text:
User:EML4500.f08.JAMAMA/FE/ex15

= HW 2.1 =

Find
By taking infinitesimal slice of the bar (shown in red in Figure 1),$$ dx $$, develop expression for elastodynamic response in Heat Problem in 1-D

Solution
Free Body Diagram:

Heat is transferred in the form of Conduction, Convection and Thermal Radiation. In this problem we will do Heat Conduction only.

Here we have consider a small infinitesimal portion of wall of thickness dx as shown in figure.

Let,The Heat Flux is represented by "q".

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

$$ \alpha_2=1, \alpha_1=0, \alpha_3=...=\alpha_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \alpha _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline b_i=\underline b_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \alpha_n=1,\ \alpha_1=...=\alpha_{n-1}=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \alpha_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline b_i=\underline b_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \alpha_1,...,\alpha_n \right \}\in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha_i\underline b_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\underline a_{i},i=1,...,n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \underline a_{j}= \delta _{ij} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \delta _{ij}=\left\{\begin{matrix} 1\ for\ i=j\\ 0\ for\ i\neq j \end{matrix}\right. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\underline a_{i} \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline v}=\sum_{i=1}^{n}\ \textrm{v}_{i}\underline a_{i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline v}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline P (\underline v)}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot\textrm{\underline P (\underline v)}=\underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \underline a_{i}\cdot \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j}=\underline a_{i}\cdot \textrm{\underline v}=\sum_{j=1}^{n}(\underline a_{i}\cdot \underline a_{j})\textrm{v}_{j}=\sum_{j=1}^{n}\ \delta_{ij}\cdot \textrm{v}_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P (\textrm{\underline v})=0 \ \forall \ \underline w\in \mathbb{R}^{n} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w = \sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w \cdot \underline P(\textrm{\underline v})=0 \ \forall \ \left \{ \beta_1,..., \beta_n \right \}\ \in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_1=1, \beta_2=...=\beta_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\beta _1, ..., \beta _n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \forall \ \left \{\beta _1, ..., \beta _n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 1,0,...,0 \right \}\underline a_i=\underline a_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_1\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_2=1, \beta_1=0, \beta_3=...=\beta_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \beta _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline a_i=\underline a_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_n=1,\ \beta_1=...=\beta_{n-1}=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \beta_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline a_i=\underline a_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \beta_1,...,\beta_n \right \}\in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

= HW 2.4 =

Problem Statement
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Do the following problems to lead this result

1) Show $$\int{log(x) }dx = xlog|x| - x $$ 2) Show $$\int{x log(x)}dx = \frac{1}{2}x^2\left[log(x)-\frac{1}{2}\right]$$ 3) Find $$\int{\frac{x^2}{1+cx}}dx$$ 4) Find $$\int{\frac{x^2}{a+bx}}dx$$ 5) Find the exact solution of $$ \displaystyle u(x) $$ for,
 $$ \displaystyle            \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$
 6) Plot $$ \displaystyle u(x) $$

1) Show $$\int{logx}dx = xlogx - x $$
Using Integration By Parts, (4.1) Let

(4.2)

Entering values from eqns 4.2 into eqn. 4.1 yields
 $$xlogx-\int 1dx$$

2) Show $$\int{x logx}dx = \frac{1}{2}x^2\left[logx-\frac{1}{2}\right]$$
Using Integration by parts, let

(4.3) Now entering values from eqns 4.3 into eqn. 4.1 yields
 $$\frac{x^2}{2}logx - \int{\left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right)}dx $$

$$\frac{x^2}{2}logx- \frac{x^2}{4}$$

3) Find $$\int \frac{x^2}{1+cx}$$
Using Integration by parts, let
 (4.4)

Now, entering values from eqns 4.4 into eqn 4.1 yields,
 (4.5)

Now, integration by parts must be done again. Let (4.6)

The equation now takes the form (4.7)

Entering values from eqns 4.6 into 4.7 yields $$ \displaystyle \frac{x^{2}}{c} log(1+cx) - \left[ {\left( \frac{x^2}{c} log(1+cx)- \frac{1}{c^3}log(1+cx)- \frac{x^2 }{2c} + \frac{x}{c^2} \right)} {\left( 2x \right)} - \int {\left(2dx \right)}\right] $$

Distributing terms,

(4.8)

Solving each integral in 4.8 individually
 $$ \displaystyle \begin{align} & {2\int{\frac{x^2}{c}log(1+cx)}\ dx} = \frac{2x^3}{3c}\log(1+cx)+\frac{2}{3c^4}\log(1+cx)-\frac{2x^3}{9c}+\frac{x^2}{3c^2} -\frac{2x}{3c^3}\\ & {-2\int{\frac{1}{c^3}log(1+cx)}\ dx} = -\frac{2x}{c^3}log(1+cx) - \frac{2}{c^4}log(1+cx) + \frac{2x}{c^3} \\ & {\int{\frac{x^2}{c}}\ dx} = \frac{x^3}{3c} \\ & {2\int{\frac{x}{c^2}}\ dx} = \frac{x^2}{c^2} \end{align} $$
 (4.9)

Entering values for eqns 4.9 into eqn. 4.8 yields

Which simplifies to

When c=1, the problem further simplifies to

4) Find $$\int \frac{x^2}{a+bx}$$
Using integration by parts, let

(4.10)

Now, entering values from eqns 4.10 into eqn 4.1 yields, (4.11)

Now, integration by parts must be done again. Let

(4.12)

The equation now takes the form

(4.13)

Entering the values from 4.12 into 4.13 yields,

Distributing terms,

(4.14)

Solving each integral in 4.14 individually,

$$ \displaystyle \begin{align} &\frac{2}{b}\int{xlog(a+bx)}dx = \frac{x^2}{b}\log(a+bx)-\frac{a^2}{b^3}\log(a+bx)-\frac{x^2}{2b}+\frac{ax}{b^2} \\ &\frac{2a}{b^2}\int{log(a+bx)}dx = \frac{2ax}{b^2}log(a+bx)+\frac{2a^2}{b^3}log(a+bx) - \frac{2ax}{b^2} \\ &\frac{2}{b}\int{x}dx = -\frac{x^2}{b} &\end{align} $$
 (4.15)

Substituting eqns 4.15 into 4.14 yields

Which simplifies to

 (4.16)

5) Find the exact solution for
$$ \displaystyle \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$  ;    $$  \forall x \in ] 0,1 [ $$

With the following Boundary Conditions:

Essential: $$ u(1)=4 $$
 (4.17) Natural:$$ -\frac{d}{dx} u(x=0) =6 $$
 (4.18)

$$ \frac{d}{du} [(2+3x)\frac{du}{dx}]=-5x $$
 (4.19) Integrate both sides of eqn. 4.19,
 $$ \int d [(2+3x)\frac{du}{dx}] = \int-5x dx$$

$$ (2+3x)\frac{du}{dx}=\frac{-5x^2}{2}+c_1$$

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{c_1}{2+3x}$$
 (4.20) Apply Natural B.C., eqn. 4.18, to find $$ c_1 $$,

$$-\frac{du}{dx}(x=0) =6 \implies -6= 0+ \frac{c_1}{2} $$

$$c_1=-12$$
 (4.21) Enter value from eqn. 4.21 into eqn 4.20

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$du=\frac{-5x^2}{2(2+3x)}dx+\frac{-12}{2+3x}dx$$
 (4.22) Integrate both sides of eqn. 4.22
 $$\int du=\int\frac{-5x^2}{2(2+3x)}dx+\int\frac{-12}{2+3x}dx$$

$$u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

$$u=\frac{-5}{2}[\frac{x^2}{6}-\frac{2x}{9}+\frac{4log(2+3x)}{27}]-12[\frac{log(2+3x)}{3}]+c_2$$

$$u=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{98}{27}log(2+3x)+c_2$$
 (4.23) Apply Essential B.C., eqn 4.17, to find

$$ u(1)=4 \implies 4=\frac{-5}{12} +\frac{5}{9}-\frac{98}{27}log(2+3x)+c_2$$

$$c_2=-1.98055$$
 (4.24)

Enter the value from eqn 4.24 into eqn. 4.23 to obtain exact solution,

= HW 2.5 =

title: User:EML4500.f08.JAMAMA/FE/ex15


text:
User:EML4500.f08.JAMAMA/FE/ex13

= HW 2.1 =

Find
By taking infinitesimal slice of the bar (shown in red in Figure 1),$$ dx $$, develop expression for elastodynamic response in Heat Problem in 1-D

Solution
Free Body Diagram:

Heat is transferred in the form of Conduction, Convection and Thermal Radiation. In this problem we will do Heat Conduction only.

Here we have consider a small infinitesimal portion of wall of thickness dx as shown in figure.

Let,The Heat Flux is represented by "q".

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

$$ \alpha_2=1, \alpha_1=0, \alpha_3=...=\alpha_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \alpha _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline b_i=\underline b_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \alpha_n=1,\ \alpha_1=...=\alpha_{n-1}=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \alpha_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline b_i=\underline b_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \alpha_1,...,\alpha_n \right \}\in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\alpha_i\underline b_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\underline a_{i},i=1,...,n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \underline a_{j}= \delta _{ij} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \delta _{ij}=\left\{\begin{matrix} 1\ for\ i=j\\ 0\ for\ i\neq j \end{matrix}\right. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\underline a_{i} \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline v}=\sum_{i=1}^{n}\ \textrm{v}_{i}\underline a_{i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline v}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \textrm{\underline P (\underline v)}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot\textrm{\underline P (\underline v)}=\underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \underline a_{i}\cdot \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j}=\underline a_{i}\cdot \textrm{\underline v}=\sum_{j=1}^{n}(\underline a_{i}\cdot \underline a_{j})\textrm{v}_{j}=\sum_{j=1}^{n}\ \delta_{ij}\cdot \textrm{v}_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P (\textrm{\underline v})=0 \ \forall \ \underline w\in \mathbb{R}^{n} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w = \sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w \cdot \underline P(\textrm{\underline v})=0 \ \forall \ \left \{ \beta_1,..., \beta_n \right \}\ \in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_1=1, \beta_2=...=\beta_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{\beta _1, ..., \beta _n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \forall \ \left \{\beta _1, ..., \beta _n \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 1,0,...,0 \right \}\underline a_i=\underline a_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_1\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_2=1, \beta_1=0, \beta_3=...=\beta_n=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \beta _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline a_i=\underline a_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \beta_n=1,\ \beta_1=...=\beta_{n-1}=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \left \{ \beta_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline a_i=\underline a_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \beta_1,...,\beta_n \right \}\in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. x.x)
 * }

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

title: User:EML4500.f08.JAMAMA/FE/ex13


text:
User:EML4500.f08.JAMAMA/FE/ex45

= HW 2.4 =

Problem Statement
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Do the following problems to lead this result

1) Show $$\int{log(x) }dx = xlog|x| - x $$

2) Show $$\int{x log(x)}dx = \frac{1}{2}x^2\left[log(x)-\frac{1}{2}\right]$$

3) Find $$\int{\frac{x^2}{1+cx}}dx$$

4) Find $$\int{\frac{x^2}{a+bx}}dx$$

5) Find the exact solution of $$ \displaystyle u(x) $$ for, $$ \displaystyle            \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$

6) Plot $$ \displaystyle u(x) $$

1) Show $$\int{logx}dx = xlogx - x $$
Using Integration By Parts, (4.1) Let

(4.2)

Entering values from eqns 4.2 into eqn. 4.1 yields

$$xlogx-\int 1dx$$

2) Show $$\int{x logx}dx = \frac{1}{2}x^2\left[logx-\frac{1}{2}\right]$$
Using Integration by parts, let

(4.3) Now entering values from eqns 4.3 into eqn. 4.1 yields

$$\frac{x^2}{2}logx - \int{\left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right)}dx $$

$$\frac{x^2}{2}logx- \frac{x^2}{4}$$

3) Find $$\int \frac{x^2}{1+cx}$$
Using Integration by parts, let (4.4)

Now, entering values from eqns 4.4 into eqn 4.1 yields, (4.5)

Now, integration by parts must be done again. Let (4.6)

The equation now takes the form (4.7)

Entering values from eqns 4.6 into 4.7 yields $$ \displaystyle \frac{x^{2}}{c} log(1+cx) - \left[ {\left( \frac{x^2}{c} log(1+cx)- \frac{1}{c^3}log(1+cx)- \frac{x^2 }{2c} + \frac{x}{c^2} \right)} {\left( 2x \right)} - \int {\left(2dx \right)}\right] $$

Distributing terms,

(4.8)

Solving each integral in 4.8 individually

$$ \displaystyle \begin{align} & {2\int{\frac{x^2}{c}log(1+cx)}\ dx} = \frac{2x^3}{3c}\log(1+cx)+\frac{2}{3c^4}\log(1+cx)-\frac{2x^3}{9c}+\frac{x^2}{3c^2} -\frac{2x}{3c^3}\\ & {-2\int{\frac{1}{c^3}log(1+cx)}\ dx} = -\frac{2x}{c^3}log(1+cx) - \frac{2}{c^4}log(1+cx) + \frac{2x}{c^3} \\ & {\int{\frac{x^2}{c}}\ dx} = \frac{x^3}{3c} \\ & {2\int{\frac{x}{c^2}}\ dx} = \frac{x^2}{c^2} \end{align} $$  (4.9)

Entering values for eqns 4.9 into eqn. 4.8 yields

Which simplifies to

When c=1, the problem further simplifies to

4) Find $$\int \frac{x^2}{a+bx}$$
Using integration by parts, let

(4.10)

Now, entering values from eqns 4.10 into eqn 4.1 yields,

(4.11)

Now, integration by parts must be done again. Let

(4.12)

The equation now takes the form

(4.13)

Entering the values from 4.12 into 4.13 yields,

Distributing terms,

(4.14)

Solving each integral in 4.14 individually,

$$ \displaystyle \begin{align} &\frac{2}{b}\int{xlog(a+bx)}dx = \frac{x^2}{b}\log(a+bx)-\frac{a^2}{b^3}\log(a+bx)-\frac{x^2}{2b}+\frac{ax}{b^2} \\ &\frac{2a}{b^2}\int{log(a+bx)}dx = \frac{2ax}{b^2}log(a+bx)+\frac{2a^2}{b^3}log(a+bx) - \frac{2ax}{b^2} \\ &\frac{2}{b}\int{x}dx = -\frac{x^2}{b} &\end{align} $$  (4.15)

Substituting eqns 4.15 into 4.14 yields

Which simplifies to

 (4.16)

5) Find the exact solution for
$$ \displaystyle \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$  ;    $$  \forall x \in ] 0,1 [ $$

With the following Boundary Conditions:

Essential: $$ u(1)=4 $$

 (4.17) Natural:$$ -\frac{d}{dx} u(x=0) =6 $$

 (4.18)

$$ \frac{d}{du} [(2+3x)\frac{du}{dx}]=-5x $$  (4.19) Integrate both sides of eqn. 4.19,

$$ \int d [(2+3x)\frac{du}{dx}] = \int-5x dx$$

$$ (2+3x)\frac{du}{dx}=\frac{-5x^2}{2}+c_1$$

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{c_1}{2+3x}$$

 (4.20) Apply Natural B.C., eqn. 4.18, to find $$ c_1 $$,

$$-\frac{du}{dx}(x=0) =6 \implies -6= 0+ \frac{c_1}{2} $$

$$c_1=-12$$  (4.21) Enter value from eqn. 4.21 into eqn 4.20

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$du=\frac{-5x^2}{2(2+3x)}dx+\frac{-12}{2+3x}dx$$

 (4.22) Integrate both sides of eqn. 4.22

$$\int du=\int\frac{-5x^2}{2(2+3x)}dx+\int\frac{-12}{2+3x}dx$$

$$u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

$$u=\frac{-5}{2}[\frac{x^2}{6}-\frac{2x}{9}+\frac{4log(2+3x)}{27}]-12[\frac{log(2+3x)}{3}]+c_2$$

$$u=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{98}{27}log(2+3x)+c_2$$  (4.23) Apply Essential B.C., eqn 4.17, to find

$$ u(1)=4 \implies 4=\frac{-5}{12} +\frac{5}{9}-\frac{98}{27}log(2+3x)+c_2$$

$$c_2=-1.98055$$  (4.24)

Enter the value from eqn 4.24 into eqn. 4.23 to obtain exact solution,

= HW 2.5 =

title: User:EML4500.f08.JAMAMA/FE/ex45


text:
User:Eml5526.s11.team3.hylon/Homework 2

=Problem2.9=

Problem Statement
Given $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$. $$1).$$ Find two equations that enforce boundary conditions for $${{u}^{h}}(x)=\sum\limits_{j=0}^{n}{{{d}_{j}}{{b}_{j}}(x)}$$. Natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$; essential boundary condition: $${{u}^{h}}(x=1)=0$$ . $$2).$$ Find one more equation to solve for $$\underline{d}=\left\{ {{d}_{j}} \right\} (j=0,1,2)$$by project the residue $$P({{u}^{h}})$$ on a basis function $${{b}_{k}}(x)$$ with $$k=0,1,2 $$, such that the additional equation is linear independent from the above two equations in $$(1)$$. $$3).$$ Display three equations in matrix form $$\underline{K}\underline{d}=\underline{F}$$, observe symmetric property of $$\underline{K}$$. $$4).$$ Solve for $$\underline{d}$$. $$5).$$ Construct $$u_{2}^{h}(x)$$ and plot $$u_{2}^{h}(x)$$ vs. $$u_{2}^ – (x)$$. $$6).$$Compute $$u_{2}^{h}(x=0.5)$$ and Error $${{e}_{2}}(x=0.5)=u(x=0.5)-{{u}^{h}}(x=0.5)$$. $$7).$$ Repeat $$1)-6)$$ for $$n=4,n=6$$. $$8).$$ Plot $${{e}_{2}}(x=0.5)$$ vs. $$n$$ Refer to lecture slide [[media:fe1.s11.mtg12.djvu|12-1]] for more information.

Solution
For $$n=2$$ : $$1).$$ Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then $${{b}_{0}}(x)=\cos (\phi ),{{b}_{1}}(x)=\cos (x+\phi ),{{b}_{2}}(x)=\cos (2x+\phi )$$. In order to make sure that $$b_{i}^{'}(x=0)\ne 0$$,we choose $$\phi =\frac{\pi }{4}$$. Hence we have From natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain Hence we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have We also have Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2$$, we have For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2, f(x)=3$$. Thus, So we can get one more equation

$$3).$$ Till now, we obtain three equations as follow: Display them in matrix form, yields

Hence,

non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

Since $$2\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+3=0$$, $$u(x=1)=0$$ and $${{u}^{'}}(x=0)=-4$$, we can obtain

Plot

Matlab Code: $$6).$$ We have already got $$u_{2}^{h}(x)=3.803\cos (\frac{\pi }{4})-0.1495\cos (x+\frac{\pi }{4})+2.9032\cos (2x+\frac{\pi }{4})$$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

For $$n=4$$ : $$1).$$ We also choose $$\phi =\frac{\pi }{4}$$ here. Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then Hence we have

From natural boundary condition:$$ -\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain

Hence, we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have We get that Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2,3,4$$, we have $$\int\limits_{0}^{1}{{{b}_{i}}(x)P({{u}^{h}})}dx=0$$. For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2,f(x)=3$$. Thus, $$\int\limits_{0}^{1}{{{b}_{i}}(x)\{\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)}\}dx=\int\limits_{0}^{1}{{{b}_{i}}(x)[2{{d}_{j}}{{b}_{j}}^{(2)}(x)+3}]dx=0$$. We can have three more equations here,

$$3).$$ Till now, we obtain five equations as follow: Display them in matrix form, yields

Hence,

non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

We have already obtained

Plot

Matlab Code: $$6).$$ We have already got $$u_{4}^{h}(x)=14.1214\cos (\frac{\pi }{4})-22.8534\cos (x+\frac{\pi }{4})+26.5637\cos (2x+\frac{\pi }{4})-12.2778\cos (3x+\frac{\pi }{4})+3.0541\cos (4x+\frac{\pi }{4})$$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

For $$n=6$$ : $$1).$$ We also choose $$\phi =\frac{\pi }{4}$$ here. Since $$\left\{ {{b}_{i}}(x)=\cos (ix+\phi ),i=0,1,2,\cdots ,n \right\}$$, then

Hence we have From natural boundary condition: $$-\frac{d{{u}^{h}}(x=0)}{dx}=4$$, we can obtain

Hence, we have

From essential boundary condition: $${{u}^{h}}(x=1)=0$$, we can have

We get that

Hence we can get the second equation, that

$$2).$$ Projecting the residue $$P({{u}^{h}})$$ on the $${{b}_{i}}(x)$$,$$i=0,1,2,3,4,5,6$$, we have For static form, we know that $$P({{u}^{h}})=\frac{\partial }{\partial x}[{{a}_{2}}(x)\frac{\partial {{u}^{h}}}{\partial x}]+f(x)$$. And it’s already given that $${{a}_{2}}(x)=2,f(x)=3$$. Thus,

We can have five more equations here,

$$3).$$ Till now, we obtain seven equations as follow:

Display them in matrix form, yields

Hence,

is non-symmetric. $$4).$$ Using Matlab, we can easily get

$$5).$$

We have already obtained

Plot

Matlab Code: $$6).$$ We have already got $$u_{6}^{h}(x)=0.6678\cos (\frac{\pi }{4})+5.2048\cos (x+\frac{\pi }{4})+6.9757\cos (2x+\frac{\pi }{4})-10.7753\cos (3x+\frac{\pi }{4})+7.5430\cos (4x+\frac{\pi }{4})-2.2711\cos (5x+\frac{\pi }{4})+0.0017\cos (6x+\frac{\pi }{4}) $$ and $$u(x)=-\frac{3}{4}{{x}^{2}}-4x+\frac{19}{4}$$. Hence

Plot $${{e}_{n}}(x=0.5)$$ vs. $$n$$:

Problem solved by Hailong Chen.

Reference
All matrix operations and integrations are solved by Matlab.

title: User:Eml5526.s11.team3.hylon/Homework 2


text:
User:Egm6341.s11.team3/sub1/a

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

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

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

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

title: User:Egm6341.s11.team3/sub1/a


title: File:CORDIC.1.A.Background.pdf


text:
User talk:Nike911

 Hello Nike911, 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://twiter.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! --Ottava Rima (talk) 20:51, 1 February 2011 (UTC)

title: User talk:Nike911


text:
User talk:Eml5526.s11.team3.hylon

 Hello Eml5526.s11.team3.hylon, 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://twiter.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! --Ottava Rima (talk) 20:52, 1 February 2011 (UTC)

title: User talk:Eml5526.s11.team3.hylon


text:
User talk:Srilalithkumar

 Hello Srilalithkumar, 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://twiter.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! --Ottava Rima (talk) 21:25, 1 February 2011 (UTC)

title: User talk:Srilalithkumar


text:
User talk:Eml5526.s11.team5.savery

 Hello Eml5526.s11.team5.savery, 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://twiter.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! --Ottava Rima (talk) 21:25, 1 February 2011 (UTC)

title: User talk:Eml5526.s11.team5.savery


text:
User:Eml5526.s11.team5.savery

Problem

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

Consider $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ on the interval $$ \Omega =[0,1] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

Find

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

1. Construct the $$ \Gamma(F)$$ matrix and observe its propeties.
 * style="width:95%" |
 * style="width:95%" |
 * }

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

 * 2. Find the determinant of $$ \Gamma(F)$$.
 * }

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

3. Conclude if F is an orthogonal family., i.e., $$ \Gamma_{ij} = \delta_{ij} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

1)

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

The Gram Matrix is defined as:
 * style="width:95%" |
 * style="width:95%" |

 * $$\Gamma_{ij}==\int_\Omega b_i(x)b_j(x)\,dx$$

$$
 *  $$ \displaystyle (eq. 1.1)
 * }

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

For the case where $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ b_1(x)=1$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

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

$$ b_2(x)=x $$ $$
 * style="width:90%" |
 * style="width:90%" |
 *  $$ \displaystyle
 * }

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

$$  b_3(x)=x^2  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

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

$$  b_4(x)=x^3  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ b_5(x)=x^4  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Now we need to calculate each element in the $$ \Gamma $$ Matrix using eqn. 1.1: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Since $$ \int_\Omega b_i(x)b_j(x)\,dx = \int_\Omega b_j(x)b_i(x)\,dx$$,  $$\Gamma_{ij}$$ will be equivalent to $$\Gamma_{ji}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,1} = \int_{0}^{1} 1*1*dx = \left [ x\right ]_{0}^{1} = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,2} = \Gamma_{2,1} = \int_{0}^{1} 1*x*dx = \left [ x^2/2\right ]_{0}^{1} = 1/2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,3} = \Gamma_{3,1} = \int_{0}^{1} 1*x^2*dx = \left [ x^3/3 \right ]_{0}^{1} = 1/3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,4} = \Gamma_{4,1} = \int_{0}^{1} 1*x^3*dx = \left [ x^4/4 \right ]_{0}^{1} = 1/4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,5} = \Gamma_{5,1} = \int_{0}^{1} 1*x^4*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,2} = \int_{0}^{1} x*x*dx = \left [ x^3/3\right ]_{0}^{1} = 1/3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,3} = \Gamma_{3,2} = \int_{0}^{1} x*x^2*dx = \left [ x^4/4\right ]_{0}^{1} = 1/4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,3} = \int_{0}^{1} x^2*x^2*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,4} = \Gamma_{4,3} = \int_{0}^{1} x^2*x^3*dx = \left [ x^6/6\right ]_{0}^{1} = 1/6 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,5} = \Gamma_{5,3} = \int_{0}^{1} x^2*x^4*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,4} = \int_{0}^{1} x^3*x^3*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,5} = \Gamma_{5,4} = \int_{0}^{1} x^3*x^4*dx = \left [ x^8/8\right ]_{0}^{1} = 1/8 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{5,5} = \Gamma_{5,5} = \int_{0}^{1} x^4*x^4*dx = \left [ x^9/9\right ]_{0}^{1} = 1/9 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Assembling the matrix we get: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

\Gamma_ \left (F \right) = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7}\\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8}\\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \end{bmatrix} $$

 * The matix is observed to only be symmetric

2)

 * Using the matlab function det(X) the determinant of $$\Gamma$$ can be found to be:

 * $$ \det \left ( \Gamma \right ) = 0 $$

3)

 * To conclude if $$\Gamma$$ is orthogonal we compare it to a matrix formed by the Kronecker Delta.

 * The Kronecker Delta is defined by:

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

$$\delta_{ij} = \begin{cases} 1, & \mbox{ for i=j} \\ 0, & \mbox{ for i≠j} \end{cases} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * Thus the general form of a the matrix is:

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

$$ \begin{bmatrix} 1    & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & 1 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$ $$
 *
 $$ \displaystyle
 * }

 * Thus the $$ \Gamma $$ Matrix is not orthogonal

title: User:Eml5526.s11.team5.savery


text:
User:Egm6341.s11.team3.rakesh/hw2

 REMARK: We did the solution on our own

 * }

Problem
Do the integration by parts to reveal three more terms  :  $$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$

Problem 2.1.2
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

Problem
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

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

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$|| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = -\frac {(x-t)^2}{2} f(t)-\int\limits_{x}^{x_0}\frac{-(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = \frac {(x-x_0)^2}{2} f(x_0)+\int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = -\frac {(x-t)^3}{2*3} f(t)-\int\limits_{x}^{x_0}\frac {-(x-t)^3}{2*3}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = \frac {(x-x_0)^3}{3!} f(t)+\int\limits_{x}^{x_0}\frac {(x-t)^3}{3!}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = -\frac {(x-t)^4}{4*3!} f'(t)-\int\limits_{x}^{x_0}-\frac {(x-t)^4}{4!}f'(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = \frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f(x_0)+ \frac {(x-x_0)^3}{3!} f'(t)+\frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f'(t)dt} $$||
 $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

APPLYING IMVT THEOREM TO ABOVE EQUATION
 * {| style="width:100%" border="0"

$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f^(4)(t)dt =  f^5(\xi)\frac {(x-x_0)^5}{5!} $$, $$   \xi  \in (x,x_0) $$
 * style="width:95%" |
 * style="width:95%" |

APPLYING IMVT TO EQUATION

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

$$\int\limits_{x}^{x_0}\frac{(x-t)^n}{n!}f^{n+1}(t)dt$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$f^{n+1}(\xi)\int\limits_{x}^{x_0}\frac {(x-t)^n}{n!}dt $$ $$ \xi \in (x,x_0)$$ $$
 *
 $$ \displaystyle
 * }

title: User:Egm6341.s11.team3.rakesh/hw2


text:
User:Eml5526.s11.team4.yang/hw2-6

=Homework 2.6=

Problem
Consider F= { 1, Cos iwx, Sin iwx } on interval [ 0,T ], i.e. i=1,2

1). Construct $$\displaystyle \underline{\Gamma }(F) $$; observe property of $$\displaystyle \underline \Gamma  (F)$$.

2). Find det $$\underline{\Gamma }(F)$$.

3). Conclude that F is orthogonal basis, i.e., $$\displaystyle {{\Gamma }_{ij}}=\left\langle {{b}_{i}},{{b}_{j}} \right\rangle ={{\delta }_{ij}}$$

Solutions
1). Calculate the value of Gram matrix, $$\displaystyle w=\frac{2\pi }{T}$$:
 * {| style="width:100%" border="0"

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

{{\Gamma }_{ij}}=\left\langle {{b}_{i}},{{b}_{j}} \right\rangle =\int\limits_{\Omega }(x){{b}_{j}}(x)dx

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

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

{{\Gamma }_{11}}=\int\limits_{\Omega }(x){{b}_{1}}(x)dx=\int_{0}^{T}{1dx}=T

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

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

{{\Gamma }_{12}}={{\Gamma }_{21}}=\int\limits_{\Omega }(x){{b}_{2}}(x)dx=\int_{0}^{T}{\cos wxdx}=\frac{\sin wT}{w}=0

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

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

\begin{align} & {{\Gamma }_{22}}=\int\limits_{\Omega }(x){{b}_{2}}(x)dx=\int_{0}^{T}{{{\left( \cos wx \right)}^{2}}dx} \\ & =\frac{\sin 2wT+2T}{4}-\frac{\sin 0+2\centerdot 0}{4}=\frac{T}{2} \\ \end{align}

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

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

{{\Gamma }_{13}}={{\Gamma }_{31}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{\cos 2wxdx}=\frac{\sin 2wT}{2w}=0

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

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

{{\Gamma }_{23}}={{\Gamma }_{32}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \cos 2wxdx}=\frac{3\sin (wx)+\sin (3wx)}{6w}|_{0}^{T}=0

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

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

{{\Gamma }_{33}}={{\Gamma }_{33}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{{{(\cos 2wx)}^{2}}dx}=\frac{\sin 4wT+4wx}{8w}|_{0}^{T}=\frac{T}{2}

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

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

{{\Gamma }_{14}}={{\Gamma }_{41}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{\sin wxdx}=\frac{-\cos wx}{w}|_{0}^{T}=0

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

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

{{\Gamma }_{24}}={{\Gamma }_{42}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \sin wxdx}=\frac{-\cos (2wx)}{4w}|_{0}^{T}=0

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

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

{{\Gamma }_{43}}={{\Gamma }_{34}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{(\cos 2wx)\centerdot \sin wxdx}=-\frac{\cos 3wx-3\cos wx}{6w}|_{0}^{T}=0

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

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

{{\Gamma }_{44}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{{{(\sin wx)}^{2}}dx}=\frac{2wx-\cos 2wx}{4w}|_{0}^{T}=\frac{T}{2}

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

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

{{\Gamma }_{15}}={{\Gamma }_{51}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{\sin 2wxdx}=\frac{-\cos 2wx}{2w}|_{0}^{T}=0

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

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

{{\Gamma }_{25}}={{\Gamma }_{52}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{\cos wx\centerdot \sin 2wxdx}=\frac{-2{{\cos }^{3}}(wx)}{3w}|_{0}^{T}=0

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

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

{{\Gamma }_{53}}={{\Gamma }_{35}}=\int\limits_{\Omega }(x){{b}_{3}}(x)dx=\int_{0}^{T}{(\cos 2wx)\centerdot \sin 2wxdx}=\frac{-\cos 4wx}{8w}|_{0}^{T}=0

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

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

{{\Gamma }_{54}}={{\Gamma }_{45}}=\int\limits_{\Omega }(x){{b}_{4}}(x)dx=\int_{0}^{T}{(\sin wx)\sin 2wxdx}=\frac{2{{\sin }^{3}}wx}{3w}|_{0}^{T}=0

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

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

{{\Gamma }_{55}}=\int\limits_{\Omega }(x){{b}_{5}}(x)dx=\int_{0}^{T}{{{(\sin 2wx)}^{2}}dx}=\frac{4wx-\cos 4wx}{8w}|_{0}^{T}=\frac{T}{2}

$$ So, the Gram matrix is:
 *
 * }
 * {| style="width:100%" border="0"

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

\Gamma =\left[ \begin{matrix} T & 0 & 0 & 0 & 0 \\ 0 & T/2 & 0 & 0 & 0 \\ 0 & 0 & T/2 & 0 & 0 \\ 0 & 0 & 0 & T/2 & 0 \\ 0 & 0 & 0 & 0 & T/2 \\ \end{matrix} \right]

$$ It is observed that this matrix is diagonal matrix, i.e. only the diagonal component is not zero.
 *

 * }

2). Calculate the determinate of the Gram matrix,
 * {| style="width:100%" border="0"

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

\det (\Gamma )=\det \left[ \begin{matrix} T & 0 & 0 & 0 & 0 \\ 0 & T/2 & 0 & 0 & 0 \\ 0 & 0 & T/2 & 0 & 0 \\ 0 & 0 & 0 & T/2 & 0 \\ 0 & 0 & 0 & 0 & T/2 \\ \end{matrix} \right]=\frac{16}$

$$
 *

 * }

3). from the matrix, it is found that only the diagonal components, i.e. i=j, have non-zero value.
 * {| style="width:100%" border="0"

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

{{\Gamma }_{ij}}=0\begin{matrix} {} & i\ne j \\ \end{matrix}

$$     (Eq )
 *

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

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

{{\Gamma }_{ij}}\ne 0\begin{matrix} {} & i=j \\ \end{matrix}

$$ This agrees with the definition of Kronecker delta function. So, the F is orthogonal basis.
 *

 * }

title: User:Eml5526.s11.team4.yang/hw2-6


text:
User:Eml5526.s11.team6.kurth/2-8

=Problem 2.8= Refer to lecture slides [[media:fe1.s11.mtg11.djvu|11-1]] and [[media:fe1.s11.mtg10.djvu|10-4]] for the problem statement.

Given
where $$\displaystyle w^h(x)$$ and $$\displaystyle u^h(x)$$ are approximations of $$\displaystyle w(x)$$ and $$\displaystyle u(x)$$, respectively, as follows

and $$\displaystyle b_i(x)$$ for $$\displaystyle i=1,...,n$$ is a family of linearly independent basis functions.

Objective
Show $$\displaystyle (8.1)\Leftrightarrow(8.2)$$.

Part 1
$$\displaystyle (8.1)\Rightarrow(8.2)$$

Since (8.1) is valid for all weighting functions $$\displaystyle w^h(x)$$, we can choose arbitrary values of $$\displaystyle c_i$$ to satisfy (8.3).

Case 1

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{1,0,0,...,0,0\}$$.

From (8.3):

Then (8.1) becomes:

Case 2

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{0,1,0,...,0,0\}$$.

From (8.3):

Then (8.1) becomes:

 * $$\displaystyle \vdots$$

 * $$\displaystyle \vdots$$

Case n

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{0,0,0,...,0,1\}$$.

From (8.3):

Then (8.1) becomes:

Hence, $$\displaystyle (8.1)\Rightarrow(8.2)$$:

Part 2
$$\displaystyle (8.2)\Rightarrow(8.1)$$

Multiplying both sides of (8.2) by $$\displaystyle c_i$$ yeilds

Since (8.12) is equivalently zero for every value of i, we can successively add (8.12) to itself for each value of i without changing its value.

Hence,

Solved by William Kurth.

title: User:Eml5526.s11.team6.kurth/2-8


text:
User:John Vandenberg/IRC logs

Subpages here will be IRC logs of various sorts, where either permission has explicitly been given, or the other parties have been warned that continuing the private discussion will result in it being made public.
 * /Ottava - misc/
 * /Ottava - misc/2011-02-01

title: User:John Vandenberg/IRC logs


text:
User:John Vandenberg/IRC logs/Ottava - misc/2011-02-01

 * TZ is AEDT; log is unedited

 Feb 02 09:09:27 	Tell your dog to leave me alone. Feb 02 09:25:04 	pls talk to me in public forums only unless otherwise agreed Feb 02 09:25:56 	No, you took hundreds of hours of my work to justify giving Jack Merridew ops at Wikisource and never bothered to inform me first that you were doing it. Feb 02 09:26:10 	And all I get is that ass hole who is one of the worst offenders wikistalking me for 3 years Feb 02 09:26:23 	Tell him to leave me alone Feb 02 09:26:30 	stop - I will post any subsequent private comms publicly - you have been warned Feb 02 09:30:50 	I even have records where you told those like Durova that you knew he would continue to harass other people. Why do you keep doing that crap to me? Feb 02 09:31:34 	to be more clear, if you want to have a private discussion about something with me, I expect that to be a conversation that we both agree to participate in. due to problems resulting from previous private discussions, I no longer extend to you the usual protocol of implicit agreement to have private conversations, and now require that the mutual agreement is explicit Feb 02 09:32:01 	The problems were you didn't forward information about my harassment to people who matter!!! Feb 02 09:32:22 	Two major incidents where you let major sock puppet harassers get by who had long term harassment dealing wihtt me Feb 02 09:32:26 	Do you hate me? Feb 02 09:32:32 	Why do you do that stuff to me? Feb 02 09:32:40 	I bent over backwards every time you needed help Feb 02 09:32:49 	I never tried to fuck you or get you harassed by the worst people around Feb 02 09:32:53 	this is all going public Ottava - if you are unhappy with that, please stop Feb 02 09:33:00 	Two elections for ArbCom I busted my ass to help you get in Feb 02 09:33:04 	Go post in public Feb 02 09:33:08 	I just want the nonsense to stop Feb 02 09:33:12 	Stop encouraging them to harass me Feb 02 09:33:26 <Ottava>	I'm tired of it. Feb 02 09:34:30 	I have no intention of responding to the implied allegations you are making, such as the accusation that I am encouraging them. Feb 02 09:34:49 <Ottava>	I don't care if you reply or not Feb 02 09:34:53 <Ottava>	I want you to get him to leave me be Feb 02 09:35:32 <Ottava>	You built him up, you gave him my stuff to make people think he did work, you knew he was still a major problem, and you got him back into a position to harass those like myself. We both know that you have the ability to get him to back off. Feb 02 09:36:13 	do you think accusations are a good framework from which to ask for my assistance with a problem that you have which I have no prior knowledge of? Feb 02 09:36:26 <Ottava>	You were involved in most of it Feb 02 09:36:31 <Ottava>	How can you say you aren't aware of it? Feb 02 09:36:32 	more accusations Feb 02 09:36:36 <Ottava>	You even nominated him for admin! Feb 02 09:36:44 <Ottava>	You invoked my name multiple times, linked to -my- work Feb 02 09:36:48 <Ottava>	and didn't even inform me first Feb 02 09:37:16 	you made it free. Feb 02 09:37:24 <Ottava>	http://en.wikisource.org/wiki/Wikisource:Administrators/Archives/Jack_Merridew Feb 02 09:37:26 <Ottava>	I made it free Feb 02 09:37:32 <Ottava>	I didn't make it for you to lie about who did the work Feb 02 09:37:43 <Ottava>	I didn't make it so you could use -my- name to get a banned user to admin Feb 02 09:37:57 <Ottava>	You don't think you had any ethical obligation to inform me? Feb 02 09:38:09 <Ottava>	Especially when it was a guy who long term stalked one of my close wiki friends? Feb 02 09:38:14 <Ottava>	and then when allowed back went after me? Feb 02 09:38:30 	if you have any dispute with the wording I used in that nomination, please raise it on my ws user talk page Feb 02 09:38:44 <Ottava>	My dispute is that you violated me Feb 02 09:38:48 <Ottava>	and he continues to Feb 02 09:38:51 <Ottava>	and you just let him Feb 02 09:39:12 <Ottava>	http://meta.wikimedia.org/wiki/Requests_for_comment/User:Ottava_Rima#Jack.27s_Solution Feb 02 09:39:15 <Ottava>	He claims I stalk him Feb 02 09:39:20 <Ottava>	and his links are all where he goes after me Feb 02 09:39:47 <Ottava>	Are you saying I shouldn't be upset? Feb 02 09:40:03 <Ottava>	That I shouldn't feel betrayed that my work was used to help a guy harass me for 2 years? Feb 02 09:40:23 <Ottava>	God dammit John. I really liked you. I really thought highly of you. Feb 02 09:40:37 <Ottava>	Why is it that you keep helping these guys who are so incredibly wretched? Feb 02 09:41:38 <Ottava>	Even Durova told me to go to you to get Jack to stop bothering me. I talked to many admin and Arbs about what Jack is doing. Most say to go to you. So its not like I am special in thinking that you can get him to stop. Feb 02 09:41:41 <Ottava>	You owe me that much. Feb 02 09:56:17 	Ottava, I may be able to help you, but I am not going to engage with you in private unless I agree to do so. I do not agree to do so at present as I am at work, and would only agree to do so if you took a very different approach. At a minimum I would require that we discuss the problems you are seeing without any accusations against anyone involved. If you cant agree to do that, then i refuse to have a Feb 02 09:56:17 	private discussion with you Feb 02 09:56:53 	as discussed, I am publicly logging. https://secure.wikimedia.org/wikiversity/en/wiki/User:John_Vandenberg/IRC_logs/Ottava_-_misc/2011-02-01 Feb 02 09:57:10 <Ottava>	I don't care if you are publicly logging, I said what I wanted to say. Feb 02 09:58:30 	I will be doing the same with any emails, unless the substance of the private comms is about setting up and agreeing to parameters around a private discussion Feb 02 09:58:41 <Ottava>	I told you I was done.


text:
User:John Vandenberg/IRC logs/Ottava - misc

 * /2011-02-01


text:
User:Egm6341.s11.team2.asdfasdf

=Homework 2.5= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|mtg-7]] for the problem statement.

Given
Gauss-legendre quadrature:

Objectives
Verify this table against NIST handbook (lecture plan).

Solution
According to the Legendre polynomial in lecture note [[media:nm1.s11.mtg8.djvu|mtg-8]], {  Xi, i=1….n }  is the root of Legendre Polynomial of order n, and the associated weight with Xi is :
 * {| style="width:100%" border="0"

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

{W_i} = \frac

$$     (Eq 5.1 ) where the Legendre Polynomial is:
 *
 * }
 * {| style="width:100%" border="0"

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

{P_n}(x) = \sum\limits_{i = 0}^{[n/2]} \frac

$$     (Eq 5.2 )
 *
 * }.
 * {| style="width:100%" border="0"

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

{P_1}(x) = x

$$     (Eq 5.3 )
 *
 * }
 * {| style="width:100%" border="0"

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

{P_2}(x) = \frac{1}{2}(3{x^2} - 1)

$$     (Eq 5.4 )
 *
 * }
 * {| style="width:100%" border="0"

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

{P_3}(x) = \frac{1}{2}(5{x^3} - 3x)

$$     (Eq 5.5)
 *
 * }
 * {| style="width:100%" border="0"

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

{P_4}(x) = \frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8}

$$     (Eq 5.6)
 *
 * }
 * {| style="width:100%" border="0"

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

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.7 )
 *
 * }
 * {| style="width:100%" border="0"

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

{P_6}(x) = \frac{x^6} - \frac{x^4} + \frac{x^2} - \frac{5}

$$     (Eq 5.8 )
 *
 * }

1). When n=1, the root of P1(x)=0 is:
 * {| style="width:100%" border="0"

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

{x_{1}} = 0

$$     (Eq 5.9 )
 *
 * }

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

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

{{W}_{1}}=\frac{-2}{(1+1)[P_{1}^{'}({{x}_{1}}){{P}_{2}}({{x}_{1}})]}=\frac{-2}{(1+1)[1\centerdot (3{{x}_{1}}^{2}-1)/2]}=\frac{-2}{(1+1)[1\centerdot (3\centerdot 0-1)/2]}=1

$$     (Eq 5.10 )
 *
 * }

2). When n=2, the root of P2(x)=0 is:
 * {| style="width:100%" border="0"

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

{x_{1,2}} = \pm 1/\sqrt 3

$$     (Eq 5.11 )
 *
 * }

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

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

{{W}_{1}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{1}}){{P}_{3}}({{x}_{1}})]}=\frac{-2}{(2+1)[(3{{x}_{1}})\centerdot (5{{x}_{1}}^{3}-3{{x}_{1}})/2]}=1

$$     (Eq 5.12 )
 *
 * }
 * {| style="width:100%" border="0"

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

{{W}_{2}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{2}}){{P}_{3}}({{x}_{2}})]}=\frac{-2}{(2+1)[(3{{x}_{2}})\centerdot (5{{x}_{2}}^{3}-3{{x}_{2}})/2]}=1

$$     (Eq 5.13 )
 *
 * }

3). When n=3, the root of P3(x)=0 is:
 * {| style="width:100%" border="0"

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

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{x_{2,3}} = \pm \sqrt {15} /5} \end{array}

$$     (Eq 5.14 )
 *
 * }

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

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

{{W}_{1}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{1}}){{P}_{4}}({{x}_{1}})]}=\frac{-2}{(3+1)[(15{{x}_{1}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{1}}^{4}-\frac{15}{4}{{x}_{1}}^{2}+\frac{3}{8})]}=\frac{8}{9}

$$     (Eq 5.15 )
 *
 * }
 * {| style="width:100%" border="0"

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

{{W}_{2}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{2}}){{P}_{4}}({{x}_{2}})]}=\frac{-2}{(3+1)[(15{{x}_{2}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{2}}^{4}-\frac{15}{4}{{x}_{2}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.16)
 *
 * }
 * {| style="width:100%" border="0"

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

{{W}_{3}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{3}}){{P}_{4}}({{x}_{3}})]}=\frac{-2}{(3+1)[(15{{x}_{3}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{3}}^{4}-\frac{15}{4}{{x}_{3}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.17)
 *

 * }

4). When n=4, the root of P4(x)=0 is:
 * {| style="width:100%" border="0"

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

\frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8} = 0

$$     (Eq 5.18)
 *

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

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

{x^2} = (3 \pm 2\sqrt {6/5} )/7

$$     (Eq 5.19)
 *

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

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

{x_{1,2}} = \pm \sqrt {(3 + 2\sqrt {6/5} )/7}

$$     (Eq 5.20 )
 *

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

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

{x_{3,4}} = \pm \sqrt {(3 - 2\sqrt {6/5} )/7}

$$     (Eq 5.21 )
 *

 * }

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

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

{{W}_{1}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{1}}){{P}_{5}}({{x}_{1}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{1}}^{3}-\frac{15}{2}{{x}_{1}})\centerdot (\frac{63}{8}{{x}_{1}}^{5}-\frac{70}{8}{{x}_{1}}^{3}+\frac{15}{8}{{x}_{1}})]}=(18-\sqrt{30})/36

$$     (Eq 5.22 )
 *

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

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

{{W}_{2}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{2}}){{P}_{5}}({{x}_{2}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{2}}^{3}-\frac{15}{2}{{x}_{2}})\centerdot (\frac{63}{8}{{x}_{2}}^{5}-\frac{70}{8}{{x}_{2}}^{3}+\frac{15}{8}{{x}_{2}})]}=(18-\sqrt{30})/36

$$     (Eq 5.23 )
 *

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

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

{{W}_{3}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{3}}){{P}_{5}}({{x}_{3}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{3}}^{3}-\frac{15}{2}{{x}_{3}})\centerdot (\frac{63}{8}{{x}_{3}}^{5}-\frac{70}{8}{{x}_{3}}^{3}+\frac{15}{8}{{x}_{3}})]}=(18+\sqrt{30})/36

$$     (Eq 5.24)
 *

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

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

{{W}_{4}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{4}}){{P}_{5}}({{x}_{4}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{4}}^{3}-\frac{15}{2}{{x}_{4}})\centerdot (\frac{63}{8}{{x}_{4}}^{5}-\frac{70}{8}{{x}_{4}}^{3}+\frac{15}{8}{{x}_{4}})]}=(18+\sqrt{30})/36

$$     (Eq 5.25)
 *

 * }

5). When n=5, the root of P5(x)=0 is:
 * {| style="width:100%" border="0"

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

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.26)
 *

 * }

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

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

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{{({x_{2,3,4,5}})}^2} = (5 \pm 2\sqrt {10/7} )/9} \end{array}

$$     (Eq 5.27)
 *

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

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

{x_{2,3}} = \pm \sqrt {(5 + 2\sqrt {10/7} )} /3

$$     (Eq 5.28)
 *

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

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

{x_{4,5}} = \pm \sqrt {(5 - 2\sqrt {10/7} )} /3

$$     (Eq 5.29)
 *

 * }

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

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

\begin{align} & {{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{1}}^{4}-\frac{210}{8}{{x}_{1}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{1}}^{6}-\frac{315}{16}{{x}_{1}}^{4}+\frac{105}{16}{{x}_{1}}^{2}-\frac{5}{16})]} \\ & =128/225 \\ \end{align}

$$     (Eq 5.30)
 *

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

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

\begin{align} & {{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{2}}^{4}-\frac{210}{8}{{x}_{2}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{2}}^{6}-\frac{315}{16}{{x}_{2}}^{4}+\frac{105}{16}{{x}_{2}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.31)
 *

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

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

\begin{align} & {{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{3}}^{4}-\frac{210}{8}{{x}_{3}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{3}}^{6}-\frac{315}{16}{{x}_{3}}^{4}+\frac{105}{16}{{x}_{3}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.32)
 *

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

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

\begin{align} & {{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{4}}^{4}-\frac{210}{8}{{x}_{4}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{4}}^{6}-\frac{315}{16}{{x}_{4}}^{4}+\frac{105}{16}{{x}_{4}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.33 )
 *

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

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

\begin{align} & {{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{5}}^{4}-\frac{210}{8}{{x}_{5}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{5}}^{6}-\frac{315}{16}{{x}_{5}}^{4}+\frac{105}{16}{{x}_{5}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.34)
 *

 * }

title: User:Egm6341.s11.team2.asdfasdf


text:
User:Egm6341.s11.team2.cleveland/HW2

=Problem 2.5 Gaussian Quadrature= Refer to lecture slide from [[media:nm1.s11.mtg7.djvu|mtg-7]] for the problem statement and table shown below.

Given
The Legendre Polynomians $$ P_{n}(x) $$ defined here
 * {| style="width:100%" border="0"

$$ \displaystyle {P}_{n}(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} $$      (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Verify
The points $$ \displaystyle x_i $$ and weights $$\displaystyle w_i $$ for $$\displaystyle n=1,2,3,4,5 $$.

The table above was found at [|Gaussian_quadrature], a companion wiki site.

Solution
As was discussed in class during [[media:nm1.s11.mtg7.djvu|mtg-7]], $$ \displaystyle x_i $$ is the root of the $$ n-th $$ order Legendre Polynomial $$ \displaystyle P_{n}(x) $$ defined in (5.1). Using (5.1) we obtain the following for $$ n=1,2,3,4,5 $$

For n = 1
The root of the Legendre Polynomial $$ P_1(x)=0 $$ is found by solving $$ \displaystyle P_1(x)=x=0 $$

And the weight, $$ w_{1} $$ is $$ \displaystyle \begin{align} w_{1} & = \frac{-2}{(1+1)[P_{1}^{'}({{x}_{1}}){{P}_{2}}({{x}_{1}})]}  \\ & = \frac{-2}{(1+1)\left[ 1\cdot \frac{3{x_1}^{2}-1}{2}\right] }  \\ & = \frac{-2}{(1+1)\left[ 1\cdot \frac{3\cdot{0}^{2}-1}{2}\right] } \\ & = 2         \end{align} $$

For n = 2
The root of the Legendre Polynomial $$ P_2(x)=0 $$ are found by solving $$ \displaystyle P_2(x)=(3x^2-1)=0 $$

Where the weights, $$ w_{1} $$ and $$ w_{2} $$ are $$ \displaystyle \begin{align} w_{1} & = \frac{-2}{(2+1)[P_2^{'}(x_1)P_3(x_1)]}  \\ & = \frac{-2}{(2+1)\left[ 3x_1 \cdot \frac{5{x_1}^{3}-3x_1}{2}\right] }  \\ & = \frac{-2}{(2+1)\left[ 3\left( {\frac{1}{\sqrt 3 }}\right)^3 \cdot \frac{5\left( {\frac{1}{\sqrt 3 }}\right)^3 -3\left(\frac{1}{\sqrt 3 }\right) }{2}\right] } \\ & = 1         \end{align} $$

$$ \displaystyle \begin{align} w_{2} & = \frac{-2}{(2+1)[P_2^{'}(x_2)P_3(x_2)]}  \\ & = \frac{-2}{(2+1)\left[ 3x_2 \cdot \frac{5{x_2}^{3}-3x_2}{2}\right] }  \\ & = \frac{-2}{(2+1)\left[ 3\left( {-\frac{1}{\sqrt 3 }}\right)^3 \cdot \frac{5\left( {-\frac{1}{\sqrt 3 }}\right)^3 -3\left( -\frac{1}{\sqrt 3 }\right) }{2}\right] } \\ & = 1         \end{align} $$

For n = 3
The root of the Legendre Polynomial $$ P_3(x)=0 $$ are found by solving $$ \displaystyle P_3(x)=(5x^3-3x)=0 $$

Where the weights, $$ w_{1}, w_{2}, $$ and $$ w_{3} $$ are $${{W}_{1}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{1}}){{P}_{4}}({{x}_{1}})]}=\frac{8}{9} $$ $${{W}_{2}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{2}}){{P}_{4}}({{x}_{2}})]}=\frac{5}{9} $$ $${{W}_{3}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{3}}){{P}_{4}}({{x}_{3}})]}=\frac{5}{9}$$

For n = 4
The root of the Legendre Polynomial $$ P_4(x)=0 $$ are found by solving $$ \displaystyle P_4(x)=(35x^4-30x^2+3)=0 $$

Where the weights, $$ w_{1}, w_{2}, w_{3}$$ and $$ w_{4} $$ are

$${{W}_{1}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{1}}){{P}_{5}}({{x}_{1}})]}=(18-\sqrt{30})/36 $$ $${{W}_{2}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{2}}){{P}_{5}}({{x}_{2}})]}=(18-\sqrt{30})/36 $$ $${{W}_{3}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{3}}){{P}_{5}}({{x}_{3}})]}=(18+\sqrt{30})/36 $$ $${{W}_{4}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{4}}){{P}_{5}}({{x}_{4}})]}=(18+\sqrt{30})/36 $$

For n = 5
The root of the Legendre Polynomial $$ P_5(x)=0 $$ are found by solving $$ \displaystyle P_5(x)=(63x^5-70x^3+15x)=0 $$

Where the weights, $$ w_{1}, w_{2}, w_{3}, w_{4} $$ and $$ w_{5} $$ are

$${{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{128}{225}$$ $${{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{322-13\sqrt{70}}{900}$$ $${{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{322-13\sqrt{70}}{900}$$ $${{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{322+13\sqrt{70}}{900}$$ $${{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{322+13\sqrt{70}}{900}$$

=Problem 2.1 Taylor's Series Proof=

Problem Statement
2.1.1) Using integration by parts show that the first five terms of the Taylor's Series expansion for $$ f(x) $$ is

2.1.2) Using the IMVT shown below in (2.2) express the remainder in terms of $$ f^5(\xi) $$ such that $$ \xi\in [x_0,x] $$. Integral Mean Value Theorem: Let $$\displaystyle w(x)$$ be a nonnegative integrable function defined on $$\displaystyle [x_0,x] $$

If $$\displaystyle f(x) $$ is a continuous function defined on $$\displaystyle [x_0,x] $$, then

for some $$\displaystyle \xi $$ such that

2.1.3) Assume that (1.5) and (1.6) defining $$ p_{n}(x) $$ and $$ R_{n+1}(x) $$ hold. Perform integration by parts again to obtain the $$ (n+1)th $$ expansion containing $$ R_{n+2}(x) $$. where $$ R_{n+1}(x) $$ will be

2.1.4) Use the IMVT shown in (2.1.2) to show

For n=1
Since $$ f(x) $$ can be written as follows, where we will define $$ {\color{red} I_1(x)} $$ as

Recall the general form for Integration By Parts

Then using integration by parts as shown in (2.9) we can let

Which provides us with the following result

Plugging (2.11) into (2.8) reveals the first term of the Taylor's Series Expansion

For n=2,3,4
Similarly we can carry out the same procedure to elucidate the $$ 2^{nd},3^{rd} and ,4^{th} $$ terms.

Putting all of these terms together results in

Solution to 2.1.2
From the previous part we know the remainder for the Taylor's Series Expansion

Using the IMVT shown in (1.2), (1.3) and (1.4) let us define $$ w(x=t) $$ and $$ f(x=t) $$ as follows

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

Solution to 2.1.3
Assume that (1.5) and (1.6) defining $$ p_{n}(x) $$ and $$ R_{n+1}(x) $$ hold. Perform integration by parts again to obtain the $$ (n+1)th $$ expansion containing $$ R_{n+2}(x) $$.

where $$ R_{n+1}(x) $$ will be

Performing the integration results in

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

Solution to 2.1.4
Therefore by applying the IMVT for $$ R_{n+2}(x) $$ will be

where $$ R_{n+2}(x) $$ will be

title: User:Egm6341.s11.team2.cleveland/HW2


text: File:Channel.png

title: File:Channel.png


text: File:Rtt-treatment-of-channel-flow.png

title: File:Rtt-treatment-of-channel-flow.png


text:
User:Eml5526.s11.t1.ab/HW2

=Problem 2=

Given
$$\left [B _{jk} \right ]=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}$$

$$\overset{v}{\rightarrow}=5\overset{a_{1}}{\rightarrow}-7\overset{a_{2}}{\rightarrow}-4\overset{a_{3}}{\rightarrow}$$ $$\displaystyle (Eq. 2.1) $$

Find
1) $$ Det\left [B _{jk} \right ]$$.

2) $$Find \Gamma (B_{1},B_{3},B_{3})=K$$.

3) $$Find \overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$.

4) Solve $$\displaystyle (Eq. 5) $$ on P 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

5) Use 1) on P 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$ What is $$\vec{K}$$ & $$\vec{F}$$?

6) Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in 4).

7) Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

1)
$$ Det\left [B _{jk} \right ]$$=

$$\begin{vmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{vmatrix}=-8.$$.

2)
$$Find \Gamma (B_{1},B_{3},B_{3})=\overset{K}{\rightarrow}$$.

According to the $$\displaystyle (Eq. no 4) $$ on P 7-1, $$B_{1},B_{3},B_{3}$$ given before we have,

$$\overset{b_{j}}{\rightarrow}.d_{j}=\overset{V}{\rightarrow}$$

Successively performing multiplication by $$B_{1},B_{3},B_{3}$$  we get,

$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$ $$\displaystyle (Eq. 2.2) $$

Substituting the given values we get,

$$d_{1}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}+d_{2}\begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix}+d_{3}\begin{bmatrix} 3\\ 2\\ 6 \end{bmatrix}=\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

This equation can also be written in the following format,

$$\begin{bmatrix} 1 &2 &3 \\ 1 &-1  &2 \\ 1 & 3 & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ 4 \end{pmatrix}$$

$$\displaystyle (Eq. 3) $$ on P 7-2 is obtained by multiplying $$\displaystyle (Eq. 2.2)  $$ by each $$\overset{b}{\rightarrow}$$, so we can write,

$$\overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{1}}{\rightarrow}.\overset{V}{\rightarrow}==F_{1}$$

$$\overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{2}$$

$$\overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{3}$$

These 3 equations can be further presented as below in the matrix form,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}$$

Which also can be written as

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix} \begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{F}{\rightarrow}$$

Substituting the numerical values we get,

$$\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \\ 1 &-1 &2\\ 1 &3  &6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

Performing the product, we get:

$$\begin{bmatrix} 3 &4 &11\\ 4 &14  &22\\ 11 &22  &49 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\overset{F}{\rightarrow}$$ $$\displaystyle (Eq. 2.3) $$

This $$\displaystyle (Eq. 2.3) $$ is equivalent to,

$$Kd=F$$, where,

$$K=\begin{bmatrix} 3 &4 &11 \\ 4 &14 &22 \\ 11&22 &49 \end{bmatrix}=Stiffness Matrix$$

Furthermore,

$$Det[K]=\begin{vmatrix} 3 &4 &11\\ 4 &14 &22 \\ 11 &22 &49 \end{vmatrix}= 64$$

3)
$$\overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$=

From $$\displaystyle (Eq. 2.3) $$

$$\overset{F}{\rightarrow}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}$$

4)
Solve $$\displaystyle (Eq. 5) $$ on P 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

Rearranging $$\displaystyle (Eq. 2.3) $$, We get,

$$d=K^{-1}F$$

$$d=\begin{bmatrix} 3 &4 &11 \\ 4 &14  &22 \\ 11 &22  &49 \end{bmatrix}^{-1}\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$

5)
Use 1) on P 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$. Further, What is $$\vec{K}$$ & $$\vec{F}$$?

In this case, $$w_{i}=a_{i}$$, where $$a_{i}$$ is the orthonormal basis.

Starting with Equation no 1 on P 7-2, we get:

$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$

If we multiply successively by $$a_{i}$$, we get the following equations,

$$\overset{a_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{1}}{\rightarrow}.\overset{V}{\rightarrow}=F_{1}$$

$$\overset{a_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{2}}{\rightarrow}.\overset{V}{\rightarrow}=F_{2}$$

$$\overset{a_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{3}}{\rightarrow}.\overset{V}{\rightarrow}=F_{3}$$ $$\displaystyle (Eq. 2.4) $$

Which can be further written as,

$$\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

As $$ a_{i}$$ is orthonormal basis, we can write the following,

$$a_{1}=\begin{bmatrix} 1 & 0 &0 \end{bmatrix};a_{2}=\begin{bmatrix} 0&1&0 \end{bmatrix};a_{3}=\begin{bmatrix} 0& 0 &1 \end{bmatrix}$$

Substituting the respective values we have,

$$\begin{bmatrix} 1 &0 &0 \\ 0 &1  &0 \\ 0 &0  &1 \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0& 1 \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Simplifying we obtain that,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Replacing the numerical values we get,

$$\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}$$ $$\displaystyle (Eq. 2.5) $$

This equation is nothing but the system

$$\overset{\vec{K}}{\rightarrow}\overset{d}{\rightarrow}=\overset{\vec{F}}{\rightarrow}$$

From where we can conclude that $$\overset{\vec{K}}{\rightarrow}$$ is a matrix whose columns are the vectors $$\overset{b_{j}}{\rightarrow}$$ and $$\overset{\vec{F}}{\rightarrow}= V$$.

6)
Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in question 4).

$$\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}^{-1}\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$

We can see that this result is identical to the result we obtained in 4).

7)
Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

We can explicitely see that $$\overset{K}{\rightarrow}$$  is always symmetric because it is obtained from the product of $$\overset{b_{j}}{\rightarrow}.\overset{b_{i}}{\rightarrow}$$; where as $$\overset{\vec{K}}{\rightarrow}$$ is not always symmetric. It will be symmetric only when the vector b is a Orthonormal basis.

The pros of Bubnov-Galerkin method is the symmetry of the stiffness matrix which is generally easy to solve, however more calculations are necessary to get the stiffness matrix. Whereas, in the Petrov-Galerking method, obtaining the stiffness matrix is easier but solving the system on other hand can be more difficult owing to its non-symmetry.

Contributing Members

 * Abhijeet Bhalerao

=Problem 3= Show that the Equation no 1 P.8-1 which states $$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$

$$\forall w=\in R^{n}$$ $$\displaystyle (Eq. 3.1) $$

is equivalent to

$$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$ where $$\overset{w}{\rightarrow}=\sum \beta _{i}.a_{i}$$

$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.2) $$

Given
$$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$

$$\forall w\in R^{n}$$ $$\displaystyle (Eq. 3.1) $$

Find
Show that $$\displaystyle (Eq. 3.1) $$ is equivalent to $$\displaystyle (Eq. 3.2) $$

Solution
For avoiding confusion we are designating the weight function as $$\overset{W}{\rightarrow}$$ in $$\displaystyle (Eq. 3.1) $$ and the weight function as $$\overset{w}{\rightarrow}$$ in $$\displaystyle (Eq. 3.2)  $$.

Considering $$\displaystyle (Eq. 3.2) $$ and Substituting for $$\overset{W}{\rightarrow}$$

$$\sum \beta _{i}a_{i}.\overset{P(v)}{\rightarrow}=0$$

$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.3) $$

As $$a_{i}$$ is an orthonormal basis, performing a successive product by $$a_{i}$$ in the $$\displaystyle (Eq. 3.3) $$, we get

$$\begin{bmatrix} 1 &0 &0  &....0 \end{bmatrix}\beta _{1}.\overset{P(v)}{\rightarrow}=0$$

In the similar way

$$\begin{bmatrix} 0 &1 &0  &....0 \end{bmatrix}\beta _{2}.\overset{P(v)}{\rightarrow}=0$$

.

.

.

$$\begin{bmatrix} 0 &0 &0  &....1 \end{bmatrix}\beta _{n}.\overset{P(v)}{\rightarrow}=0$$;

Furthermore all the equations can be combined and written as

$$ \begin{bmatrix} 1 &0 &...0 \\ 0 &1  &...0 \\ 0 &0  &...1 \end{bmatrix}\bigl(\begin{smallmatrix} \beta_{1}\\ .\\ .\\ \beta_{n} \end{smallmatrix}\bigr).\overset{P(v)}{\rightarrow}=\bigl(\begin{smallmatrix} 0\\ .\\ .\\ 0 \end{smallmatrix}\bigr) ... \forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.4) $$

We can cleraly observe that the Matrix on the left hand side is the identity matrix $$\begin{bmatrix} I \end{bmatrix}$$; Moreover we have also written $$(\beta _{1},\beta _{2}..\beta _{n})$$ as a vector which further can be called as $$\overset{\beta}{\rightarrow}$$. and $$\displaystyle (Eq. 3.4) $$ becomes

$$\overset{\beta}{\rightarrow}.\overset{P(v)}{\rightarrow}=0; \forall \overset{\beta}{\rightarrow} \in R^{n}$$

Which is Identical to $$\displaystyle (Eq. 3.4) $$.

Contributing Members

 * Abhijeet Bhalerao

title: User:Eml5526.s11.t1.ab/HW2


title: File:Fe1.s11.team6.HW2.6.fig1.JPG


text: File:Egm6341.s11.hw2 4a.jpg

title: File:Egm6341.s11.hw2 4a.jpg


text: File:Rtt-treatment-of-pipe-flow.png

title: File:Rtt-treatment-of-pipe-flow.png