}{128}$$ $$\displaystyle +\frac{3465\,{\mu}^{4}\,{\rho}^{8}}{64}-\frac{315\,{\mu}^{2}\,{\rho}^{8}}{32}+\frac{35\,{\rho}^{8}}{128}-\frac{693\,{\mu}^{5}\,{\rho}^{7}}{16}+\frac{315\,{\mu}^{3}\,{\rho}^{7}}{16}-\frac{35\,\mu\,{\rho}^{7}}{16}$$ $$\displaystyle +\frac{231\,{\mu}^{6}\,{\rho}^{6}}{16}-\frac{315\,{\mu}^{4}\,{\rho}^{6}}{16}+\frac{105\,{\mu}^{2}\,{\rho}^{6}}{16}-\frac{5\,{\rho}^{6}}{16}+\frac{63\,{\mu}^{5}\,{\rho}^{5}}{8}-\frac{35\,{\mu}^{3}\,{\rho}^{5}}{4}+\frac{15\,\mu\,{\rho}^{5}}{8}$$ $$\displaystyle +\frac{35\,{\mu}^{4}\,{\rho}^{4}}{8}-\frac{15\,{\mu}^{2}\,{\rho}^{4}}{4}+\frac{3\,{\rho}^{4}}{8}+\frac{5\,{\mu}^{3}\,{\rho}^{3}}{2}-\frac{3\,\mu\,{\rho}^{3}}{2}+\frac{3\,{\mu}^{2}\,{\rho}^{2}}{2}-\frac{{\rho}^{2}}{2}+\mu\,\rho+1$$ Since the highest order we consider is 6，then $$\displaystyle \mathcal{G}_{L}(\mu,\rho):=1+\mu\rho+\frac{1}{2}(3\mu^{2}-1)\rho^{2}+\frac{1}{2}(5\mu^{3}-3\mu)\rho^{3}+\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\rho^{4}$$ $$\displaystyle +\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\rho^{5}+\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5)\rho^{6}$$ Therefore $$ \displaystyle \begin{align} P_0 &= 1\\ P_1 &= \mu \\ P_2 &= \frac{1}{2}(3\mu^2-1) \\ P_3 &= \frac{1}{2}(5\mu^3-3\mu) \\ P_4 &= \frac{1}{8}(35\mu^4-30\mu^2+3) \\ P_5 &= \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \\ P_6 &= \frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5) \end{align} $$ So, using two different methods, we obtained the same result.

=Contributing Members=

Members Signatures
Egm6322.s12.team2.Xia, I solved R9.4 and R*9.8 without others assistance and proofread R9.2 and R9.6. 10:55, 10 February 2012 (UTC) Egm6322.s12.sungsik 02:07, 10 February 2012 (UTC)

= References for Report 9 = King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) [[media:Pea2.s12.sec43.a.djvu|Mtg 43a]] 24 Jan 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) [[media:Pea2.s12.sec43.b.djvu|Mtg 43b]] 24 Jan 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43c) [[media:Pea2.s12.sec43.c.djvu|Mtg 43c]] 31 Jan 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) [[media:Pea2.s12.sec43.d.djvu|Mtg 43d]] 31 Jan 2011. http://en.wikipedia.org/wiki/Binomial_theorem http://mathworld.wolfram.com/InverseHyperbolicTangent.html

title: User:Egm6322.s12.team2/HW9


title: File:L0.jpg


text:
Talk:Panspermia/a critique of NASA researchers make first discovery of life's building block in comet

Glycine could have been created separately on Earth as well
The glycine found in the comet was composed with carbon-13, so yes, there is glycine in space. However, this does not prove that all glycine on earth originated in space. It could have been created separately on Earth using carbon-12. PolkaDots576 01:54, 1 February 2012 (UTC)

Actually, it does prove that glycine on the earth originated in space. The carbon 13 atoms in the molecules of glycine from space were heavier than the ones found on Earth. This is how NASA scientists discovered glycine in a cometDesi diva007 (talk) 02:51, 21 November 2012 (UTC).


text:
User:Egm4313.s12.team19.cc

R1.4

Report 1: Problem R1.4

R2.5
Contributed by Charles Chiamchittrong.

Problem Statement:
Kreyszig 2011 p.59 pbs. 16-17

We are asked to find a second-order homogeneous linear ODE of the form

 * {| style = "width: 73%"

! $$ \ y'' + ay' + by= 0 $$ !!  (Eq. 1) for a given basis of solutions.
 * }

Problem 16
 * {| style = "width: 73%"

! $$ \ e^{2.6x} \, \ e^{-4.3x} $$ !!
 * }

Problem 17
 * {| style = "width: 73%"

! $$ \ e^{-\sqrt{5}x} \, \ xe^{-\sqrt{5}x} $$ !!
 * }

Solution:
Problem 16

Examining the given basis, we can see that this is a case of two distinct real roots.

The general solution in this case is:
 * {| style = "width: 73%"

! $$ \ y = c_1e^{\lambda_1x} + c_2e^{\lambda_2x} $$ !!  (Eq. 2)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 73%"

! $$ \ y = c_1e^{2.6x} + c_2e^{-4.3x} $$ !!  (Eq. 3)
 * }

Comparing Eq. 2 and Eq. 3 we can determine the roots $$\lambda_1$$ and $$\lambda_2$$:

 * {| style = "width: 73%"

! $$ \ \lambda_1 = 2.6 \, \ \lambda_2 = -4.3 $$ !!  (Eq. 4a, 4b)
 * }

Now we can write the characteristic equation:

 * {| style = "width: 73%"

! $$ \ (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$ !!  (Eq. 5)
 * }

 * {| style = "width: 73%"

! $$ \ (\lambda - 2.6)(\lambda + 4.3) = 0 $$ !!  (Eq. 6)
 * }

 * {| style = "width: 73%"

! $$ \ \lambda^2 + 1.7\lambda - 11.18 = 0 $$ !!  (Eq. 7)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:

 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ y'' + 1.7y' - 11.18y = 0 $$  (Eq. 8)
 * }

Problem 17

Examining the given basis, we can see that this is a case of a real double root.

The general solution in this case is:
 * {| style = "width: 73%"

! $$ \ y = (c_1 + c_2x)e^{\lambda x} $$ !!  (Eq. 9)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 73%"

! $$ \ y = (c_1 + c_2x)e^{-\sqrt{5} x} $$ !!  (Eq. 10)
 * }

Comparing Eq. 8 and Eq. 9 we can determine the root $$\lambda_1$$:

 * {| style = "width: 73%"

! $$ \ \lambda_1 = -\sqrt{5} $$ !!  (Eq. 11)
 * }

Now we can write the characteristic equation:

 * {| style = "width: 73%"

! $$ \ (\lambda - \lambda_1)^2 = 0 $$ !!  (Eq. 12)
 * }

 * {| style = "width: 73%"

! $$ \ (\lambda + \sqrt{5})^2 = 0 $$ !!  (Eq. 13)
 * }

 * {| style = "width: 73%"

! $$ \ \lambda^2 + 2\sqrt{5}\lambda + 5 = 0 $$ !!  (Eq. 14)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:

 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ y'' + 2\sqrt{5}y' + 5y = 0 $$ <p style="text-align: right;"> (Eq. 15)
 * }

title: User:Egm4313.s12.team19.cc


text:
Category:Egm6322.s12.TEAM1.HW

Welcome to our team's gallery.

title: Category:Egm6322.s12.TEAM1.HW


text:
User:Mikahugh

Mikah J. Pritchard

I am currently pursuing my M.S. Ed. in Instructional Systems Technology through the distance education. I am an Instructional Designer at Depauw University. I have a passion for Graphic Design and aim to incorporate it in all of my instructional projects. I aim to bridge the gap between technological possibilities and educational reality.

I can be found on Google+ and on Twitter @mikahjenae.

Mikahugh 02:13, 1 February 2012 (UTC)

title: User:Mikahugh


text:
User:Egm6322.s12.team2/HW10

= R 10.1 - Proof of Recurrence Relation RR2 =

Given

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

$$  \displaystyle (n+1)P_{n+1}-(2n+1)\mu P_n + nP_{n-1} = 0 $$     (10.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \mu[P_0\rho^0 + P_1\rho^1 +\sum_{n=2}P_n{\color{red}\rho^n}]-[P_0\rho^1 + \sum_{n=2}P_{n-1}{\color{red}\rho^n}]=[P_1\rho^0 + P_2\cdot2\cdot\rho^1 +\sum_{n=2}P_{n+1}(n+1){\color{red}\rho^n}]-2\mu[P_1\cdot1\cdot\rho + \sum_{n=2}P_nn{\color{red}\rho^n}] +[\sum_{n=2}P_{n-1}(n-1){\color{red}\rho^n}]$$ (10.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
Show that Eq(10.1.2) satisfy Eq(10.1.1)

Solution
the author may mistake the meaning of this problems,so the solution maybe incorrectproofread by Lang Xia

Ignore initial terms and then extract rest terms

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

$$  \displaystyle \sum_{n=2}P_n{\rho^n}-\sum_{n=2}P_{n-1}{\rho^n}=\sum_{n=2}P_{n+1}(n+1){\rho^n} + \sum_{n=2}P_nn{\rho^n} +\sum_{n=2}P_{n-1}(n-1){\rho^n} $$     (10.1.3) Because every term is summated from n=2, Rearrage it then,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

we can get

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

$$  \displaystyle (n+1)\sum_{n=2}P_{n+1}-2\mu n \sum_{n=2}P_{n} + \mu P_{n} + (n-1)P_{n-1} + P_{n-1} = 0$$ (10.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Where $$\rho^n$$ is cancelled out

Taking off $$\sum $$

Rearrage

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

$$  \displaystyle (n+1)P_{n+1}-(2n+1)\mu P_{n} + nP_{n-1} = 0$$ (10.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Therefore Eq(10.1.2) satisfy Eq(10.1.1)

Egm6322.s12.sungsik 15:49, 22 February 2012 (UTC)

= R 10.2 - Generating Legendre polynomials using RR2 =

Given
Recurrence relation 2 (RR2):
 * {| style="width:100%" border="0"

$$  \displaystyle (n+1)P_{n+1} - (2n+1)\mu P_n + nP_{n-1} = 0 $$     (10.2.1) Legendre polynomials $$P_0$$ and $$P_1$$:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} P_0(\mu) &= 1 \\ P_1(\mu) &= \mu \end{align} $$     (10.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
Use RR2 to generate $$\left\{P_n,n=3,\cdots, 6\right\}$$ based on the knowledge of $$\left\{P_0,P_1\right\}$$.

Solution
This problem is quite straightforward to solve; it can be simplified slightly by the method used in the [[Media:Pea2.s12.sec44.a.djvu|class notes]] to change the index of the polynomials. Defining $$m=n+1$$ and rearranging, Eq. 10.2.1 becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle P_m = \frac{(2m+1)}{m}\mu P_{m-1} - \frac{(m-1)}{m}P_{m-2} $$     (10.2.3) Since m is a dummy index, it can be replaced by n to yield RR2 in a form more compatible with generating the next Legendre polynomial given the two previous ones:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle P_n = \frac{(2n+1)}{n}\mu P_{n-1} - \frac{(n-1)}{n}P_{n-2} $$     (10.2.3) The Legendre polynomial $$P_2$$ must be generated first:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} P_2 &= \frac{(2(2)-1)}{(2)}\mu P_{(1)} - \frac{((2)-1)}{(2)}P_{(0)}\\ &= \frac{3}{2}\mu (\mu) - \frac{1}{2}(1)\\ &= \frac{1}{2}\left(3\mu^2 - 1\right) \end{align} $$     (10.2.4) Similarly, the Legendre polynomials $$P_3, P_4, P_5$$ and $$P_6$$ are generated as:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} P_3 &= \frac{(2(3)-1)}{(3)}\mu P_{(2)} - \frac{((3)-1)}{(3)}P_{(1)} = \frac{5}{3}\mu (\frac{1}{2}\left(3\mu^2 - 1\right)) - \frac{2}{3}(\mu)\\ &= \frac{1}{2}\left(5\mu^3 - 3\mu\right)\\ P_4 &= \frac{(2(4)-1)}{(4)}\mu P_{(3)} - \frac{((4)-1)}{(4)}P_{(2)} = \frac{7}{4}\mu (\frac{1}{2}\left(5\mu^3 - 3\mu\right)) - \frac{3}{4}(\frac{1}{2}\left(3\mu^2 - 1\right))\\ &= \frac{1}{8}\left(35\mu^4 - 30\mu^2 +3\right)\\ P_5 &= \frac{(2(5)-1)}{(5)}\mu P_{(4)} - \frac{((5)-1)}{(5)}P_{(3)} = \frac{9}{5}\mu (\frac{1}{8}\left(35\mu^4 - 30\mu^2 +3\right)) - \frac{4}{5}(\frac{1}{2}\left(5\mu^3 - 3\mu\right))\\ &= \frac{1}{8}\left(63\mu^5 - 70\mu^3 + 15\mu\right)\\ P_6 &= \frac{(2(6)-1)}{(6)}\mu P_{(5)} - \frac{((6)-1)}{(6)}P_{(4)} = \frac{11}{6}\mu (\frac{1}{8}\left(63\mu^5 - 70\mu^3 + 15\mu\right)) - \frac{5}{6}(\frac{1}{8}\left(35\mu^4 - 30\mu^2 +3\right))\\ &= \frac{1}{16}\left(231\mu^6 - 315\mu^4 + 105\mu^2 - 5\right)\\ \end{align} $$     (10.2.5) The Legendre polynomials generated in Eq. 10.2.5 can be compared to those given in the [[Media:Pea1.f11.mtg41.djvu|class notes]] and are seen to be identical
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

=R10.3 Orthogonality of Legendre Functions by Recurrence Relations=

Given
We are given the following recurrence relation from 44-11. $$h_n = \frac{2n-1}{2n+1} \, h_{n-1}$$

Find
(1) The parameter $$h_0$$ by direct integration and (2) Derive the orthogonality of Legendre Functions from (3) on 44-12. $$h_n := \left \langle P_n,P_n \right \rangle = \frac{2}{2n+1}$$ From lecture notes [[media:pea2.f12.sec44.b.djvu|Mtg 44b]]

Solution
The first step is to define the integration for $$h_0$$. The key is to apply the method from King pp. 30 - 42. Section 2.5 of King defines "Orthogonality of Legendre Polynomials." The method starts with Rodrigues' Formula. Apply the derivatives and integration by parts. The nth derivative of an mth order polynomial is 0 for m n.  $$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} {(x^2-1)^n}$$  $$\begin{align} \int_{-1}^1 f(x)P_n(x)dx &= \frac{1}{2^n n!}\int_{-1}^1f(x)\frac{d^n}{dx^n} {(x^2-1)^n} dx\\ &= \frac{-1^n}{2^n n!}\int_{-1}^1{(x^2-1)^n}P_m^{(n)}(x)dx \\ \end{align}$$ In summary, the integral is 0 for m not equal to n.  $$\int_{-1}^1 P_m(x)P_n(x)dx = 0$$ When n = m, the integral becomes  $$\begin{align} \int_{-1}^1 P_n(x)P_n(x)dx &= \frac{-1^n}{2^n n!}\int_{-1}^1{(x^2-1)^n} \frac{1}{2^n n!} (2n)!dx \\ &= \frac{-1^n}{2^n n!} \frac{1}{2^n n!} (2n)! \int_{-1}^1{(x^2-1)^n} dx \end{align}$$ Note that the following integral becomes  $$\int_{-1}^1 (x^2 - 1)^ndx = \frac{2^{2n+2}n!(n+1)!(-1)^n}{(2n+2)!}$$ In summary, the orthogonality of Legendre polynomials is
 $$\int_{-1}^1 P_n^2(x)dx = \frac{2}{2n+1}$$ So to answer part 1 of the question, let n = 0 to give the solution by direct integration.
 $$h_0 = <P_0,P_0> = \int_{-1}^1 P_0^2(x)dx = \frac{2}{2(0)+1} = 2$$ Next, we apply the following relationship to deduce (3) p. 43-12 by induction.
 $$h_n = \frac{2n-1}{2n+1} \, h_{n-1}$$
 $$h_1 = \frac{2(1) - 1}{2(1)+1} h_0 = \frac{2}{3}$$
 $$h_2 = \frac{2(2) - 1}{2(2)+1} h_1 = \frac{2}{5}$$
 $$h_3 = \frac{2(3) - 1}{2(3)+1} h_2 = \frac{2}{7}$$
 $$h_4 = \frac{2(4) - 1}{2(4)+1} h_3 = \frac{2}{9}$$ Hence, by induction, the solution to part 2 shows
 $$h_n = <P_n,P_n> = \frac{2}{2n+1}$$

=R*10.4 -Weight Function =

Given
The following equation: $$\displaystyle w_{j}=\frac{-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}$$
(10.4.1)

where the Legendre polynomial is $$\displaystyle P_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)!$$.
(10.4.2)

Find
Verify that $$\displaystyle w_{2}=1$$

Solution
According to the 10.4.2, we have $$\displaystyle {\displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1)}$$
(10.4.3) $$\displaystyle P_{3}(x)=\frac{1}{2}(5x^{3}-3x)$$ thus $$\displaystyle P_{2}^{'}(x)=3x$$ and the root of 10.4.3 are $$\displaystyle x_{1}=\frac{1}{\sqrt{3}},\quad x_{2}=-\frac{1}{\sqrt{3}} $$ Interms of 10.4.1 $$\displaystyle w_{2}=\frac{-2}{(2+1)P_{2}'(x_{2})P_{2+1}(x_{2})}$$ then $$\displaystyle w_{2}=\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(5\cdot(\frac{-1}{\sqrt{3}})^{3}-3\cdot\frac{-1}{\sqrt{3}}\right)}$$ $$\displaystyle =\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(\frac{4\sqrt{3}}{9}\right)}$$ $$\displaystyle =1$$

= R 10.5 - Gauss Legendre Quadrature Weight Function =

Given

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

$$  \displaystyle w_i=\frac{2}{(1-x_i^2)[P'_n(x_i)]^2}$$ (10.5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle w_j=\frac{-2}{(n+1)P'_n(x_j)P_{n+1}(x_j)}$$ (10.5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Discuss the relative merits of using exact vs using the numerical values Extend the table to n=6 verify the table using Eq(10.5.1) and Eq(10.5.2)

Solution
a. When we use exact value in Gauss Legendre Quadrature, we can reduce the order of error.

Therefore, we can get more precise integration value.

b. First Find the roots of Legendre Polynomial when n = 6

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

$$  \displaystyle P_{6} = \frac{1}{16}(231x^6 - 315x^4 + 105x^2 -5) $$     (10.5.3) where $$P'_{6}=\frac{21}{8}x(33x^4 - 30x^2 +5)$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute $$ x_i $$ into Eq(10.5.1) when n= 6 $$ x_i= \pm 0.239, w_i =\pm1.4327$$ $$ x_i= \pm 0.661, w_i =\pm1.1054$$ $$ x_i= \pm 0.932, w_i =\pm0.5282$$ c. Legendre Polynomials $$P_{1}=x$$ $$P'_{1}=1$$ $$P_{2}=\frac{1}{2}(3x^2 -1)$$ $$P'_{2}=3x$$ $$P_{3}=\frac{1}{2}(5x^3 -3x)$$ $$P'_{3}=\frac{3}{2}(5x^2 -1)$$ $$P_{4}=\frac{1}{8}(35x^4 -30x^2+3)$$ $$P'_{4}=\frac{5}{2}x(7x^2 -3)$$ $$P_{5}=\frac{1}{8}(63x^5 -70x^3 +15x)$$ $$P'_{5}=\frac{15}{8}(21x^4 -14x^2 +1)$$ $$P_{6} = \frac{1}{16}(231x^6 - 315x^4 + 105x^2 -5)$$ Build Mathlab Code Subsitute into Eq(10.5.1) and Eq(10.5.2) Egm6322.s12.sungsik 17:09, 22 February 2012 (UTC)

= R 10.6 - Weight function for Gauss-Legendre Quadrature =

Given
Simplified weight formula:
 * {| style="width:100%" border="0"

$$  \displaystyle w_j = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$     (10.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2}\frac{f^{(2n)}(\eta)}{(2n)!},\ \eta \in [-1,1] $$     (10.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
In the NIST DLMF Sec 3.5v, the weight function shown in 10.6.1 is given by the formula:

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

$$  \displaystyle w_j = \int_{-1}^1\frac{P_n(x)}{(x-x_j)P_n'(x_j)}w(x)dx $$     (10.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

with w(x) = 1. Show that 10.6.1 can be obtained from 10.6.3. Also verify the formula for the error in 10.6.2 against the corresponding formula in the DLMF.

Solution
The first step in this problem is considering the Christoffel-Darboux formula, given by Isaacson and Keller in Analysis of Numerical Methods:

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

$$  \displaystyle \frac{a_n}{a_{n+1}}[P_{n+1}(x)P_n(\xi) - P_{n+1}(\xi)P_n(x)] = (x-\xi)\sum_{k=0}^n P_k(x)P_k(\xi) $$     (10.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We substitute $$x_j$$ for $$\xi$$ and multiply both sides by $$w(x)/(x-x_j)$$:

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

$$  \displaystyle \frac{a_nw(x)}{a_{n+1}(x-x_j)}[P_{n+1}(x)P_n(x_j) - P_{n+1}(x_j)P_n(x)] = w(x)\sum_{k=0}^n P_k(x)P_k(x_j) $$     (10.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We note that, since $$x_j$$ are the roots of the polynomial $$P_n$$, the first term inside the brackets on the left hand side of Eq. 10.6.5 goes to zero. We then integrate both sides between a and b:

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

$$  \displaystyle \begin{align} \int_a^b\frac{a_nw(x)}{a_{n+1}(x-x_j)}[P_{n+1}(x_j)P_n(x)]dx &= \int_a^bw(x)\sum_{k=0}^n P_k(x)P_k(x_j)dx\\ -\frac{a_n}{a_{n+1}}P_{n+1}(x_j)\int_a^b \frac{P_n(x)}{x-x_j}w(x)dx &= \int_a^bw(x)\sum_{k=0}^n P_k(x)P_k(x_j)dx \end{align} $$     (10.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

In which we can move $$P_{n+1}(x_j)$$ can be moved outside the integral as $$x_j$$ is a constant. With some manipulation, the right hand side can somehow be written as

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

$$  \displaystyle \int_a^bw(x)\sum_{k=0}^n P_k(x)P_k(x_j)dx = \int_a^b w(x)[P_n(x)]^2 dx $$ (10.6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If $$P_n(x)$$ is a Legendre polynomial, then w(x)=1, a=-1,  and b=1. Using the orthogonality principle of Legendre polynomials:

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

$$  \displaystyle \begin{align} \int_a^b w(x)[P_n(x)]^2 dx &= \int_{-1}^1 (1)[P_n(x)]^2 dx\\ &= \frac{2}{2n+1} \end{align} $$     (10.6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting 10.6.7 and 10.6.8 into 10.6.6 yields:

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

$$  \displaystyle -\frac{a_n}{a_{n+1}}P_{n+1}(x_j)\int_a^b \frac{P_n(x)}{x-x_j}w(x)dx = \frac{2}{2n+1} $$     (10.6.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Rearranging:

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

$$  \displaystyle \int_a^b \frac{P_n(x)}{x-x_j}w(x)dx = \frac{2}{2n+1}\frac{-a_{n+1}}{a_nP_{n+1}(x_j)} $$     (10.6.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Weisstein, the coefficients of the $$x^n$$ term of the polynomial are shown to be

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

$$  \displaystyle \frac{a_{n+1}}{a_n} = \frac{2n+1}{n+1} $$     (10.6.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting 10.6.11 into 10.6.10 yields

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

$$  \displaystyle \int_a^b \frac{P_n(x)}{x-x_j}w(x)dx = \frac{-2}{P_{n+1}(x_j)} $$     (10.6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally, we recognize that the left hand side of 10.6.12 is equivalent to the integral term in 10.6.3 (in which the $$P_n'(x_j)$$ term can be moved outside the integral as it is not a function of x). Hence, 10.6.12 can be substituted into 10.6.3 to yield the desired formula:

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

$$  \displaystyle w_j = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$     (10.6.13)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

It can also be noted by inspection that the error formula given in 10.6.2 is equivalent to that given by NIST DLMF Sec 3.5v. Matt Shields 17:57, 22 February 2012 (UTC)

=R10.7 Apply Gauss-Legendre Quadrature to Boundary of Heat Problem=

Given
We are given the following recurrence relation from 45-10. $$f(\mu) = T_O \sinh(1 - \mu^2)$$

Find
Compute the first 3 non-zero coefficients using Gauss-Legendre quadrature to within 5% error. From lecture notes [[media:pea2.f12.sec45.c.djvu|Mtg 45c]]

Solution
The overall solution has three large steps: (1) Use Gauss-Quadrature parameters for 10 roots and respective weights, (2) Apply Gauss-Quadrature summation to integral solution of nonzero coefficients, and (3) check degree of accuracy. The problem to find the first three coefficients with integral methods has been analyzed in 9.1 by our team, so that solution will be used as a cross-reference. In 9.1, the solution established the fact that the first three non-zero coefficients of $$T_0 sinh(1-x^2)$$ are below.
$$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}d(x)$$ In 9.1, the following non-zero coefficients were established for the given function.
$$A_0 = 0.7459 T_0$$
$$A_2 = -0.7997 T_0$$
$$A_4 = 0.06671 T_0$$ Next, the identities of the Gauss-Quadruture formulas must be defined from 45. A given integral, I(f), can be approximated by a summation and error term. The identities are given on 45-6 to 45-7.
$$I(f) = I_n(f) + E_n(f)$$ The summation is defined below.
$$I_n(f) = \sum_{j=1}^n w_jf(x_j)$$ The argument of the summation, $$x_j$$, is defined by the roots of the Legendre polynomials.
$$\{x_j, j=1,...,n \} = \text{roots of Legendre polynomial}$$ The weights of the summation of the Gauss-Quadruture as defined below.
$$w_j = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)}$$ The error term is defined by this complicated fraction from 45-7.
$$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{2n}(\eta)}{(2n)}, \eta \in [-1,1]$$ Next, the method of Gauss-Quadruture must be applied to the integral for $$A_j$$. The key is to apply the relationship below.
$$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}d(x) \approx I_n(f) = \sum_{j=1}^n w_jf(x_j)$$ In order to achieve greater than 95% accuracy, the appropriate range of j = 1 to j = n must be chosen. Here, we choose to do the summation from n = 1 to n = 10. The choice gives the following roots and weights from http://dlmf.nist.gov/3.5#v.
$$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}d(x) \approx I_n(f) = \sum_{j=1}^{10} w_jf(x_j)$$ The roots and weights must be applied to each Legendre polynomial for the respective non-zero coefficients.
$$P_0 = 1$$
$$P_2 = \frac{1}{2} (3x^2 - 1)$$
$$P_4 = \frac{35}{8}x^4 -\frac{15}{4}x^2 + \frac{3}{8}$$
$$A_0 = \frac{1}{2}\int_{-1}^1P_0(x)F(x)dx = \sum_{j=1}^{10} w_jP_0 = \sum_{j=1}^{10} w_j(1)$$
$$A_2 = \frac{5}{2}\int_{-1}^1P_2(x)F(x)dx =\sum_{j=1}^{10} w_jP_1 = \sum_{j=1}^{10} w_j \frac{1}{2} (3x^2 - 1)$$
$$A_4 = \frac{9}{2}\int_{-1}^1P_4(x)F(x)dx =\sum_{j=1}^{10} w_jP_2 = \sum_{j=1}^{10} w_j \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$ The next step is to do the summations above with a spreadsheet on MS Excel. For $$A_0$$ we have the following table transcribed from the spreadsheet.

The key to solve the problem is sum all terms in the "all" column to get the coefficient.
$$A_0$$ = sum of numbers in "all" column * $$T_0$$ = 0.745999*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.7459/0.745999*100, which is almost 100%. This meets the criteria for an approximation with error < 5%. In addition, for $$A_2$$ we have the following table transcribed from the spreadsheet.

Again, the key to solve the problem is sum all terms in the "all" column to get the coefficient.
$$A_2$$ = sum of numbers in "all" column * $$T_0$$ = -0.7997*$$T_0$$ This value above matches the value calculated in solution 9.1 by 100%, so the error is negligible with a 100% match.

Finally, for $$A_4$$ we have the following table transcribed from the spreadsheet.

Once again we sum all terms in the "all" column to get the coefficient.
$$A_4$$ = sum of numbers in "all" column * $$T_0$$ = 0.066715*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.066715/0.06671*100, which is almost 100%. The close match makes errors negligible and fulfills the criteria of an error < 5%. In summary, we have successfully calculated the coefficients by using the Gauss-Quadrature formulations.

=R 10.8 -Lagrange Interpolation =

Given
The following Lagrange interpolation: $$\displaystyle p(x)\thickapprox p^{h}(x):=\sum_{j=1}^{n}p(x_{j})\ell_{j}(x)\in\mathcal{P}_{n-1} $$
(10.8.1)

where Lanrage basis function is $$\displaystyle \ell_{j}(x):=\frac{\displaystyle \prod_{k=1,k\neq j}^{n}(x-x_{k})}{\displaystyle \prod_{k=1,k\neq j}^{n}(x_{j}-x_{k})}\in\mathcal{P}_{n-1}$$.
(10.8.2) Define $$\displaystyle d(x):=p(x)-p^{h}(x)\in\mathcal{P}_{n-1}$$. $$\displaystyle $$

Find
1.Verify $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$. 2.Show that: $$\displaystyle d(x_{i})=0$$, for $$\displaystyle i=1,\ldots,n$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$

Solution
1, From the equation 10.8.2, when $$\displaystyle i\neq j$$, we have $$\displaystyle {\displaystyle \ell_{j}(x_{i}):=\frac}=\displaystyle \prod_{k\neq j}\frac{x_{i}-x_{k}}{x_{j}-x_{k}}=\frac{(x_{i}-x_{0})}{(x_{j}-x_{0})}\cdots\frac{\overbrace{(x_{i}-x_{i})}^{=0}}{(x_{j}-x_{i})}\cdots\frac{(x_{i}-x_{k})}{(x_{j}-x_{k})}=0$$ an when $$\displaystyle i=j$$ $$\displaystyle \ell_{i}(x_{i})=\ell_{j}(x_{j}):=\prod_{k\neq j}\frac{x_{j}-x_{k}}{x_{j}-x_{k}}=1$$ Therefore $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$ here the Kronecker delta is
 * $$\delta_{ij} = \left\{\begin{matrix}

1, & \mbox{if } i=j  \\ 0, & \mbox{if } i \ne j  \end{matrix}\right.$$ 2, We know that $$\displaystyle d(x):=p(x)-p^{h}(x)$$ Therefore $$\displaystyle d(x_{i}):=p(x_{i})-p^{h}(x_{i})=p(x_{i})-\sum_{j=1}^{n}p(x_{j})\ell_{j}(x_{i})$$ From above solution, we know that $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$,thus $$\displaystyle d(x_{i})=p(x_{i})-\sum_{j=1}^{n}\underbrace{p(x_{j})\delta_{ji}}_{=0,\, if\, i\neq j}=p(x_{i})-p(x_{i})=0$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$ $$\displaystyle $$

=Contributing Members=

Egm6322.s12.team2.steele.m2 19:26, 22 February 2012 (UTC) Egm6322.s12.team2.Xia 10:55, 21 February 2012 (UTC) Egm6322.s12.sungsik 17:11, 22 February 2012 (UTC)

= References for Report 10 = King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003. http://dlmf.nist.gov/3.5#v Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 44b) [[media:Pea2.s12.sec44.b.djvu|Mtg 44b]] 7 Feb 2012 to 9 Feb 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45c) [[media:Pea2.s12.sec45.c.djvu|Mtg 45c]] 16 Feb 2012.

title: User:Egm6322.s12.team2/HW10


title: File:Egm6322.s12.TEAM1.Legen1.jpg


title: File:Egm6322.s12.TEAM1.Legen2.jpg


title: File:Egm6322.s12.TEAM1.Legen3.jpg


title: File:Egm6322.s12.TEAM1.Legen4.jpg