text:
User:Eml5526.s11.team6/hwk2

Homework 2

=Problem 2.1=

Find
For the heat problem, derive the following partial differential equation by performing a balance of heat:

$$ \frac{\partial}{\partial x}\left [ A(x)k(x)\frac{\partial u}{\partial x} \right ]+f(x,t)=A(x) \rho(x)c\cdot \frac{\partial u}{\partial t}\ $$

Solution
Consider the FBD of an infinitesimally small section of length dx of the 1-D conductive bar with varying material properties at a distance x from the origin. Where, A(x) - Varying cross section of the bar, k(x) - conductivity of the material, rho (x) - Mass density, u - Temperature constrained only to x- axis (1-D bar), t - Time.

Where, heat flux at $$\displaystyle x=q(x)$$,

heat flux at $$\displaystyle x+dx=q(x+dx)$$

outward heat flow through A(x)=Q(x), at n(x)=-1,

$$\displaystyle Q(x)=q(x)A(x)n(x)= -q(x)A(x)$$ $$\displaystyle (Eq. 2.1.1) $$

Similarly, outward heat flow through $$\displaystyle A(x+dx)=Q(x+dx), at n(x+dx)=1,$$

$$\displaystyle Q(x+dx)=q(x+dx)A(x+dx)n(x+dx)= q(x+dx)A(x+dx) $$ $$\displaystyle (Eq. 2.1.2) $$

Also, internal heat source per unit volume= r(x,t)

Hence, heat source per unit length= r(x,t)A(x) = f(x,t)$$\displaystyle (Eq. 2.1.3) $$

Now, balance of heat can be done as follows:

$$\displaystyle H_{1}= $$heat flow in control volume

$$\displaystyle =-[Q(x)+Q(x+dx)]$$ $$\displaystyle =-[-q(x)A(x)+q(x+dx)A(x+dx)]$$ $$\displaystyle (Eq. 2.1.4) $$

$$\displaystyle H_{2}= $$Heat by internal source r(x,t) $$\displaystyle =f(x,t)dx$$$$\displaystyle (Eq. 2.1.5) $$

$$\displaystyle H_{3}= $$heat by temperature difference $$\displaystyle=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.6) $$

Now, for balance of heat

$$\displaystyle H_{1}+H_{2}=H_{3}$$$$\displaystyle (Eq. 2.1.7) $$

Therefore, from $$\displaystyle (Eq. 2.1.4) $$, $$\displaystyle (Eq. 2.1.5) $$,  $$\displaystyle (Eq. 2.1.6) $$,  $$\displaystyle (Eq. 2.1.7) $$ we can write,

$$\displaystyle q(x)A(x)-q(x+dx)A(x+dx)+f(x,t)dx=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.8) $$

put equations $$\displaystyle (Eq. 2.1.1) $$ and $$\displaystyle (Eq. 2.1.2) $$ in $$\displaystyle (Eq. 2.1.8) $$

$$\displaystyle Q(x,t)-Q(x+dx,t)+f(x,t)dx=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$

Including Taylor series expansion for $$Q(x+dx,t)$$ and neglecting the higher order terms (h.o.t) yields:

$$ \cancel {Q(x,t)}-[\cancel {Q(x,t)}+\frac {\partial }{\partial x}(Q(x,t))+(h.o.t)]+f(x,t)=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$

Hence, $$ -\frac {\partial }{\partial x}(Q(x,t))+(h.o.t)]+f(x,t)=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.9) $$ where, $$\displaystyle Q(x,t)= -q(x)A(x)$$ and $$\displaystyle q(x)=k(x)\frac {\partial u}{\partial x}$$

Therefore we get,

=Problem 2. 2=

Given
Reference for all formulae is Mtg 7 $$ \{ {{\mathbf{b}}_i},i = 1, \cdot \cdot  \cdot ,n\} $$ are basis vectors for $$ {{\mathbf{R}}^n} $$ which are not necessarily orthonormal.

In order to find $$ \{ {{\mathbf{v}}_i} \} $$ such that $$ {\mathbf{v}} = \sum\limits_{i = 1}^n $$ , we consider $$ \{ {{\mathbf{a}}_i},i = 1, \cdot \cdot  \cdot ,n\} $$ are orthonormal basis in Cartesian coordinates, and $$ {{\mathbf{b}}_j} = b{_{jk}}{{\mathbf{a}}_k} $$ , $$ {{\mathbf{v}}=5{{\mathbf{a}}_1}-7{{\mathbf{a}}_2}-4{{\mathbf{a}}_3}} $$

where $$ \left[ \right] = \left[ {\begin{array}{ccccccccccccccc} 1&1&1 \\  2&{ - 1}&3 \\   3&2&6 \end{array}} \right] $$

Find
A) Find $$ \det \left[ \right] $$

B) Find $$ \det \Gamma $$ where $$ {\mathbf{\Gamma }} = \left\{ \right\} = {\mathbf{K}} $$

C) Find $$ {\mathbf{F}} = \left\{ \right\} = \left\{ {{{\mathbf{b}}_{\mathbf{i}}} \cdot {\mathbf{v}}} \right\}$$

D) Solve equation (5) on page 2 of Mtg 7 for $$ {\mathbf{d}} = \left\{ \right\} $$

E) Use equation (1) on page 4 of Mtg 7 to find $$ \overline {\mathbf{K}} {\mathbf{d} = \overline {\mathbf{F}} }$$ and what is $$ \overline {\mathbf{K}}, \overline {\mathbf{F}} $$

F) Solve for $$ {\mathbf{d}}$$ and compare of $$ {\mathbf{K}}$$ and $$ \overline {\mathbf{K}} $$

G) Observe symmetric properties of $$ {\mathbf{K}}$$ and $$ \overline {\mathbf{K}} $$ and discuss pros and cons of these two methods

Solution
A) The determinant of $$ \left[ \right] $$ is as following:

B) Write $$ \left[ \Gamma \right] = \left[ K \right] $$ in matrix form:

Then the determinant is as following

C) Write $$ \left[ F \right] $$ in matrix form:

D) Use equation (2.2.5)on page 2 of Mtg 7 and write it in matrix form:

$$ \left[ {\begin{array}{ccccccccccccccc} 3&4&{11} \\  4&{14}&{22} \\   {11}&{22}&{49} \end{array}} \right] \times \left[ d \right] = \left[ {\begin{array}{ccccccccccccccc} { - 6} \\  5 \\   { - 23} \end{array}} \right] $$

Multiply the above equation with the inverse of matrix K, we can get d in matrix form:

E) Consider $$ {w_i} = {a_i},i = 1, \cdot \cdot  \cdot ,n $$ and follow the steps above, it is easy to get K and F in matrix form:

This is as $$ {w_i} = {a_i},i = 1, \cdot \cdot  \cdot ,n $$ form an identity matrix and this is a multiplication involving an identity matrix

Similar is the case about Matrix F

F) Similar with step D, we can get d in matrix form with finding inverse of $$ \overline {\mathbf{K}} $$, using Matlab

Note that matrix d obtained in step D and step F are the same.

G)$$ {\mathbf{K}}$$ is always a symmetric matrix because it is obtained from the product of $$\underline{b}_{j}.\underline{b}_{i}$$. However $$ \overline {\mathbf{K}} $$ is unsymmetrical and will be symmetric only if the basis are orthonormal and not otherwise.

Bubnov-Galerkin method gives slightly more difficult to compute the symmetric stiffness matrix K, but can choose arbitrary vector to find the answer.

Petrov Galerkin method gives easier way to compute the unsymmetrical stiffness matrix K but with unsymmetrical nature of the stiffness matrix and without orthonormal vectors it becomes difficult to solve the system of equations.

=Problem 2. 3 =

Find
Show that (Eq. 2.3.2) is equivalent to

Solution
(Eq. 2.3.2) is equivalent to (Eq. 2.3.3) given arbitrary choices of $$\beta_n$$. For example,

Choice 1
 * Choosing $$\beta_1 = 1, \beta_2 = ... = \beta_n = 0$$, and substituting into (Eq. 2.3.2) yields
 * $$ a_1 \cdot \underline P \left ( \underline{v} \right )=0$$

Choice 2
 * Choosing $$\beta_2 = 1, \beta_1 = \beta_3 = ... = \beta_n = 0$$, and substituting into (Eq. 2.3.2) yields
 * $$ a_2 \cdot \underline P \left ( \underline{v} \right )=0$$

Choice n
 * Choosing $$\beta_n = 1, \beta_1 = ... = \beta$$n-1 $$= 0$$, and substituting into (Eq. 2.3.2) yields
 * $$ a_n \cdot \underline P \left ( \underline{v} \right )=0$$

Therefore, given a specific choice of $$ \beta_n $$ within the real domain,

=Problem 2.4=

Find
To Prove Solution provided by Wolfram Alpha

Prove Following steps and use them to get the final proof of the above eqution:

Part 1: $$ \int log(x)dx = x*log(x) - x$$ Solution provided by Wolfram Alpha using integration by parts.

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

Part 3: $$ \int \frac {x^2 dx}{1+C*x} = \frac {2*log(C*x + 1) + x*(C*x-2)*C}{2*C^3} $$

Part 4: $$ \int \frac {x^2 dx}{A+C*x} = \frac {2*A^2*log(C*x + A) + C*x*(C*x-2*A)}{2*C^3} $$

Part 5: Now refer to meeting 9 [[media:fe1.s11.mtg9.djvu|Mtg 9 (c)]] and find exact solution $$\displaystyle u(x)$$ for problem (3)

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

Solution
Part 1

Prove Solution provided by Wolfram Alpha

using integration by parts, i.e.,

Part 2

Prove Solution provided by Wolfram Alpha

using integration by parts, (Eq. 2.4.3)

Part 3

Prove Solution provided by Wolfram Alpha

Using integration by parts, (Eq. 2.4.3)

Continuing with the integration by parts with the latter part of the RHS of (Eq. 2.4.15), Given (Eq. 2.4.7) and the substitution given in (Eq. 2.4.14),

Applying integration by parts again, we obtain the following relations:

Finally, the integral in the RHS of (Eq. 2.4.19) can be solved for and simplified using (Eq. 2.4.12) and the integration by substitution given (Eq. 2.4.14), i.e., Combining the former portion of the RHS of (Eq. 2.4.15), the former portion of the RHS of (Eq. 2.4.19), and the latter portion of the RHS of (Eq. 2.4.20), and then simplifying, the final solution to the integral in (Eq. 2.4.13) can be found to be

Part 4

Prove Solution provided by Wolfram Alpha

and the first integration by parts given in Step 3, (Eq. 2.4.15),

Continuing with the integration by parts with the latter part of the RHS of (Eq. 2.4.24),

and given the second integration by parts in Step 3, (Eq. 2.4.19)

Finally, the integral in the RHS of (Eq. 2.4.27) can be solved for and simplified using (Eq. 2.4.12) and the integration by substitution given in (Eq. 2.4.23), i.e.,

Combining the former portion of the RHS of (Eq. 2.4.24), the former portion of the RHS of (Eq. 2.4.27), the latter portion of the RHS of (Eq. 2.4.28), and then simplifying, the final solution to the integral in (Eq. 2.4.22) can be found in a similar fashion to (Eq. 2.4.20) of Step 3, i.e.,

Final Step

Given (Eq. 2.4.29) from Step 4, if A = C = 1, then

Thus, required result stated in ' Find' section Eq 2.4.1 is achieved.

Part 5

Given:

$$\frac{d}{dx}[(2+3x)*(\frac{du}{dx})]+5x = 0 \forall x \in \left]0,1 \right[  $$

$$ u(1)=4, -(\frac{du}{dx}) (x=0) = 6 $$

To find:

Find the exact solution for u(x):

Solution:

$$ \frac{d}{dx}[(2+3x)(\frac{du}{dx})]+5x=0 $$

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

Integrating with respect to x

$$ (2+3x) \frac{(du)}{(dx)}= \frac{-5x^2}{2}+k1 $$

$$ \frac{du}{dx}= \frac{-5x^2}{2(2+3x)}+\frac{k1}{(2+3x)} $$

Integrating again with respect to x and using eq.2.4.22

$$ u(x)= \frac{-5}{2}[\frac{log(3x+2)+3x(x-4)}{(3^3*2)}]+k1[\frac{ln(2+3x)}{3}]+k2 $$

To find the constants, we use the given boundary conditions:

$$ \frac{du}{dx}= \frac{-5*x^2}{2(2+3x)}+\frac{k1}{(2+3x)} $$

$$ but \frac{-du}{dx}(x=0)=6 $$

Substituting x=1 in the above equation,

$$ k1= -12 $$

Substituting $$ k1=-12 $$ and simplifying, we get:

$$ u(x)= \frac{-118}{27}*log(2+3x)-\frac{5*x^2}{12}+\frac{5}{9}x+k2 $$

Also u(x)=4 at x=1

Substituting x=1 in the expression for u(x)

$$ k2= 10.896 $$

Thus final expression for u(x) is

Part 6

=Problem 2.5=

Given
We have Eq. (2) from page 7-2 of mtg 7 ,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

where  i =  1, 2 ,..., n

Find
Show that $$ \underline{w} .\underline{P}(\underline{v})=0 $$ ,

where $$  \underline{w} = \sum_{i=1}^{n}\alpha_{i} \underline{b} _{i} $$ ,

and $$ \forall (\alpha _{1},\alpha _{2},...,\alpha _{n})\in R^{n} $$

Solution
We have,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

With i = 1 to n, we can then write -

$$ \underline{b}_{1}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

$$ \underline{b} _{n}.\underline{P}(\underline{v})=0 $$

Now consider a collection of arbitrary numbers $$  (  \alpha _{1}, \alpha _{2} ,..., \alpha _{n}  ) $$ i.e.   \alpha _{i}      with i = 1 to n

Now successively multiplying above n equations, on both sides by corresponding  \alpha _{i}, we get the following new set of n equations -

With i = 1 to n, we can then write -

$$ \alpha _{1}. \underline{b}_{1} .\underline{P}(\underline{v})=0 $$

$$ \alpha _{2}. \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \alpha _{3}. \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

$$ \alpha _{n}. \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

Adding LHS and RHS respectively for above n equations, we get the following -

$$ \alpha _{1} \underline{b}_{1}.\underline{P}(\underline{v})+ $$ $$ \alpha _{2} \underline{b}_{2}.\underline{P}(\underline{v})+...+ $$ $$ \alpha _{n} \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

which in index notation can be expressed as the following:

$$ \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i} .\underline{P}(\underline{v})=0 $$

But as per previous declaration in 'Find' tab, we have stated -

$$ \underline{w} = \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i}$$ ,

Hence we can now say that,

$$ \underline{w}.\underline{P}(\underline{v})=0 $$

This shows that required equation (2) of page 7- 2 mtg 7 can be used to get equation (2) of page 8-2 mtg 7

= Problem 2.6=

Given
$$F=\left \{ 1,cos(\omega x),cos(2\omega x),sin(\omega x),sin(2\omega x) \right \} , \Omega =[0,T], i=1,2$$

Find
1) Construct Γ5x5(F) and observe the proportionality of Γ.

2) Find det Γ(F)

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

Solution
We have from reference of Mtg 10_Page 2 1) $$ \mathbf{\Gamma} _{5x5}=\int_{\Omega }{}b_{i}\cdot b_{j}\;dx=\int_{0}^{T}b_{i}\cdot b_{j}\;dx $$

Constructing a matrix of bi ˑ bj yields:

$$ \Rightarrow \begin{bmatrix} 1\cdot 1 & cos(\omega x)\cdot 1 & cos(2\omega x)\cdot 1 & sin(\omega x)\cdot 1 & sin(2\omega x)\cdot 1\\ 1\cdot cos(\omega x) & cos(\omega x)\cdot cos(\omega x) & cos(2\omega x)\cdot cos(\omega x) & sin(\omega x)\cdot cos(\omega x) & sin(2\omega x)\cdot cos(\omega x)\\ 1\cdot cos(2\omega x) & cos(\omega x)\cdot cos(2\omega x) & cos(2\omega x)\cdot cos(2\omega x) & sin(\omega x)\cdot cos(2\omega x) & sin(2\omega x)\cdot cos(2\omega x)\\ 1\cdot sin(\omega x) & cos(\omega x)\cdot sin(\omega x) & cos(2\omega x)\cdot sin(\omega x) & sin(\omega x)\cdot sin(\omega x) & sin(2\omega x)\cdot sin(\omega x)\\ 1\cdot sin(2\omega x) & cos(\omega x)\cdot sin(2\omega x) & cos(2\omega x)\cdot sin(2\omega x) & sin(\omega x)\cdot sin(2\omega x) & sin(2\omega x)\cdot sin(2\omega x) \end{bmatrix} $$

Multiplying out the above matrix we integrate the resulting terms within limits from 0 to T to get the $$ \mathbf{\Gamma} _{5x5} $$ :

$$\Rightarrow \begin{bmatrix} x & \frac{sin(\omega x)}{\omega } & \frac{sin(2\omega x)}{2\omega } & -\frac{cos(\omega x)}{\omega } & -\frac{1}{2\omega }cos(2\omega x)\\

\frac{sin(\omega x)}{\omega }) & \frac{1}{2}(x+\frac{sin(2\omega x)}{2\omega }) & \frac{1}{6\omega }(3sin(\omega x)+sin(3\omega x)) & -\frac{1}{2\omega }cos^2(\omega x) & -\frac{2}{3\omega }cos^3(\omega x)\\

\frac{sin(2\omega x)}{2\omega } & \frac{1}{6\omega }(3sin(\omega x)+sin(3\omega x)) & \frac{1}{8\omega }(4x+sin(4\omega x)) & \frac{1}{6\omega }(cos(3\omega x)-3cos(\omega x)) & -\frac{1}{8\omega }cos(4\omega x))\\

-\frac{cos(\omega x)}{\omega }) & -\frac{1}{2\omega }cos^2(\omega x) & \frac{1}{6\omega }(cos(3\omega x)-3cos(\omega x)) & \frac{1}{2\omega}(x-sin(\omega x)cos(\omega x)) & \frac{2}{3\omega }sin^3(\omega x)\\

-\frac{1}{2\omega }cos(2\omega x) & -\frac{2}{3\omega }cos^3(\omega x) & -\frac{1}{8\omega }cos(4\omega x)) & \frac{2}{3\omega}sin^3(\omega x) & \frac{1}{8\omega}(4x-sin(4\omega x)) \end{bmatrix} $$

By definition of Angular Frequency

$$2\pi =\omega T\rightarrow \omega=\frac{2\pi}{T}$$

Above is the resulting matrix to be integrated within limits of 0 to T.

And now integrating the results using a table of integrals yielded the following terms for the final integrated matrix.

$$\begin{array}{l} \int_0^T {1 \cdot 1dx = x} |_0^T = T \\ \int_0^T {1 \cdot \cos (\omega x)dx = \frac{1}{\omega }\sin (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \cos (2\omega x)dx = \frac{1}\sin (2\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (\omega x)dx = - \frac{1}{\omega }\cos (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (2\omega x)dx = - \frac{1}\cos (2\omega x)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \cos (\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}}\\ \int_0^T {\cos (\omega x) \cdot \cos (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \cos (2\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\cos (2\omega x) \cdot \sin (\omega x)dx = \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\sin (\omega x) \cdot \sin (\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\sin (\omega x) \cdot \sin (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\sin (2\omega x) \cdot \sin (2\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \end{array}$$

Above integrals results were verified with Wolframalpha.

The $$ \mathbf{\Gamma} _{5x5} $$ matrix is thus :

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

 * }
 * }

We note that this Gram matrix is a diagonal matrix.

2 ) Determinant of the above gram matrix $$ \mathbf{\Gamma} _{5x5} $$   :

The determinate of the above Г matrix was calculated using MATLAB as shown below:

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

 * }
 * }

We note that the determinant is non- zero and hence we can say that the given functions form a family of Linearly Independent Basis Functions

3) We see that the $$ \mathbf{\Gamma} _{5x5} $$ derived is not equivalent to The Kronecker Delta.

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

 * }
 * }

=Problem 2.7=

Given
$$F=\left \{ 1,x,x^2,x^3,x^4 \right \},

\Omega =[0,1]$$

Find
1) Construct Γ5x5(F) and observe the proportionality of Γ.

2) Find det Γ(F)

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

Solution
1) $$ \mathbf{\Gamma} _{5x5}=\int_{\Omega }{}b_{i}\cdot b_{j}\;dx=\int_{0}^{1}b_{i}\cdot b_{j}\;dx $$

Constructing a matrix of bi ˑ bj yields:

$$ \Rightarrow \begin{bmatrix} 1\cdot 1 & x\cdot 1 & x^2\cdot 1 & x^3\cdot 1 & x^4\cdot 1\\ 1\cdot x & x\cdot x & x^2\cdot x & x^3\cdot x & x^4\cdot x\\ 1\cdot x^2 & x\cdot x^2 & x^2\cdot x^2 & x^3\cdot x^2 & x^4\cdot x^2\\ 1\cdot x^3 & x\cdot x^3 & x^2\cdot x^3 & x^3\cdot x^3 & x^4\cdot x^3\\ 1\cdot x^4 & x\cdot x^4 & x^2\cdot x^4 & x^3\cdot x^4 & x^4\cdot x^4 \end{bmatrix} $$

Multiplying out the above matrix yields:

$$ \Rightarrow \begin{bmatrix} 1 & x & x^2 & x^3 & x^4\\ x & x^2 & x^3 & x^4 & x^5\\ x^2 & x^3 & x^4 & x^5 & x^6\\ x^3 & x^4 & x^5 & x^6 & x^7\\ x^4 & x^5 & x^6 & x^7 & x^8 \end{bmatrix} $$

Integrating the above matrix with respect to x yields:

$$ \Rightarrow \begin{bmatrix} x & \frac{x^2}{2} & \frac{x^2}{2} & \frac{x^3}{3} & \frac{x^4}{4}\\ \frac{x^2}{2} & \frac{x^3}{3} & \frac{x^4}{4} & \frac{x^5}{5} & \frac{x^6}{6}\\ \frac{x^3}{3} & \frac{x^4}{4} & \frac{x^5}{5} & \frac{x^6}{6} & \frac{x^7}{7}\\ \frac{x^4}{4} & \frac{x^5}{5} & \frac{x^6}{6} & \frac{x^7}{7} & \frac{x^8}{8}\\ \frac{x^5}{5} & \frac{x^6}{6} & \frac{x^7}{7} & \frac{x^8}{8} & \frac{x^9}{9} \end{bmatrix}_{0}^{1} $$

Evaluating the integrated matrix with the limits of [0,1] yielded:

$$ \Rightarrow \begin{bmatrix} 1-0 & \frac{1^2}{2}-0 & \frac{1^3}{3}-0 & \frac{1^4}{4}-0 & \frac{1^5}{5}-0\\ \frac{1^2}{2}-0 & \frac{1^3}{3}-0 & \frac{1^4}{4}-0 & \frac{1^5}{5}-0& \frac{1^6}{6}-0\\ \frac{1^3}{3}-0 & \frac{1^4}{4}-0 & \frac{1^5}{5}-0 & \frac{1^6}{6}-0& \frac{1^7}{7}-0\\ \frac{1^4}{4}-0 & \frac{1^5}{5}-0 & \frac{1^6}{6}-0 & \frac{1^7}{7}-0& \frac{1^8}{8}-0\\ \frac{1^5}{5}-0 & \frac{1^6}{6}-0 & \frac{1^7}{7}-0 & \frac{1^8}{8}-0& \frac{1^9}{9}-0 \end{bmatrix} $$

The family of linearly independent basis functions matrix is thus:

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

 * }
 * }

Г is a symmetric matrix about the diagonal.

2) The determinate of the above Г matrix was computed using MATLAB as shown.

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

 * }
 * }

3) An orthogonal matrix is a matrix such that the transpose of the matrix multiplied by the original matrix should yield the identity matrix I, a matrix with only 1's on the diagonal and zeros everywhere else.

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

 * }
 * }

=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.

Find
Show $$\displaystyle (Eq  2.8.1)\Leftrightarrow(Eq   2.8.2)$$.

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

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

Case 1

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

From (Eq 8.3):

Then (Eq 8.1) becomes:

Case 2

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

From (Eq 8.3):

Then (Eq 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 (Eq 8.3):

Then (Eq 8.1) becomes:

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

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

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

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

Hence,

=Problem 2. 9=

Given
General one dimensional Model 1.0/Data set 2 on page 1 of Mtg 12:

where $$ {a_2} = 2,f = 3 $$

Consider $$ {b_j}\left( x \right) = \cos \left( {jx + \phi } \right) $$ and $$ \phi = \frac{1}{4}\pi $$ or $$ \phi  = \frac{3}{4}\pi $$

Find
A) Let n=2, What is the ndof of this problem?

B) Find 2 equations that enforce boundary conditions for $$ {u^h}\left( x \right) = \sum\limits_{j = 0}^n $$

C) Find 1 more equation to solve for $$ {\mathbf{d}} = {\left\{ \right\}_{3 \times 1}} $$ by projecting the residue $$ {\mathbf{P}}\left(  \right) $$ on a basis function $$ {b_k}(x)   with  k = 0,1,2  $$ s.t. the additional equation is linear independence from the above 2 equations

D) Display 3 equations in matrix form $$ {\mathbf{Kd}} = {\mathbf{F}} $$ and observe symmetric property of $$ {\mathbf{K}} $$

E) Solve for $$ {\mathbf{d}} $$

F) Construct $$ u_n^h\left( x \right) $$ and plot $$ u_n^h\left( x \right) \ vs. \ u\left( x \right) $$ ($$ u\left( x \right) $$ is the exact solution)

G) Repeat (1) to (6) for n=4 and n=6

H) Compare $$ u_n^h\left( {x = 0.5} \right)forn = 2,4,6 $$. If error $$ {e_n}\left( {0.5} \right) = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$, plot $$ {e_n}\left( {0.5} \right) \ vs. \ n $$

Solution
A) Because n begins from 0, so the degree of freedom is 3 when n=2.

B) From (Eq 9.2) we can get $$ \sum\limits_{j = 0}^2 {{d_j}{b_j}\left( 1 \right)} = 0 $$

Then rewrite the above equation in matrix form:

where

From (9.3) we can get $$ \sum\limits_{j = 0}^2 {{d_j}b_j^{\left( 1 \right)}\left( 0 \right)} =  - 4 $$

Then rewrite the above equation in matrix form:

where

C) Because the residue $$ P = \frac{d}\left( {{a_2}\frac{d}{b_j}\left( x \right)} \right) - f\left( x \right)$$, after projecting the residue we get:

$$ \sum\limits_{j = 0}^2 {\left\{ {\int_0^1 {{b_k}\left( x \right)\left[ {\frac{d}\left( {{a_2}\frac{d}{b_j}\left( x \right)} \right)} \right]dx} } \right\}} {d_j} = - \int_0^1 {{b_k}\left( x \right)f\left( x \right)dx} $$

Then rewrite the above equation in matrix form:

where

E)-H) In order to simplify the computation procedure, the following Fortran code based on (Eq 9.8), (Eq 9.9) and (Eq 9.10) was used to calculate the results, when n=2,4,6 and $$\phi = \frac{1}{4}\pi,\frac{3}{4}\pi$$:

program main use lib implicit none integer :: n,i,j,k,m,l real,allocatable :: KK,d,F,b real :: fi(2),pi,x(1001),y(1001),yh(1001),e(3,2) n=6; pi=3.1415926; fi(1)=pi/4.0; fi(2)=pi*3.0/4.0 open (unit=11, file="egm6611.s11.team6.hw2.9.1.plt", status="replace") open (unit=12, file="egm6611.s11.team6.hw2.9.2.plt", status="replace") open (unit=13, file="egm6611.s11.team6.hw2.9.3.plt", status="replace") do l=2,n,2 write(*,*) "When the degree of freedom is ", l+1 do i=1,2 allocate(KK(l+1,l+1),d(l+1),F(l+1),b(2,l+1)) do j=0,l b(1,j+1)=cos(j*1.0+fi(i)) b(2,j+1)=-1.0*j*sin(j*0.0+fi(i)) end do      do k=0,l do j=0,l if (k==j) then KK(k+1,j+1)=-1.0*k/2.0*(2.0*k-sin(2*fi(i))+sin(2*fi(i)+2*k)) else KK(k+1,j+1)=2*j*j*j*(cos(fi(i))*sin(fi(i))-cos(fi(i)+k)*sin(fi(i)+j))-k*(cos(fi(i))*sin(fi(i))-cos(fi(i)+j)*sin(fi(i)+k)))/(j*j-k*k)           end if             if (k==0) then               F(k+1)=-3.0*cos(fi(i))            else                F(k+1)=-3.0*(sin(fi(i)+k)-sin(fi(i)))/k            end if                      end do      end do       write(*,*) "when phi= ",fi(i)      do j=1,l+1      KK(1,j)=b(1,j)      KK(2,j)=b(2,j)      end do      F(1)=0     F(2)=-4     call invert(KK)     do k=1,l+1        d(k)=0.0        do j=1,l+1           d(k)=d(k)+KK(K,j)*F(j)        end do     end do     write(*,*) "solution vector d"     write(*,*) d     x=0.0; y=0.0; yh=0.0     do m=1,1001     x(m)=(m-1)/1000.0     y(m)=-3.0/4.0*x(m)*x(m)-4.0*x(m)+19.0/4.0     do k=1,l+1        yh(m)=yh(m)+d(k)*cos((k-1)*x(m)+fi(i))     end do   end do   do k=1,l+1      e(l/2,i)=e(l/2,i)+d(k)*cos((k-1)*0.5+fi(i)) end do  e(l/2,i)=(-3.0/4.0*0.5*0.5-4.0*0.5+19.0/4.0)-e(l/2,i) if (i==1) then write(11,*) "zone" do m=1,1001 write(11,"(3F12.4)") x(m),y(m),yh(m) end do  else write(12,*) "zone" do m=1,1001 write(12,"(3F12.4)") x(m),y(m),yh(m) end do  end if   deallocate(KK,d,F,b) end do !for i=1-2 end do !for n=1-8 write(13,"(A,2F12.4)") "2",e(1,1),e(1,2) write(13,"(A,2F12.4)") "4",e(2,1),e(2,2) write(13,"(A,2F12.4)") "6",e(3,1),e(3,2) end program module lib implicit none contains !  subroutine invert(matrix) ! invert a small square matrix onto itself implicit none real,intent(in out)::matrix integer::i,k,n; real::con ; n= ubound(matrix,1) do k=1,n con=matrix(k,k); matrix(k,k)=1. matrix(k,:)=matrix(k,:)/con do i=1,n if(i/=k) then con=matrix(i,k); matrix(i,k)=0.0 matrix(i,:)=matrix(i,:) - matrix(k,:)*con end if     end do   end do   return end subroutine invert end module lib

Using this code we get the output result for all the situations as following:

With the degree of freedom =           3 & with phi=  0.7853981    OR  pi /4 solution vector d      1.142170       7.209867     -0.7765067 With phi=   2.356194    OR 3 pi / 4 solution vector d     -17.24394       13.96153      -4.152339

With the degree of freedom =          5 With phi=  0.7853981 OR      pi /4 solution vector d     -2.044705       13.62609      -7.081253       2.811028     -0.5599513 when phi=   2.356194      OR 3 pi /4 solution vector d     -16.73928       16.00889      -7.965427       2.459494     -0.4499143

When the degree of freedom is           7 when phi=  0.7853981      OR pi /4 solution vector dc     -2.067435       14.06402      -8.531017       4.714233      -1.883228 c       0.4795609     -5.8918476E-02 when phi=   2.356194     OR 3 pi /4 solution vector d     -19.44370       26.57353      -25.56540       18.85248      -9.806686 3.187393    -0.5081997

For n=2,4,6 and

A ) $$ \phi = \frac{1}{4}\pi $$,

we compare the results with exact solution in the following image:

For n=2,4,6,

B) $$ \phi = \frac{3}{4}\pi $$,

we compare the results with exact solution in the following image:

In both the images, we can observe that the results become more close compared to exact solution when n increases.

For different shift phases, we compare the how the error's convergence situation in the following image:

It seems that both the shift phase can make a convergence around n=6.

=Contributing Team Members=
 * Eml5526.s11.team6.joglekar 19:52, 2 February 2011 (UTC)
 * Eml5526.s11.team6.kurth 00:57, 2 February 2011 (UTC)
 * Eml5526.s11.team6.gravois 00:59, 2 February 2011 (UTC)
 * Eml5526.s11.team6.tupsakhare 01:11, 2 February 2011 (UTC)
 * Eml5526.s11.team6.deshpande 03:37, 2 February 2011 (UTC)
 * Eml5526.s11.team6.vork 12:23, 2 February 2011 (UTC)

title: User:Eml5526.s11.team6/hwk2


text:
User:Eml5526.s11.team6.tupsakhare/hwk1

=Elastic bar PDE, Dynamic case=

Find
Derive the following equation by performing a balance of forces:

$$ \frac{\partial}{\partial x}\left [ A(x)E(x)\frac{\partial u}{\partial x} \right ]+f(x,t)=A(x) \rho(x)\cdot \frac{\partial ^{2}u}{\partial t^{2}}\ $$

Given
An infinitesimally small section of length dx of the 1-D elastic bar with varying material properties at a distance x from the origin

A(x) - Varying cross section of the elastic bar

E(x) - Modulus of elasticity

$$\rho(x) $$ - Mass density

u - Displacement constrained only to x- axis (1-D bar)

t - Time

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

Where, $$\; {A}'=A(x) (dx)\cdot \rho (x)\cdot\frac{\partial^{2}u }{\partial t^{2}} =mass\cdot acceleration=Inertia\, Force$$
 * $$\displaystyle (Eq. 1.1) $$
 * }
 * }

Solution
As per Newton's law of motion, the involved forces should contribute to inertia force.

Force equilibrium in the x-direction yields:

$$\sum F_{x}=0 $$

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

$$\sum F_{x}=-N(x,t)+N(x+dx,t)+f(x,t) dx-m(x)\cdot \frac{\partial ^{2}u}{\partial t^{2}}=0$$
 * $$\displaystyle (Eq. 1.2) $$
 * }
 * }

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

Where, $$\;\; m(x)=A(x)(dx)\rho (x)$$
 * $$\displaystyle (Eq. 1.3) $$
 * }
 * }

Including Taylor series expansion for $$N(x+dx,t)$$ and neglecting the higher order terms (h.o.t) yields:

$$ \cancel {-N(x,t)}+[\cancel {N(x,t)}+\frac {\partial }{\partial x}(N(x,t))+(h.o.t)]+f(x,t)=\frac{m(x)}{dx}\cdot\frac{\partial ^{2} u}{\partial t^{2}}$$

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

$$\frac{\partial }{\partial x}(N(x,t))+f(x,t)=\frac{m(x)}{dx}\cdot\frac{\partial ^{2} u}{\partial t^{2}}$$
 * $$\displaystyle (Eq. 1.4) $$
 * }
 * }

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

$$Also,\; N(x,t)=A(x) \cdot \sigma (x)=A(x) \cdot [E(x)\cdot \epsilon (x,t)] $$
 * $$\displaystyle (Eq. 1.5) $$
 * }
 * }

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

$$\;\;\;\;\;\;\;\;\;\;\;\Rightarrow N(x,t)= A(x)\cdot E(x) \cdot\frac{\partial u}{\partial x}(x,t)$$
 * $$\displaystyle (Eq. 1.6) $$
 * }
 * }

Thus proving:

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

 * $$\displaystyle (Eq. 1.7) $$
 * }
 * }

= Modified Case=

Find
Discuss the case in which the bar has a rectangular cross section shown below.

Solution
We have from reference of class notes for mtg 2, eq(3) pg.2-3,

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

$$m=\rho (x) \cdot A(x) $$
 * $$\displaystyle (Eq. 2.1) $$
 * }
 * }

Also with reference to class notes for mtg 6, pg 6-1, point 2), for A(x)=bˑh(h),

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

$$m=\rho \left ( x+\frac{dx}{2} \right )\cdot \frac{1}{2}\left [ h(x)+h(x+dx) \right ]b $$
 * $$\displaystyle (Eq. 2.2) $$
 * }
 * }

Substituting Eq 2.1 and Eq 2.2 into Eq 1.7 for case of A(x)=bˑh(x),
 * {| style="width:100%" border="0"

$$\frac{\partial}{\partial x}\left [ \left ( b\cdot h(x)\cdot E(x)\cdot \frac{\partial u}{\partial x}\right ) \right ]+f(x,t)=\frac{1}{2}\left [ h(x)+h(x+dx) \right ]\rho (x+dx)\cdot \frac{\partial^{2} u}{\partial t^{2}}\cdot b$$
 * $$\displaystyle (Eq. 2.3) $$
 * }
 * }

Note:

1) $$\left [ \frac{1}{2}\left [ h(x)+h(x+dx) \right ]\rho (x+dx) \right ]$$ denotes mass per unit length at a distance of (x+dx/2), which is the middle of the elementary section considered in Figure 1.2.

2) $$\frac {1}{2}\left [h(x)+h(x+dx) \right ]$$ denotes average height of elementary section of figure 1.2.

3) Since we are considering an infinitesimally small section (dx) of the entire bar, we can write,

$$h(x)\approx h(x+dx)$$

$$\Rightarrow \frac{1}{2}\left [ h(x)+h(x+dx) \right ]\approx h(x)$$

4) Also, b is the width of the cross section and is constant and ρ(x+dx) is approximately equal to ρ(x) for a small section as in 3) above.

Eq 2.3 converts to the following:

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

 * $$\displaystyle (Eq. 2.4) $$
 * }
 * }

Eq 2.4 is the modified form of Eq 1.7 for the specific case of A(x)=bˑh(x).

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

title: User:Eml5526.s11.team6.tupsakhare/hwk1


text:
Wiki culture

Topics

 * Entry point
 * Newcomer welcoming
 * Sustaining participation
 * Core values
 * Is there an umbrella wiki culture?
 * Is wiki culture unique? (Or is it just social psychological processes playing out in a wiki environment?)
 * Sock puppetry
 * Blocking/Banning
 * Is wiki culture North American? Is there room for multiple cultures e.g., in different human cultures/languages?
 * Forking
 * http://wikiindex.org - Index of wikis
 * How tos
 * What are the boundaries between wikis?
 * What is the higher purpose? (Get this clear and come back to it)
 * Why run a small wiki (and struggle)? Why not use a larger wiki with funding etc.?
 * Wiki culture is by default Wikimedia Foundation based because its the largest - how to work in this space?

Jtneill's thoughts
In this conversation, there was discussion about struggling with people issues on wiki - how to get them involved, how to influence their involvement around core values, how to be inclusive but how to deal with disruptive editing.

Wow, the tension rose and got stuck into how to deal with "troublesome" people!

I prefer: The Library of Babel. Fork away.

Much of the tension discussion about managing unwanted editing for me also relates to 5 goods needed per bad (Gottman's 5 to 1 ratio: ) for sustainable relationships.

This discussion is like a Project: namespace talk page! http://www.mediawiki.org/wiki/Help:Namespaces#4:_Project

For me: Wikis + fun (without this :

Wiki is just quick editing of web pages - that's all! (it is light, easy, fast)

Joakims thoughts
I leave my thoughts about Wiki culture here, however, in Swedish. It's based on situative cognition, such as Greenfield & Lave (1982), Rogoff's view on socio-culture (2003):

Undervisning via Wiki sker via informella upplägg. De som deltar i Wikikulturen tar del av och förändrar en eller flera artiklar. I wikikulturen är artikeln det kulturella verktyget. Det kollaborativt utvecklade kulturella verktyget förändrar i sin tur de som deltar. Den kollaborativa framväxten förändrar både deltagare och artikel. Varje artikel på en Wiki bildar en micro-kultur av de personer som deltar.

Wiki = informell undervisning. Där användarna är de lärande och själva ansvariga för att ta till sig kunskap och skicklighet. Ofta sker detta kollaborativt eftersom att flera personer jobbar på en och samma artikel, de diskuterar och förbättrar artiklarna. Förändringar av artikeln kommenteras av och det är en social interaktion till de andra som läser historien bakom artikelns kollaborativa framväxt.

title: Wiki culture


text:
User talk:Ernie.cordell

 Hello Ernie.cordell, 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! --Leutha 00:59, 28 January 2011 (UTC)

Hi
Hi Ernie. The project i am concentrating just now is SPIR608 Political Simulation and Gaming/2011. This is the first time this course has been run. It is based at the University of Westminster and I am working with User:Richard Barbrook to use wikiversity in conjunction with classroom lessons/seminars. It is proving very interesting.Leutha 21:06, 28 January 2011 (UTC)

Some help on the cs department page
Hello! I have been sending out some requests for help - the main department page appears to be abandoned. This could serve as a resource to find courses and generally advertise the field. I have been trying to fix it up but I appear to be alone in my efforts. If you have some free time I would love to have a coworker. I notice you are interested in Linux - there are several operating system related pages that are in dire need of adoption Topic:Operating systems or Topic:Kernel internals to name two off the top of my head. -- Azotochtli (talk) 13:38, 8 August 2012 (UTC)

title: User talk:Ernie.cordell


text: Category:Advanced Level Mathematics

title: Category:Advanced Level Mathematics


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

=--Problem 2.4--=

Problem Statement
To solve for a G1DM1.0/D1 given in (3) in P 9-2

Given

An elastic bar of unit length is applied with a distributed load of $$f(x)=5x$$  along the axis of the bar, and linear variation of product of elastic modulus and area of the bar is given as  $$a_{2}(x)=2+3x$$

Boundary conditions

1) Essential boundary condition $$\gamma _{g}={1}, g=4$$ , which implies  $$ u(1)=4$$

2) Natural boundary condition $$\gamma _{h}={1}, h=6$$ , which implies  $$-\frac{\partial u(x=0)}{\partial x} =6 $$

Hence the above problem reduces to finding exact solution to the differential equation given by,

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

to solve the above differential equation we first solve the following integrals

1) $$\int lnxdx$$ 2) $$\int xlnxdx$$ 3) $$\int \frac{x^2}{1+cx}$$ 4) $$\int \frac{x^2}{a+bx}$$ 5) Find the exact solution of

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

$$ u(1)=4, -\frac{\partial u(x=0)}{\partial x} =6 $$

6) Plot u(x) 7) Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Solution
1) To the solve the above integral we use, Integration by parts {| style="width:100%" border="0" $$ \int{u}dv =uv-\int{v}du $$ {| style="width:100%" border="0" where,
 * style="width:100%" |
 * style="width:100%" |
 * style="width:100%" |
 * style="width:100%" |

$$\int lnxdx=xlnx-\int \frac{x}{x}dx$$

$$=xlnx-\int 1dx$$

2)

we again use Integration by parts

here,

$$\int xlnxdx=\frac{x^2}{2}lnx-\int \frac{x^2}{2x}dx$$

$$=\frac{x^2}{2}lnx-\int \frac{x}{2}dx$$

$$=\frac{x^2}{2}lnx- \frac{x^2}{4}$$

3) here,

hence,

$$\int \frac{x^2}{1+cx}= \frac{x^2}{c}ln(1+cx)-\int \frac{2}{c}ln(1+cx)xdx$$

for, $$\int ln(1+cx)xdx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2}{c}[[\frac{x}{c}(1+cx)ln(1+cx)-\frac{x}{c}(1+cx)]-\int \frac{((1+cx)ln(1+cx)-(1+cx))}{c}dx]$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}\int (1+cx)ln(1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

for, $$\int (1+cx)ln(1+cx)dx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}[\frac{(1+cx)^2ln(1+cx)}{2c}-\int \frac{(1+cx)^2cdx}{2c(1+cx)}]-\frac{2}{c^2}\int (1+cx)dx$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}ln(1+cx)-\frac{1}{c^2}\int (1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}-\frac{2x^2c}{c^2}+\frac{1}{c^3}+\frac{c^2x^2}{c^3}+\frac{2cx}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}[\frac{(1+cx)^2}{2c}]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}[1+c^2x^2+2cx]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}-\frac{3x^2}{2c}-\frac{3x}{c^2}$$

$$=\frac{1}{c^3}ln(1+cx)+\frac{x^2}{2c}-\frac{x}{c^2}-\frac{3}{2c^3}$$

$$=\frac{2ln(1+cx)+x^2c^2-2cx}{2c^3}+constant$$

4)

here,

hence,

$$\int \frac{x^2}{a+bx}= \frac{x^2}{b}ln(a+bx)-\int \frac{2}{b}ln(a+bx)xdx$$

for, $$\int ln(a+bx)xdx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2}{b}[[\frac{x}{b}(a+bx)ln(a+bx)-\frac{x}{b}(a+bx)]-\int \frac{((a+bx)ln(a+bx)-(a+bx))}{b}dx]$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}\int (a+bx)ln(a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

for, $$\int (a+bx)ln(a+bx)dx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}[\frac{(a+bx)^2ln(a+bx)}{2b}-\int \frac{(a+bx)^2bdx}{2b(a+bx)}]-\frac{2}{b^2}\int (a+bx)dx$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}ln(a+bx)-\frac{1}{b^2}\int (a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2ax}{b}-\frac{2x^2b}{b^2}+\frac{a^2}{b^3}+\frac{b^2x^2}{b^3}+\frac{2abx}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}[\frac{(a+bx)^2}{2b}]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3}{2b^3}[a^2+b^2x^2+2abx]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3a^2}{2b^3}-\frac{3x^2}{2b}-\frac{3ax}{b^2}$$

$$=\frac{a^2}{b^3}ln(a+bx)+\frac{x^2}{2b}-\frac{ax}{b^2}-\frac{3a^2}{2b^3}$$

$$=\frac{2a^2ln(a+bx)+x^2b^2-2bax}{2b^3}+constant$$

5)

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}=-5x $$

Integrating the above expression we get,

$$ (2+3x)\frac{\partial u}{\partial x}=\frac{-5x^2}{2}+k1$$

$$\frac{\partial u}{\partial x}=\frac{-5x^2}{2(2+3x)}+\frac{k1}{2+3x}$$

given,

$$-\frac{\partial u(x=0)}{\partial x} =6$$

$$-6= 0+ \frac{k1}{2}$$

hence $$k1=-12$$

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

$$\partial u=\frac{-5x^2}{2(2+3x)}\partial x+\frac{k1}{2+3x}\partial x$$

Integrating the above equation

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

from (Eq.2) $$\int \frac{x^2}{a+bx}dx=\frac{2a^2ln(a+bx)+bx(bx-2a)}{2b^3}$$

hence $$\int \frac{x^2}{2+3x}dx=\frac{8ln(2+3x)+3x(3x-4)}{54}$$

$$u=\frac{-5}{2}[\frac{8ln(2+3x)+3x(3x-4)}{54}]-12[\frac{ln(2+3x)}{3}]+k2$$

$$u=\frac{-118}{27}ln(2+3x)-\frac{5x}{36}(3x-4)+k2$$

given $$ u(1)=4$$

$$4=\frac{-118}{27}ln(5)-\frac{5}{36}(-1)+k2$$

$$k2=10.89495$$

hence the exact solution is

6) Plot of u(x)

[[File:Fe1.s11.team3.HW2.fig2.jpg|frame|center|Fe1.s11.team3.HW2.fig2]

7)

The above integral is a special case if (Eq.2) where  $$a=b=1$$

hence,

$$\int \frac{x^2}{1+x}= \frac{2*1^2ln(1+x)+x(x-2)}{2*1^3}+constant$$

Problem Solved by akj.

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


text:
Wikis for education

Wiki for education
Jtneill asked to hear about how people use wikis with students.

Diane Phillips used wikis with students in marketing/social media:
 * 2 out of 90 students objected to going public with their content (for these two they did a private journal marked by their tutor)
 * 8 companies contacted about internships
 * Students were very self-managing
 * 3rd year UG students in Bachelor of Commerce

Tom Worthington:
 * Taught two courses: Writing for the Web and Electronic Records
 * Had them edit Wikipedia articles because it was easier to edit existing material than create their own text due to wiki markup etc.
 * Public service employees
 * Edited the pages about their organisation
 * What about Wikiversity for educators?

Rob Fitzgerald
 * What tools do we have for examining the process (e.g., qualitative and quantitative contributions) e.g. for assessment? We currently don't have particularly useful tools.
 * Face to face editing class (despite edit conflicts) got things flowing

Ben Rattray
 * This is not dissimilar to group assignments - we now have a collaboration tool, that's all.
 * Groups of students creating resources on different topics.
 * Make sure in tutorial environment that they understand recent changes, history etc. so they understand what's going on.

James Neill
 * Dedicated lecture early in semester on "how to use wiki" - emphasis on "leap in", you can't break it, have a go, then you'll learn and work it out - hit the edit button.

Kirsty - Tasmanian Polytechnic
 * Building student portfolios
 * Turn around time much quicker for remote students (compared to printed workbooks - send, mark, send back etc.)
 * A surprise: Creation of a single community between remote and on-campus students
 * Used wikispaces with subpages and a templating system - wikispaces is almost WYSIWIG but not quite. Some discussion about Mediawiki not being WYSIWIG.

Guy with a beard - Uni of SA
 * Problem with adding teaching content on Wikiversity because employer wants to own content.

Oops battery ran out - went to handwritten notes.

title: Wikis for education


text:
User:Eml5526.s11.team2.stewart/HW2

=Problem 2.1: "Derive the First Order Differential Heat Equation for a 1-D Finite Element"=

Problem Statement:
To derive the first order differential equation for heat was introduced in Meeting 6 Page 2 Using a free-body diagram apply the appropriate heat balance and verify that the following differential equation holds for the case of a 1-D finite element with heat applied:

The following figure is the free body diagram of the finite element heat problem:

Solution:
The first order differential equation can derived by summing the heat flux, heat flow, and heat outflow:

where

Summing all heat leaving and entering the finite element:

Expanding and dividing through by dx

From initial and boundary conditions, the final solution is equal to equation 1.1:

CHANGE EQUATION NUMBERS, NEATEN UP

=Problem 2.5: "Proof of Equivalent Equations"=

Problem Statement:
Complete part B of proof of equation

Such that:

Proof:
Equation 2 from Lecture 7-2 states that:

Choice 1: $$\displaystyle \alpha_1=1$$
Therefore:

Choice 2: $$\displaystyle \alpha_2=1$$
Therefore:

Choice n: $$\displaystyle \alpha_n=1$$
Therefore:

Identity
Multiplying equation 5.9 through by $$\alpha_n$$ :

Repeating for $$\alpha_1$$ and $$\alpha_2$$, etc and adding their respective equations:

Since equation 5.2 states that $$\underline w$$ is the sum of the product of $$\alpha _i$$ and $$\underline b _i$$, the following is proven:

title: User:Eml5526.s11.team2.stewart/HW2


text:
TESOL/Probability

"What is a 30% chance in English?"

English provides many ways to express different levels of probability, or chance. This is only about the objective chances of something happening, not the speaker's expectation, bias, or belief about it happening. For more information on that, see Belief in English.

Note that likely is usually an adjective even though it ends in -ly.

Quiz
{It will - rain. (80%) + probably - probable - bound to - certainly not
 * type=""}

{That person - be the president. (0%) - might - can - certainty + can't
 * type=""}

{It's - that the trains will stop tomorrow. (30%) - impossible - needn't + unlikely - off chance
 * type=""}

{Winning would be a - with this low score. (10%) + long shot - certainty - sure thing - no way
 * type=""}

{There's - you can get me to come to your birthday party. (0%) - almost certain - might be - possible + no way
 * type=""}

title: TESOL/Probability


text:
TESOL/Belief

"How do you say you believe something 80% in English?"

English has several ways to express our subjective belief in an event. They include I think that or I doubt that. (For more information about expressing the objective chance of an event, such as It's likely that, see Probability in English.)

Under the impression
We use under the impression if someone believes something, but in reality, they were wrong.
 * I was under the impression that it would rain today. (It was sunny.)
 * She was under the impression that he was coming at 5 PM. (He came at 6 PM.)

Suspect
We use suspect when we believe something about 80%, but we do not have much or any evidence. Often it is used to talk about negative things.
 * I suspect you are lying. (I think so but I can't prove it.)
 * I suspect she is coming. (I think so but she has not called me yet.)
 * So far my boss doesn't suspect anything. (My boss doesn't think I did anything, and he or she doesn't have any evidence to believe so.)

Disagreeing
We can use words indicating belief under 50% to disagree.

Doubt
We use doubt to disagree.
 * Will they finish on time? - I doubt it. (They will be early or late.)
 * Bill Clinton was the best president ever. - I doubt it. (I disagree.)

I don't know, I'm not sure
We can use I don't know (about it/that), I'm not sure (about it/that), or similar expressions to politely disagree. The real meaning is "I know and I don't think so," not "I don't know."

 * When you eat at a restaurant in Japan, you should pay a tip. - I don't know about that. I think tips are considered insulting in Japan.
 * I'm going to go swimming after eating a big meal. - I'm not sure that's a good idea. You might get sick.
 * Do you think you will get the job? - I don't know about it. It was a week ago and they haven't called me back.

Ellipsis
Often "I doubt it" undergoes ellipsis and becomes "doubt it" which is a short response that means we disagree.
 * Are you going back home this year? - Doubt it. (I don't think so.)

The same thing happens with "I don't know" and "I'm not sure."
 * Has the meeting finished? - Dunno. Nobody is back yet.
 * Is it going to rain? - Not sure. There's not a cloud in the sky.

In the extreme, "I don't know" becomes "uh-uh-uh" where each "uh" only retains the intonation of its origin in the phrase "I don't know."

Mechanics
Usually we express belief in English using verbs. If we are talking about probability, we usually do not use verbs.

Quiz
{ 30% I { doubt } that she is coming.
 * type="{}"}

{ 100% I am { sure|convinced|certain } that you are correct.
 * type="{}"}

{ 80% I { guess|think|suppose|expect|suspect } it will rain tomorrow.
 * type="{}"}

title: TESOL/Belief


text:
User:Paulgjanssen/Churlatan

churlatan. n.  An invented word, compounded of “churl” and “charlatan,” denoting an individual who justifies his or her opinion on a contested and controversial subject by crediting only the evidence that supports his side of the argument, and masks the dishonesty of his position by aggressively attacking proponents of the opposition. Example: “The churlatan railed against anyone who dared to suggest that severe variations in weather patterns are consistent with the models of global warming.”  Coined by Paul Janssen, Park Ridge, New Jersey, 2011.

title: User:Paulgjanssen/Churlatan


text:
User talk:Paulgjanssen/Churlatan

Alternate example
The churlatan railed against anyone who dared to suggest that 'excess heat' in Cold Fusion experiments were an error in measurement or a mundane effect explained by principles of classical physics that have been well known since the days of Antoine Lavoisier and Michael Faraday.

Moulton 13:50, 28 January 2011 (UTC)

title: User talk:Paulgjanssen/Churlatan


text:
User:Ali11~enwikiversity

Welcome to my Wikiversity page!
RPA

EG

title: User:Ali11~enwikiversity


text:
User talk:Ali11~enwikiversity

 Hello Ali11, 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) 15:12, 28 January 2011 (UTC)

Hi
Hi Ali, I hope you are enjoying the course. If you have any problems with wikiversity you can leave message for me here. Leutha 12:49, 9 February 2011 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Ali11. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Ali11~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:27, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:02, 19 April 2015 (UTC)

title: User talk:Ali11~enwikiversity


text:
User talk:Paulgjanssen

 Hello Paulgjanssen, 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) 15:12, 28 January 2011 (UTC)

title: User talk:Paulgjanssen


text:
Wikiversity talk:Community Review/Abd

A Man of Constant Comment
Title: I Am a Man of Constant Comment Artist: Abd Lomax and The Soggy Tea Bag Boyz Composer: Soggy Tea Bag Boyz and Barsoom Tork Associates YouTube: I Am a Man of Constant Sorrow — Soggy Bottom Boys (Oh Brother, Where Art Thou)  I am a man of Constant Comment I've seen trouble all my day. I bid farewell to Wikipedia The place where I was scorned and chased.  The place where he was scorned and chased.  For six long years I've been in trouble No treasures here my search hath found For in these threads I'm bound to ramble I have no friends to slap me down.
 He has no friends to slap him down.

 It's fare thee well my trollish bugger
 I never expect to see you win
 For I'm bound to ride that wretched railroad
 Perhaps I'll wreck upon this train.

 Perhaps he'll wreck upon this train.

 You can bury me in wordy blizzards
 For many years where I may lie
 Then you may learn to nudge a brother
 While I am ranting in my rave.

 While he is ranting in his rave.

 Maybe your friends think I'm just a wanker
 My trace you'll never see no more.
 But there is one promise that is given
 I'll meet you on Cold Fusion's lore.

 He'll meet you on Cold Fusion's lore.

 CopyClef 2011 The Soggy Tea Bag Boyz and Barsoom Tork Associates.
 Resurrection Hackware. All songs abused.

 * He'll meet you, sure, on Cold fusion. Applause! Moar atrocious song parodies! Moar! --Abd 19:46, 28 January 2011 (UTC)

Junk Science
Title: Junk Science Artist: Moulton

Composer: Lady Gaga and Barsoom Tork Associates

YouTube: Bad Romance — Lady Gaga

YouTube: Bad Project — Lady Science

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Caught in a bad defense

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Caught in a bad defense

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! Check your junk science

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! Chuck your junk science

I want your fusion I want your disease I want your excess heat As long as it’s free I want your heat Heat-heat-heat I want your heat

I want your drama Mere words are too bland I want your irritated fist pounding sand And I want your heat Heat-heat-heat I want your heat Excess heat I want your heat

You know that I troll you And you know that I feed you I want some junk, your junk science

I want your anger And I want your revenge You and me perform a tragic dance I want your anger and All your words spell revenge You and me perform a tragic dance

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Caught in a bad defense

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Check out your junk science

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! I want your junk science

I want your horror I want your design ‘Cause you’re subliminal As long as you whine I want your heat Heat-heat-heat I want your heat

I want your scope trace Your vertical blips Get thee behind me 'Cause Abd you're sick I want your heat Heat-heat-heat I want your heat Excess heat I want your heat

You know that I troll you And you know that I feed you I want some junk, some junk science

I want your anger And I want your revenge You and me perform a tragic dance I want your anger and All your words spell revenge You and me perform a tragic dance

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Check out your junk science

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Chuck out your junk science

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! I want your junk science

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! I want your junk science

Talk talk fusion, maybe Juice it Move that itch crazy Talk talk fusion, maybe Juice it Move that itch crazy Talk talk fusion, maybe Juice it Move that itch crazy Talk talk fusion, maybe Juice it Frog legs twitch, maybe

I want your anger And I want your revenge I want your heat Cuz we'll never be friends

Ich will dein Zorn Und ich will deine Rache Ich will dein Zorn And I feel your heartbreak

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Check out your junk science

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Chuck out your junk science

I want your anger And I want your revenge You and me perform a tragic dance I want your heat and All your words spell revenge You and me perform a tragic dance

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Check out your junk science

Oh-oh-oh-oh-oooh! Oh-oh-oooh-oh-oh! Chuck out your junk science

Dah-dah-ah-ah-ah! Blah-Blah-Dah-ah! Ga-ga-ooh-la-la! Want your junk science

CopyClef 2011 Lady Gaga and Barsoom Tork Associates. Resurrection Hackware. All songs abused.
 * The singer is clearly insane but hot. I'm convinced. Ottava Rima (talk) 22:52, 28 January 2011 (UTC)

There's a Kind of Spritz
Title: There's a Kind of Spritz
 Artist: Abd, Moulton, and Dieter Britz
 Composer: Les Reed, Geoff Stephens, and Barsoom Tork Associates
 YouTube: There's a Kind of Hush - Herman's Hermits (Rare Clip-Synced Version)

 There's a kind of spritz all over the fusion cell.
 All over the cell you can hear the noise of bubbles in fizz.
 You computed the mean.
 Just the buzzing hiss that nobody else could hear.
 There's nobody else and Kirk's the one who's breaking your spell.

 So listen very carefully.
 Closer now and you will measure the mean.
 It's all been a dream.

 La-la-la-la-la-la-la
 La-la-la-la-la-la-la-la
 La-la-la-la-la-la-la-la-la-la
 La-la-la-la-la

 So listen very carefully.
 Closer now and you will measure the mean.
 It's all been a dream.
 The crackling sound that you will hear
 Is popping bubbles in your ear
 Fizzing there forever and ever.

 There's a kind of spritz all over the fusion cell.
 All over the cell you can hear the noise of bubbles in fizz.
 Yeah, they're popping in hiss.
 Spritz! "They're crackling," says Britz.
 Spritz.

 CopyClef 2011 Les Reed, Geoff Stephens, and Barsoom Tork Associates.
 Resurrection Hackware. All songs abused.

How Long Must This Keep Going On?
Title: How Long Must This Keep Going On? Artist: Moulton Composer: George Gershwin, Ira Gershwin, and Barsoom Tork Associates YouTube: How Long Has This Been Going On? — Jon Bon Jovi

As a lad, when I read about The Blue Carbuncle, I was dissed by my brother, my cousin, and my uncle. Sad to say, life was gray, so dull one would fall into a funkle.

So dear Abd, I swore, "Never, never more!" On my list, I insisted that dissing must be silly. Now, I find, I was too kind, so began punning willy nilly!

I could try pounding sand; Throwing tantrums, standing grand. Little use, too much abuse: How long must this keep going on?

There were hopes for a time, And some tropes I can't define. I blog and tweet, again repeat: How long must this keep going on?

I reckon that Pd could melt; Into excess heat I'm hurled! I know how Joseph Priestley felt, Alone in his Phlogiston world.

Ask me once, then once more. What a dunce I was before. Let's take a break! For Freedom's sake! How long must this keep going on?

When into your notes I peek, That unending rendezvous, Don't bug me if I'm a geek, Let me find out if it's true!

Ask me twice, then once more. That makes thrice, let's make it four! Let's take a break! For Freedom's sake! How long must this keep going on?

How long must this ... keep going ... on?

CopyClef 2010 George Gershwin, Ira Gershwin, and Barsoom Tork Associates Resurrection Hackware. All songs abused.

Steampunk Fusion Dream
Title: Steampunk Fusion Dream

Artist: Moulton

Composer: John Lennon, Paul McCartney, and Barsoom Tork Associates

Midi: Yellow Submarine

In the town of Santa Fe Lives a man who wrote to me And he told me of his strife In the land of science dreams

So we rallied up some puns Till we found the steampunk theme And we dived beneath the raves In our steampunk fusion dream

We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream

And our friends are in a cult Many more of them need an adult And the band begins to play

We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream

As we live a life of tease Everyone of us has too much greed Bubbling gas and steampunk theme In our steampunk fusion dream.

We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream

We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream We all believe in our steampunk fusion dream, Steampunk fusion dream, steampunk fusion dream

CopyClef 2011 John Lennon, Paul McCartney, and Barsoom Tork Associates Resurrection Hackware. All songs abused.

I am honored
I am honored that this Community Review Drama Queen Contest is now accompanied by music and lyrics written by the Official Wikiversity Master of Atrocious Song Parodies, recently freed from Global Block Jail, through the dedicated efforts of the Minions of All that is True and Good, or at least Acceptable by Local Consensus.

His opinions about Cold fusion suck, but after all, nobody's perfect. Once again, Master Moulton, welcome back, and may you live long and in good health!

I propose a Toast. Unfortunately, my Cold Fusion Toaster is on the fritz, lack of spritz, and there just isn't enough noise in this Kepco Constant Current Power supply to do the trick, and my Constant Comments don't quite get up to temperature, so ... we'll have to do with bagels and scream cheese. JWS, please provide the screams. Your cue! --Abd 20:17, 29 January 2011 (UTC)

title: Wikiversity talk:Community Review/Abd


text:
User:Newseerr

My list page create

 * 1) User:Newseerr/Aaqib biggest news letter

title: User:Newseerr


text:
User:Dkabbhodofbsd

Hello people,

I`m existing in: User:Newseerr/Aaqib biggest news letter

title: User:Dkabbhodofbsd


text:
University of Florida/Eml5526/s11.team2.reiss.HW

Given
Consider the family of functions on the interval [0,T], where T=$$2\Pi/\omega$$

Find
A) Construct $$\Gamma (\Im )$$ and observe its properties B) Find $$\det [\Gamma (\Im )]$$ C) Is $$\Im$$ an orthogonal basis

Solution
Construct $$\Gamma (\Im )$$:

where

In order to construct the matrix we must first define $$ < {b_i}, {b_j} > $$

Because multiplication of continuous functions is communicative it can be shown from equation 4.3 that

And therefore $$\Gamma (\Im )$$ is a symmetric matrix

We must now evaluate the terms of the matrix All values were checked with Wolframalpha The Gram matrix then becomes

As we can see the Gram matrix based constructed from this set of functions is a diagonal matrix

Finding $$\det [\Gamma (\Im )]$$

The determinant of a diagonal matrix is Where Based on equation 4.6

For the set to be an orthogonal basis the Gram matrix must be a diagonal matrix with a non-zero determinant. As we can see from equations 4.5 and 4.7 both of these criteria are satisfied. Thus the set of functions is an orthogonal basis.


text:
User:Ritnor

You have bin progressed with existing of a page: Aaqib page

title: User:Ritnor


text:
User:Bac~enwikiversity

I`m a gummibar fan:

And your existing in User:Newseerr/Aaqib biggest news letter

title: User:Bac~enwikiversity


text:
User:J.amdjfkakaasdd

I`m existing with: User:Newseerr/Aaqib biggest news letter

title: User:J.amdjfkakaasdd


text:
User talk:Atcovi/List of presidents of Maybeury Elementary School and soccer

I have bin attract to be in this page. J.amdjfkakaasdd

watched
Watched for you! Now nobody can delete this!!!. From Ritnor

LAA! Now having table
This is great having the complex

??

What? This can`t be true If your in the test of the complex the returning of the test

From User:Newseerr

Direct educational value
Your page Newseerr is a educational value that can make people

Learn new trust about new stories on rebuild contacts on really good part?

So the real truth is really good.

So you and me are best friends. From Newseerr-- Recreate Draubb 17:42, 28 January 2011 (UTC)

Welcome to wikiversity, folks. Please be careful, and I'll be careful too.
You may create fun pages in user space, like this page in the space of Newseerr, and work on them and learn about how to use the wiki software. I am concerned about possible violation of Privacy policy, about using the real names of minors without parental permission, but if you don't identify personal names of children, with particular Wikiversity users, yourself or anyone else, it may be okay. (Other Wikiversity administrators may have different ideas about this.) Please be careful about this, and I will try to avoid using deletion and blocking. Okay? Fair trade? Thanks. --Abd 18:14, 28 January 2011 (UTC)

Burck!
I have the real temps of soccer for the list of president site for more educational values

Because I don`t get it. But the most important thing is!, Try your best -- From Draubb

I`ll careful
I`ll be careful about myself because I can change from the blocks.! So then I`m going to b-be good. From Denial

 * If you register new accounts because you are blocked, your internet access to Wikiversity can be blocked. Please don't do that. Instead, if you are blocked and you think it isn't right, ask to be unblocked on your user page. If you continue to create accounts, it will be out of my hands, and the internet access, for editing or writing here, from Maybeury Elementary School might be blocked. Please talk about what you want to do when other users ask about it. I hope to be able to help you, I've been trying to, but if continual moves and creations of pages in mainspace (pages without the User:UserName/ beginning), it's a problem for the rest of us. Thanks. --Abd 20:00, 3 February 2011 (UTC)

I`ll careful
I know! I have bin changing the temps for a good temp because I`m am dot. From Mabuabsdd

Rewrite
The talk has bin not working so far! I`ll fix the talk, Ritnor--Ritnor 14:47, 2 February 2011 (UTC)

New Zedong
You can see the Zedong or your republic on Wikipedia as the List Sri Lanka.

Transform in www.Wikipedia.com

www.Wikisource.com. --Draubb


text:
User:Atcovi

I'm a huge psychology lover, so my extensive projects apart from school work focus on topics in School:Medicine, School:Biology and School:Psychology.

My activity is high at the moment but may fluctuate due to life circumstances. Reach out to my talk page for any inquiries.

Links
I've left an arrangement of random links for me to easily access if I so desire at any given time.

 * 1) Help:Project boxes - For projects/pages.
 * 2) Help:Quiz - This is also important.
 * 3) Special:CentralAuth/Atcovi
 * Category:Atcovi's Work, User:Atcovi/Science & User:Atcovi/History
 * 1) https://tools.wmflabs.org/meta/crossactivity/Atcovi
 * 2) https://tools.wmflabs.org/topviews/?project=en.wikiversity.org&platform=all-access&date=yesterday&excludes=
 * 3) Recent changes to medical-related articles on en.wiki
 * 4) Psychology stubs and Health stubs
 * 5) User:Atcovi/Essays
 * Category:Featured resources
 * Category:Featured resources

User:Atcovi/To merge: Pages needing to be merged Pages under "User:Atcovi"

title: User:Atcovi


text:
User:STEMskills

My interest is in making it developing programs to make it easy for Students, Teachers and Educators to optimize STEM teaching and learning via the Cloud.

title: User:STEMskills


text:
User talk:STEMskills

 Hello STEMskills, 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) 18:15, 28 January 2011 (UTC)

title: User talk:STEMskills