text:
User:Eml5526.s11.team7.jin/HW1

=Problem 1.1=

Problem Description

 * Part 1: Derive the one-dimensional partial differential equation (PDE) for an elastic bar under dynamic loading conditions. (Refer to equation 3 of lecture page 2-3.)

 * Part 2: Discuss the case in which the bar has a rectangular cross-section.

Part 1

From lecture page 2-3 equation 3 ,the PDE for an elastic bar under dynamic loading conditions is:
 * {| style="width:100%" border="0"

$$ \displaystyle
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\frac{\partial }\left[ {E\left( x \right)A\left( x \right)\frac} \right] + f\left( {x,t} \right) = m\left( x \right)\frac

$$     (Eq 1.1) Since the elastic bar is loaded dynamically, the applied force, f, is a function of distance, x , as well as time, t. This is shown in the free body diagram to the right. Details of the remaining forces will be discussed in the solution below.
 *
 * }

Part 2
The cross section is shown in the following figure

Part 1
Assumptions :
 * The rod is in stress equilibrium.
 * The rod satisfies Hooke's Law, $$\sigma (x) = E(x)\epsilon (x) $$.
 * Compatible displacement field.
 * The rod obeys the strain displacement equation.

Following Newton's Second Law:

 * $$\displaystyle F=m*a $$

Where:

 * F=Force

 * m=Mass

 * a=Acceleration

Additionally,

 * $$\displaystyle m=\rho * V$$

Where $$\displaystyle\rho = $$ density and V=Volume.

and,

 * $$a=\frac{d^2u}{dt^2}$$

Where u is displacement and t is time.

Performing a force balance on the body in the x-axis direction
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$$ \displaystyle
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N\left( {x + \Delta x} \right) + N\left( x \right) + f\left( {x + \frac{2},t} \right)\Delta x = \rho \left( x \right)A\left( x \right)\Delta x\frac

$$     (Eq 1.2) Where
 *
 * }
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$$ \displaystyle
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N\left( x \right) = \sigma \left( x \right)A\left( x \right)n\left( x \right) = - \sigma \left( x \right)A\left( x \right)

$$     (Eq 1.3) And
 *
 * }
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$$ \displaystyle
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N\left( {x + \Delta x} \right) = \sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right)n\left( {x + \Delta x} \right) = \sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right)

$$     (Eq 1.4) Plugging in Equation 1.2
 *
 * }
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$$ \displaystyle
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\sigma \left( {x + \Delta x} \right)A\left( {x + \Delta x} \right) - \sigma \left( x \right)A\left( x \right) + f\left( {x + \frac{2},t} \right)\Delta x = \rho \left( x \right)A\left( x \right)\Delta x\frac

$$     (Eq 1.5) Dividing by $$\Delta x$$ on each side
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
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\frac + f\left( {x + \frac{2},t} \right) = \rho \left( x \right)A\left( x \right)\frac

$$     (Eq 1.6) If we take the limit as $$\Delta x \to 0$$
 *
 * }
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$$ \displaystyle
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\frac{\partial }\left[ {\sigma \left( x \right)A\left( x \right)} \right] + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac

$$     (Eq 1.7) The strain-displacement equation is
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (Eq 1.8) By Hooke ‘ s Law
 *
 * }
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (Eq 1.9) Substituting Equation 1.9 into Equation 1.7
 *
 * }
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$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

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

$$     (Eq 1.10) That 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" |

\frac{\partial }\left[ {E\left( x \right)A\left( x \right)\frac} \right] + f\left( {x,t} \right) = m\left( x \right)\frac

$$     (Eq 1.11) Where
 *
 * }
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$$ \displaystyle
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m\left( x \right) = \rho \left( x \right)A\left( x \right)

$$     (Eq 1.12)
 *
 * }

Part 2
The cross section area
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$$ \displaystyle
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A\left( x \right) = b \cdot h\left( x \right)

$$     (Eq 1.13) The mass
 *
 * }
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$$ \displaystyle
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m\left( {x + \frac{2}} \right) = \rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + \Delta x} \right)} \right]b

$$     (Eq 1.14) Take the limit $$\Delta x \to 0$$
 *
 * }
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$$ \displaystyle
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m\left( x \right) = \rho \left( x \right)h\left( x \right)b

$$     (Eq 1.15) Substituting Equation 1.13 and 1.14 into Equation 1.10
 *
 * }
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$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\frac{\partial }\left[ {E\left( x \right)h\left( x \right)b\frac} \right] + f\left( {x,t} \right) = \rho \left( x \right)h\left( x \right)b\frac

$$     (Eq 1.16)
 *
 * }

Author
Braden Snook, Jiang Jin, Bullard Johnathan, Philip Flater, Brandon Hua, Zongyi Yang, Srilalithkumar Swaminathan

title: User:Eml5526.s11.team7.jin/HW1


text:
User:Egm6341.s11.team3/sub1

= HW 1.1 =

Problem
Find the limit of function:

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$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x}$$ $$
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 * style="width:95%" |
 *  $$ \displaystyle
 * }

Find

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$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.1)
 * }

Solution
Take the limit of the function:

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$$ \lim_{x\to 0} f(x) = \frac {e^0 - 1}{0} = \frac {0}{0}$$ $$
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 * style="width:95%" |
 * $$\displaystyle
 * }
 * }

Hence we can apply L-Hopital Rule:

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$$ \lim_{x\to 0} f(n) = \frac {f'(n)}{g'(n)} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

By above rule:

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$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x} =\lim_{x\to 0} \frac {f(x)}{g(x)} \Rightarrow
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\begin{matrix} f'(x) = e^x -0 \\ g'(x) = 1 \end{matrix} $$

$$
 * $$\displaystyle
 * }
 * }

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$$ \Rightarrow \lim_{x\to 0} f(0) = \frac {f'(0)}{g'(0)} = \frac {e^0 - 0}{1} = 1  $$ $$
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 * $$\displaystyle
 * }
 * }

$$ Ans.: \lim_{x\to 0} f(x) = \frac {e^x - 1}{x} = 1 $$ $$
 *  $$ \displaystyle (Eq. 1.3)

Extra Solution
From Taylor series :

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + ... $$

Then the original fuction could be expressed as:

$$ f(x) = 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \frac{x^4}{120} + ... $$

$$ \Rightarrow \lim_{x\to 0} f(x) = 1 + \frac{0}{2} + \frac{0^2}{6} + \frac{0^3}{24} + \frac{0^4}{120} + ... =1 $$

The original function is plotted from 0 to 1. For comparison, serie-form of f(x) of order 2nd and 3rd are also shown in the following plot.

= HW 1.2 =

Problem
Given:

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$$ a) \qquad f(x)= e^x $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.1)
 * }

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$$ b) \qquad f(x)= \frac {e^x - 1}{x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2)
 * }

Find

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$$ P_n(x), R_{n+1}(x) \quad at \quad x_0 = 0 $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Solution

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$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + .... + \frac{({x-{x}_{0}})^{n}}{n!}$$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3)
 * }

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$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad \xi \in [x_0, x] $$ $$
 * style="width:95%" |
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 *
 $$ \displaystyle (Eq. 2.4)
 * }

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$$ f^{n+1}(x) = \frac{d^{n+1}}{dx^{n+1}}f(x) $$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

$$ a) \mathbf{\quad f(x) = e^x} $$

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$$
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\begin{matrix} f(x) = e^x & f(0) = e^0 = 1 \\ f'(x) = e^x & f'(0) = e^0 = 1 \\ \vdots \\ f^n(x) = e^x & f(0) = e^0 = 1 \end{matrix} $$ $$
 *
 $$ \displaystyle
 * }

From Eq. 2.3:
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$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + \frac{(x-{x}_{0})^2}{2!}f''({x}_{0})+.... + \frac{({x-{x}_{0}})^{n}}{1!}$$ $$
 * style="width:95%" |
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 *
 $$ \displaystyle
 * }

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$$ = \frac{x^0}{0!}(1) + \frac{x^1}{1!}(1) + \frac{x^2}{2!}(1) +....+ \frac{x^n}{n!}(1) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = \sum_{n=0}^n (\frac{x^n}{n!}) \Rightarrow P_n(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5)
 * }

From Eq. 2.4:
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$${R}_{n+1}(x)= \frac{1}{n!}\int_{{x}_{0}}^{x}(x-t)^n f^{n+1}(t)\,dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{n+1}(\xi), \quad where \quad f^{n+1}(\xi) = \frac {d^{n+1}}{d \xi ^{n+1}}(e^\xi) = e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$${R}_{n+1}(x)= \frac{x^{n+1}}{(n+1)!} e^\xi $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.6)
 * }

Therefore:

$$ f(x) = {P}_{n}(x) + {R}_{n+1}(x) = \sum_{n=0}^n (\frac{x^n}{n!}) + \frac{x^{n+1}}{(n+1)!} e^\xi $$ $$
 *
 $$ \displaystyle (Eq. 2.7)

$$ b) \mathbf{ \quad f(x) = \frac {e^x - 1}{x} }$$

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$$ \begin{matrix} f(x) = {P}_{n}(x) + {R}_{n+1}(x) \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \quad f(x) = \frac {e^x - 1}{x} \Rightarrow f(x_0) = f(0) = \frac {0}{0} = 1 \quad (from \quad equation \quad 1.3) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \quad f'(x) = \frac{d}{dx}((e^x - 1)x^{-1}) = e^x(e^x - 1)^0x^{-1}+(-1(e^x - 1)x^{-2}) = e^xx^{-1} - (e^x - 1)x^{-2} = \frac {e^x(x-1)+1}{x^2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ f'(x) = \frac {e^x(x-1)+1}{x^2} = \frac {f(x)}{g(x)}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = \frac {f'(x)}{g'(x)} = \frac {e^x(x-1)+e^x}{2x^1} = \frac {xe^x}{2x^1} \Rightarrow \frac {f'(0)}{g'(0)}= \frac{0}{0} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ = \frac {f(x)}{g(x)} = \frac {e^x + xe^x}{2} \Rightarrow \frac {f(0)}{g(0)}= \frac{1}{2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

In the same fashion we differentiated $$ f^{(2)}, f^{(3)},.... $$, then by superimposing multiple L'Hopital Rules for $$ x -> 0 $$ on existing derivatives we observed a number to approach a finite value.

Since passed $$ f''(x) $$ differentiation and L'Hoptal Rule application/differentiation became tedious, we generated Matlab code to differentiate $$ f(x) $$ for us and to superimpose multiple L'Hopital's rules in order to generate finite number as $$ x -> 0 $$

From Matlab code:

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$$ \lim_{x\to 0} f(x) = 1$$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \lim_{x\to 0} f^{(1)}(x) = \frac{1}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \lim_{x\to 0} f^{(2)}(x) = \frac{1}{3}$$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \lim_{x\to 0} f^{(3)}(x) = \frac{1}{4}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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$$ \lim_{x\to 0} f^{(4)}(x) = \frac{1}{5}$$ $$
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 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

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$$ \lim_{x\to 0} f^{(n)}(x) = \frac{1}{n+1}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

From Eq. 2.3:
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$${P}_{n}(x) = f({x}_{0}) + \frac{(x-{x}_{0})}{1!}f'({x}_{0}) + .... + \frac{({x-{x}_{0}})^{n}}{1!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$${P}_{n}(x) = 1 + \frac{(x)}{1!}\frac{1}{2} + \frac{(x^2)}{2!}\frac{1}{3} + \frac{(x^3)}{3!}\frac{1}{4}+ .... + \frac{({x-{x}_{0}})^{n}}{n!} \frac{1}{n+1} \Rightarrow $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$${P}_{n}(x) = 1 + \sum_1^n \frac{(x^n)}{(n+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8)
 * }

See the following extra solution for $$ R_n(x) $$

Extra Solution for HW1.2(b)
From Taylor series :

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$$ e^x = 1 + \sum_{i=1}^n \frac{x^i}{i!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Then the original fuction could be expressed as:

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

$$ f(x) = \frac{e^{x}-1}{x} = \sum_{i=1}^n \frac{x^{i-1}}{i!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow P_n(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.12)
 * }

To find $$ R_n(x) $$, we need to solve for $$ f^{(n+1)}(x) $$ which is difficult. So, here we compute $$ R_n(x) $$ from the previous function $$ e^{x} $$. From (Eq. 2.7), we could write $$ e^{x} $$ as:

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

$$ e^{x} = P(x) + R(x) = 1+ \sum_{i=1}^{n+1} \frac{x^i}{i!} + \frac{x^{n+2}}{(n+2)!} e^\xi = \sum_{i=0}^{n} \frac{x^i}{(i)!} +\frac{x^{n+1}}{(n+1)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle $$
 * }

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

$$ \Rightarrow f(x) = \frac{e^{x}-1}{x} = \sum_{i=0}^{n} \frac{x^i}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.13)

$$
 * }

Comparing (Eq. 2.12) and (Eq. 2.13) yields,

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

$$ R_n(x) = \frac{x^{n+1}}{(n+2)!} e^\xi, \quad \xi \in [0, x] $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle

$$
 * }

$$ f(x) = {P}_{n}(x) + {R}_{n}(x) = \sum_{i=0}^{n} \frac{x^{i}}{(i+1)!} + \frac{x^{n+1}}{(n+2)!} e^\xi $$ $$
 *
 $$ \displaystyle (Eq. 2.14)

= HW 1.3 =

Problem:
Given a smooth function that has maximum, minimum, and is solely all positive or all negative:

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$$ m\le f(x)\le M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 3.1)
 * }
 * }

Find
Show at different weighting factors (positive, negative, and zero) that there exists such value in function over interval [a,b], such that area under the curve could be represented with a product of that value and the run of a function over abscissa axes.

Solution:
a) Consider a continuous function $$ f(x) $$ on $$ [a,b] $$, and its maximum and minimum value on this interval are respectively given by $$ M $$ and $$ m $$, we have

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

$$ m\le f(x)\le M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 3.1)
 * }
 * }

If the weight function $$ w(x)\ge 0 $$ on $$ [a,b] $$, then we have

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

$$ w(x)m\le w(x)f(x)\le w(x)M $$, $$ x\in [a,b] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 3.2)
 * }
 * }

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

$$ \Rightarrow \int_a^b{w(x)m}dx\le\int_a^b{w(x)f(x)}dx\le\int_a^b{w(x)M}dx  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

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

$$ \Rightarrow m\int_a^b{w(x)}dx \le \int_a^b{w(x)f(x)}dx \le M\int_a^b{w(x)}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

i. if $$ \int_a^b{w(x)}dx\ne 0$$, this inequality can be rewritten as

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$$ m\le\dfrac{\int_a^b{w(x)f(x)}dx}{\int_a^b{w(x)}dx}\le M $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

Hence, we can always find a $$ \xi \in [a,b] $$ that satisfies

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$$ f(\xi)=\dfrac{\int_a^b{w(x)f(x)}dx}{\int_a^b{w(x)}dx}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 3.3)
 * }
 * }

Therefore

$$ \int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x)> 0 $$ $$
 *
 $$ \displaystyle (Eq. 3.4)

ii. if $$ \int_a^b{w(x)} = 0$$, then $$ w(x)\equiv 0 $$ on $$ [a,b] $$

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

$$ \int_a^b{w(x)f(x)}dx\equiv 0$$, and $$ f(\xi)\int_a^b{w(x)}dx\equiv 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

Therefore

$$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x) = 0 $$ $$
 *
 $$ \displaystyle (Eq. 3.6)

---

b) If the weight function $$ w(x)< 0 $$ on $$ [a,b] $$, we let $$ W(x)=-w(x)$$, then we know $$ W(x)> 0 $$ on $$ [a,b] $$. As been proven in $$ Eq. 3.5$$, we have

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

$$\int_a^b{W(x)f(x)}dx=f(\xi)\int_a^b{W(x)}dx$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

which is

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

$$ \int_a^b{-w(x)f(x)}dx=f(\xi)\int_a^b{-w(x)}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }
 * }

Therefore

$$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x)< 0 $$ $$\int_a^b{w(x)f(x)}dx=f(\xi)\int_a^b{w(x)}dx,\ \ w(x) = 0 $$ $$
 *
 $$ \displaystyle (Eq. 3.5)

Egm6341.s11.team3.ren 01:07, 22 January 2011 (UTC)

= HW 1.4 =

Given:

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

$$f(x)=3x-2x^3\ $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.1)
 * }
 * {| style="width:100%" border="0"

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.2)
 * }

Find
A) Plot $$ f(x) $$ and $$ g(x) $$

B) Find the following norms:

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

$$||f(x)||_{\infty}, ||g(x)||_{\infty}, ||f(x) - g(x)||_{\infty}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution
A) Equations (4.1) and (4.2) are plotted over the domain of x=[0,1] as shown below:

The difference between these two functions will be called Y(x). It is defined below as:

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

$$Y(x) = f(x)-g(x)=[3x-2x^3]-[sinh(x)]=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.3)
 * }

The plot of equation (4.3) over the domain of x=[0,1] is shown below:

B) The infinity norms of these three equations using the restricted domain of x=[0,1] are found below:

The infinity norm of a function is simply the maximum value the function achieves over a specified domain. All three of the functions plotted above will approach infinity over the unrestricted domain of $$x=(-\infty,+\infty)$$.

However, since we are restricting the domain to x=[0,1], these norms must be calculated using techniques discussed below:

B.1) Upon observing the plot of equation (4.1), it is obvious that the maximum value of f(x) occurs at the vertex.  So, the infinity norm will be found by solving for the vertex.  This is accomplished by finding the point of the function where the derivative is zero, since f(x) is not changing at the vertex:

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

$$f(x)=3x-2x^3\ $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.4)
 * }
 * {| style="width:100%" border="0"

$$ \frac{df(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} = 3-6x^2$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.5)
 * }

Solving equation (4.5) for x when the derivative of f(x) is zero yields the location of the vertex:
 * {| style="width:100%" border="0"

$$ \frac{df(x_{vertex})}{dx} = 3-6x_{vertex}^2=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.6)
 * }
 * {| style="width:100%" border="0"

$$ x_{vertex} = \sqrt{\frac{(0 - 3)}{-6}} = \frac{\sqrt{2}}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Substituting this value of x back into equation (4.1) will yield the value of f(x) at the vertex, which is also the infinity norm on x=[0,1]:

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

$$ ||f(x)||_{\infty} = f(x_{vertex}) = 3\left(\frac{\sqrt{2}}{2}\right)-2\left(\frac{\sqrt{2}}{2}\right)^3 = \sqrt{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

B.2) Upon observing the plot of equation (4.2), it is obvious that the maximum value over the prescribed domain occurs at the right boundary of the domain, or when x=1.  So, the infinity norm may be found simply by substituting this value of x directly into the function g(x):

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

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.7)
 * }

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

$$ ||g(x)||_{\infty} = g(1) = sinh(1) = \frac{1}{2}\left(e^{(1)}-e^{-(1)}\right) = 1.1752$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

B.3) Upon observing the plot of Y(x), which is defined above by equation (4.3), it is once again obvious that the maximum value of Y(x) occurs at the vertex.  So, the infinity norm will again be found by solving for the vertex:

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

$$Y(x)=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.8)
 * }

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

$$ \frac{dY(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} - \frac{1}{2}\left((1)e^{x}-(-1)e^{-x}\right) = 3-6x^2 - \frac{1}{2}\left(e^{x}+e^{-x}\right)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.9)
 * }

Solving equation (4.9) for x when the derivative of Y(x) equals zero yields the location of the vertex:
 * {| style="width:100%" border="0"

$$ \frac{dY(x_{vertex})}{dx} = 3-6x_{vertex}^2 -\frac{1}{2}\left(e^{x_{vertex}}+e^{-x_{vertex}}\right)=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.10)
 * }

This equation cannot be solved analytically. An iterative approach was used, and the solution verified by observation of the plot above:

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

$$ x_{vertex} = 0.5541\ $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Substituting this value of x back into equation (4.3) will yield the value of Y(x) at the vertex, which is also the infinity norm on x=[0,1]:

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

$$ ||f(x)-g(x)||_{\infty}=||Y(x)||_{\infty} = Y(x_{vertex}) = 3(0.5541)-2(0.5541)^3-\frac{1}{2}\left(e^{(0.5541)}-e^{-(0.5541)}\right) = 0.7392$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.11)
 * }

Thus, each norm has been found and the problem is complete:

When the domain is unrestricted, $$x=(-\infty,+\infty)$$ and the norms are:
 * {| style="width:100%" border="0"

$$||f(x)||_{\infty} = ||g(x)||_{\infty} = ||f(x) - g(x)||_{\infty} = \infty $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 4.12)
 * }

When the domain is restricted to $$x=[0,1]\ $$, the norms become:

Egm6341.s11.team3.russo 21:55, 21 January 2011 (UTC)

=Signatures=

Solved problem 1 -- Egm6341.s11.team3.rakesh 23:39, 26 January 2011 (UTC)

Solved problem 2 -- Egm6341.s11.team3.sungsikkim

Solved problem 3 -- Egm6341.s11.team3.ren 01:03, 22 January 2011 (UTC)

Solved problem 4 --Egm6341.s11.team3.russo 21:56, 21 January 2011 (UTC)

Reviewed problem 1 -- Egm6341.s11.team3.XiaEgm6341.s11.team3.elango 20:00, 22 January 2011 (UTC)

Reviewed problem 2 -- Egm6341.s11.team3.elangoEgm6341.s11.team3.Xia 20:00, 23 January 2011 (UTC)

Reviewed problem 3 -- Egm6341.s11.team3.Xia 20:10, 22 January 2011 (UTC)

Reviewed problem 4 --Egm6341.s11.team3.Xia 20:20, 22 January 2011 (UTC)

Wiki editing and reviewing problems: --Egm6341.s11.team3.JA 21:54, 24 January 2011 (UTC)

=Solution Assignments and Reviewers=

=References=

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

title: User:Egm6341.s11.team3/sub1


text:
User talk:Eml5526.s11.team6.deshpande

hi....

title: User talk:Eml5526.s11.team6.deshpande


text:
User:Richard Barbrook

Richard Barbrook on Wikipedia

title: User:Richard Barbrook


text:
SPIR608 Political Simulations and Gaming/2011/Week 1

Friday 21st January
 * Week 1 Discussion of Charles Darrow and Elizabeth Magie Phillips, Monopoly.

* Is the design of the game's mechanics (board, pieces, cards, etc.) fit for purpose?
 * Self-explanatory;
 * Information on cards and board;
 * Can visually see what's going on: Monopoly is a game of perfect information;
 * Includes lots of luck;
 * People can conspire with each other;

* Are the rules of the game easy to learn and follow?
 * Monopoly is easy to play - it had thirty years of development before publication.

* Is the game enjoyable and sociable to play?
 * It's fun!

* What techniques does the game use to model its chosen subject?
 * Greed is good - get rich or die trying!
 * Everyone starts with the same opportunity;
 * Risk - you get punished if you spend all your money on building hotels;
 * Encourages consumerism;
 * If you have to sell, you only get half your money back.

* How does the game combine abstraction and realism in its workings?
 * Quite realistic: look at how Brooklyn in New York is being developed;
 * Realism should be judged against how people behave in the game - Monopoly recreates the emotional state of being a capitalist; ::Realism is about how the game affects the players when playing;
 * Game engages players by getting peiople to trade with each other to build up groups of properties;
 * Game in three acts:
 * Getting property;
 * Trading property;
 * Deciding who has won.
 * Its is an innovation from a race game with a beginning and an end, i.e. you go round and round the same track until someone wins.

* How accurately does the game simulate the decision-making processes faced by the real-life protagonists of its chosen subject?
 * It works on abstraction.

* What political lessons can people learn by playing the game?
 * Is it a socialist game? It seems like a capitalist game;
 * It warns people to be careful with their money and avoid bankruptcy;
 * The lesson is that a free market leads to a monopoly as one capitalist owns everything, and the other players are impoverished;
 * It was students playing it on leftist economics courses who changed the name to Monopoly;
 * Why is Monopoly seen as a capitalist game?
 * It encourages egoism, screw everyone else to win;
 * People become so involved in game they forget critical message - people identify with impoverishing others;
 * Message has become hidden by game mechanics;
 * Elizabeth Magie Phillips airbrushed out of history when it was mass-marketed and became a family game of 1950s Cold War USA;
 * Anti-Monopoly is a game critique of the game;
 * Class Struggle tried to be a critical variant, but game does not play very well;
 * During the Cold War, Monopoly identifed by both Americans and Russians as a capitalist game;

Like Machiavelli's The Prince, Monopoly can be taken at face value rather than realising that it's a critique of capitalism.

* How would you improve the structure and mechanics of the game?
 * Credit card version;
 * Versions available for people to use place names from their own locality;
 * Can we make it that the person with least money wins?
 * Philosophers could use Monopoly as a way of judging human nature;
 * We can all identify with bad characters in movies and we can become one when playing Monopoly;
 * Games suck you into the Huizinga's 'magic circle'. Everyone wants to win, but only one person does;
 * People want to keep playing even when it is obvious who has won;
 * You can end up with most of the players no longer involved as two last players slug it out - could have better end game;
 * Luck is an element of the game. Good to understand probability of two dice. Also lots of luck means that inexperienced players have a chance to win;
 * Gaming the game:when you understand how the rules work you can manipulate the rules e.g. the strip of properties following the jail, the limit in the number of houses. People focus more on the rules than what they are meant to represent.

Links

 * Photos of Week 1 game playing
 * Monopoly on BoardGameGeek
 * Monopoly on Wikipedia
 * Landlord's Game on Wikipedia
 * Landlord's Game board
 * Henry George, Progress & Poverty.
 * Monopoly Live - NYT article on new computerised version.
 * Scott Nicholson, Ending the Monopoly elements of Monopoly found in other games.
 * Mosqopoly from Terrorbull Games


text:
User talk:Egm5526.s11.team2.sandhu

 Hello Egm5526.s11.team2.sandhu, 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.

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title: User talk:Egm5526.s11.team2.sandhu


text:
User talk:Egm6341.s11.team4.KTH

HW 1.3 HW 2.12, 2.14 HW 3.7, 3.8 HW 4.3, 4.5 HW 5.3 HW 6.4 HW 7.2

File:Hw2 12 11.jpg & File:Hw2 12 21.jpg
Hi! Thanks for uploading the files File:Hw2 12 11.jpg and File:Hw2 12 21.jpg. Unfortunately, they aren't currently in use and don't have any licensing information, so as part of a cleanup I've tagged them to be removed. If you would like to keep the files here, though, that would be great - all you need to do is add an appropriate license if the copyright belongs to you, or if it is otherwise available under a compatible free license. (As they are just code, this may not be an issue, but if you could help clarify this it would be great). If you need any help, just ask - I'm very happy to assist. - Bilby 01:56, 5 October 2011 (UTC)

title: User talk:Egm6341.s11.team4.KTH


text:
User:Eml5526.s11.team5.SRakesh

rakesh...

title: User:Eml5526.s11.team5.SRakesh


text:
User:EML5526.S11.Team3.vnarayanan

 * User:EML5526.S11.Team3.vnarayanan/Homework1
 * User:EML5526.S11.Team3.vnarayanan/Homework2
 * User:EML5526.S11.Team3.vnarayanan/Homework3
 * User:EML5526.S11.Team3.vnarayanan/Homework4
 * User:EML5526.S11.Team3.vnarayanan/Homework5
 * User:EML5526.S11.Team3.vnarayanan/Homework6
 * User:EML5526.S11.Team3.vnarayanan/Homework7

title: User:EML5526.S11.Team3.vnarayanan


text: File:F(x).jpg

title: File:F(x).jpg


text: File:G(x).jpg

title: File:G(x).jpg


text: File:F(x)-g(x).jpg

title: File:F(x)-g(x).jpg


text:
User:Eml6341.s11.team4

Please delete

title: User:Eml6341.s11.team4


text:
User talk:Dawu~enwikiversity

 Hello Dawu, 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.

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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! --Kevin Rutherford (talk) 22:58, 23 January 2011 (UTC) Kevin Rutherford (talk) 22:58, 23 January 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 Dawu. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Dawu~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 23:30, 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) 06:05, 19 April 2015 (UTC)

title: User talk:Dawu~enwikiversity


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

 Hello Eml5526.s11.team5.SRakesh, 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

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 * 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! --Kevin Rutherford (talk) 22:58, 23 January 2011 (UTC) Kevin Rutherford (talk) 22:58, 23 January 2011 (UTC)

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


text:
User:Egm6341.s11.team4

Numerical Methods for Engineering Analysis 1, Spring 2011 Welcome to team 4's Wikiversity page!

Team Members

 * William Kurth
 * Erle Fields
 * Taeho Kim
 * Brendan Mahon
 * Shengfeng Yang
 * Hailong Chen

Useful Links

 * Course Website
 * Homework Table

title: User:Egm6341.s11.team4


text:
User:Egm6341.s11.team4.fields

=Team 4 Wiki=

Team4_Wiki

=Homework=

HW2

HW3

HW4

HW5

HW6

HW7

title: User:Egm6341.s11.team4.fields


text:
User:Eml5526.s11.team7.jb

\displaystyle

\mathbf{K^{(1)}_{FE}} = \left\{ {\mathbf{K^{(1)}_{0j}};j = 1,...,n} \right\} =

$$i

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

title: User:Eml5526.s11.team7.jb


text:
User:SarahStewart/EdD

I have just started my EdD at Otago University, Dunedin. The story os my adventures will be told mostly in my blog and on Twitter.

title: User:SarahStewart/EdD


text:
User:Egm6341.s11.team4.KTH/HW1

Find
1-4 a) Prove IMVT P.4-1 for nonnegative function, $$ w(\cdot ) $$. i.e., $$ w(\cdot)\geq 0,  \forall x \in [a,b]  $$

1-4 b) Another version of IMVT $$ w(\cdot ) \neq 0, \forall x \in [a,b] $$. Either $$ w(\cdot)>0 $$ or $$ w(\cdot)<0 $$

Given
Continuous function

Solution
1-4 a)

Integrate both sides over closed interval a,b.

Let,

According to Intermediate Value Theorem

1-4 b)

Integrate both sides over closed interval a,b.

Let,

According to Intermediate Value Theorem

From 4-1 a) From 4-1 b)

title: User:Egm6341.s11.team4.KTH/HW1


text: File:Egm6341 s11 team5 hw1 graph1.jpg

title: File:Egm6341 s11 team5 hw1 graph1.jpg


text:
User:Eml5526.s11.team5/sub1

= HW 1.1 =

Problem
Given the generic non-uniform section bar under force as shown below

 Figure 1

Find
a) By taking an infinitesimal slice of the bar (shown in red in Figure 1),$$ dx $$, develop an expression for the elastodynamic response.

b) Assume the cross sectional area to be rectangular

Solution
Free Body Diagram:

 Figure 2

 Theory:

In elastodynamic case, there are 2 major contributors that affect body's motion:

 * Static Forces
 * Body's Inertial effects (resistance to motion caused by Static forces)

In accord to conservation laws as well as Newton's 3rd law, these 2 opposing contributors must be equal to each other.

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

$$F_{static} = F_{dynamic} \Rightarrow \vartriangle F = ma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.1)
 * }

where:

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

$$ F = force \quad (N) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

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

$$ m = mass \quad (kg) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ a = acceleration \quad(\frac{m}{s^2}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

The assumed differential element shape is of trapezoid with force acting on both of its bases. Dynamic equation that resulted is shown below:

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

$$ -N(x) + f(x + \frac {\vartriangle x}{2})\vartriangle x + N(x+\vartriangle x) = ( \hat m(x) * \vartriangle x) a(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.2)
 * }

where:

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

$$ \hat m = mass \ per \ unit \ length \ (\frac{kg}{m}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

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

$$ \frac {N(x+\vartriangle x) - N(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \frac {( \hat m(x) *\vartriangle x)}{\vartriangle x} a(t) = \hat m(x) a(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

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

$$ \Rightarrow a(t) = \frac{\vartriangle v(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.3)
 * }

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

$$ \Rightarrow \vartriangle v(t) = \frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.4)
 * }

Therefore:

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

$$ \frac {N(x+\vartriangle x) - N(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \hat m(x) \frac{1}{\vartriangle t}\frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.5)
 * }

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

$$ \lim_{\vartriangle x \to 0} \frac {N(x+\vartriangle x) - N(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \lim_{\vartriangle t \to 0} \hat m(x) \frac{1}{\vartriangle t}\frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow \lim_{\vartriangle x \to 0} \frac {N(x+\vartriangle x) - N(x)}{\vartriangle x} = \frac {\partial N}{\partial x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow \lim_{\vartriangle t \to 0} \frac{1}{\vartriangle t} = \frac{1}{\partial t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Rightarrow \lim_{\vartriangle t \to 0} \hat m \frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} = \hat m(x) \frac {\partial u}{\partial t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Therefore, substituting back into equation we get:

$$ \frac {\partial N}{\partial x} + f(x) = \hat m(x) \frac{1}{\partial t} \frac {\partial u}{\partial t} \Rightarrow \frac {\partial N}{\partial x} + f(x) = \hat m(x) \frac {\partial^2 u}{\partial t^2} $$ $$ $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle (Eq. 1.6)

Force = Stress * Area,

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

$$ \begin{matrix} N(x) = \sigma(x) * A(x) \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

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

Stress = Strain * Modulus of Elasticity,

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

$$ \begin{matrix} \sigma(x) = \epsilon(x) * E(x) \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.8)
 * }

where

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

$$ \epsilon(x) = \frac {\partial u}{\partial x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.9)
 * }

From Equations 1.7-1.9:

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

$$ N(x) = E(x) * A(x) * \frac {\partial u}{\partial x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.10)
 * }

Mass / Unit Length = Density * Area

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

$$ \hat m(x) = \rho(x) * A(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 1.11)
 * }

Substituting the above relations we get

$$ \frac {\partial}{\partial x} \left[E(x)A(x)\frac{\partial u}{\partial x}\right] + f(x) = \rho(x) A(x) \frac {\partial^2 u} {\partial t^2} $$ $$ $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle (Eq. 1.12)

b) If we consider a case in which the bar has a rectangular cross section

 Figure 3

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

$$ \begin{matrix} A(x) = h(x) * b \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

where

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

$$ h(x) = height \ of \ bar \ at \ x \ (m), $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ b = width \ of \ the \ bar \ (constant)\ (m) \ [i.e. \ Independent \ of \ x,t] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Replacing $$ A(x) $$ with $$ h(x)*b $$ in our equation we get:

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

$$ \frac {\partial}{\partial x} \left[E(x)h(x)*b*\frac{\partial u}{\partial x}\right] + f(x) = \rho(x) h(x)*b* \frac {\partial^2 u} {\partial t^2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

$$ b $$ is independent and can be brought outside the partial derivative of the first term

The equation is divided by $$ b $$ to give

$$ \frac {\partial}{\partial x} \left[E(x)h(x)\frac{\partial u}{\partial x}\right] + \frac{f(x)}{b} = \rho(x) h(x) \frac {\partial^2 u} {\partial t^2} $$ $$ $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle (Eq. 1.13)

Contribution
--Egm6341.s11.team3.JA 22:05, 24 January 2011 (UTC)

--Eml5526.s11.team5.vijay 06:07, 21 January 2011 (UTC)

Lokesh Dahiya

--Vignesh Solai RameshbabuEml5526.s11.team5.srv 05:22, 25 January 2011 (UTC)

--Eml5526.s11.team5.savery 03:20, 26 January 2011 (UTC)

--Eml5526.s11.team5.smith 19:20, 26 January 2011 (UTC)

--Eml5526.s11.team5.mcdaniel 20:04, 26 January 2011 (UTC)

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Hey Team
We can use this "Discuss" page on the team site to communicate and leave notes to each other. I have started putting together HW1 and made a table of HW assignments so that we can navigate to them easily. I will let everyone know when I have pretty much finished putting everything together for HW1 then we all need to sign it via four tildes (~) under the contributing members part. I will probably also include a little note under each problem saying who helped with each problem. I will be out for a few hours this afternoon but will get back to this later tonight.

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Psycholinguistics/Aphasia

Introduction
Aphasia is a language disorder caused by damage suffered to certain portions of the brain that are involved in language perception and production. Depending upon the area that is affected, a person suffering from Aphasia may be able to speak fluently but not coherently, and vice versa. Aphasiology is the study of these linguistic deficits resulting from brain damage and through this study, much can be learned and inferred about the normal function of language and its organization within the brain. There are a variety of modern assessment tools through which aphasia can be studied in patients. The prognosis varies greatly and is dependent upon such factors as, site of the stroke, severity of the lesion, the age of the patient, and the type of aphasia acquired.

History
The word “Aphasia” is derived from the Greek word “aphatos”, meaning speechless. Mentions of Aphasia in Greek Medicine are evident in which speechlessness accompanied convulsions are documented as resulting in paralysis of the right side of the body. Aphasia is again shown in literature from the Roman times in which authors such as Soranus of Ephesus noted the impairment of speech following paralysis of one part of the body. Great advancements occurred from the 1800s to the 1860s in which clinical knowledge, theoretical formulation and neuropathology were developed and a working knowledge of the brain as associated with speech disorders was beginning to be understood more scientifically. The most recent widespread acceptance of the brain's functioning is in the language localization theory in which different sections of the brain are responsible for the functioning of different bodily functions. The birth of this concept came with Franz Joseph Gall's theory of phrenology. Gall proposed that all mental functioning could be located on different sections of the brain and the advancement of a particular area resulted in a visible difference on a person's head. Although Franz Gall's theory of phrenology is now considered a pseudoscience, his localizationist assumptions were evidently working in the right direction as it is now possible to scientifically prove that certain sections of the brain are involved in different bodily functions. The theory of language localization gained furthur credit with significant findings attributed to Paul Broca in the 1860s. The beginning of comprehensive Aphasia understanding came with Paul Broca’s research and subsequent description of his patient Laborgne's brain. In 1861, Broca published “Remarques sur le siège de la faculté du langage articulé: suivies d' une observation d'aphémie” in which he evidences for the localization of articulate speech in the frontal lobe. Upon Leborgne’s death, Broca performed an autopsy and determined that the damage was suffered to the third convolution of the left frontal lobe, which is now commonly referred to as Broca’s area. Stemming from his influential findings, was a revolution in medical and physioogical thinking as it pertained to the brain and the establishment of cerebral localization. Less than a decade later, Wernicke identified “sensory” aphasia as being localized to the temporal lobe. Ludwig Lichtheim then branched off of Wernicke’s model, naming five other types of aphasia, pure word deafness, conduction aphasia, apraxia of speech, transcortical motor aphasia, and transcortical sensory aphasia.

As the mid 20th century approached, professionals specializing in language began searching for a revised model of understanding normal and abnormal language functioning. One professional by the name of Norman Geschwind formed the Geschwind model. Revisiting language localization theories, the model describes the interconnecting functions of a normally working human brain to produce speech and language comprehension. Aphasias were viewed as occurring along these interconnecting lines, disrupting spoken speech or comprehension, resulting in various symptoms. Although the Geschwind model was a great contribution to the understanding of language, problems with it have been uncovered in recent years and a straying away from this understanding of language functioning has occurred.

Advancements in imaging technology has propelled our understanding of the brain as it pertains to language and disorders such as aphasia. Voxel-based Lesion Symptom Mapping (VLSM) in particular, has allowed for medical professionals to determine more specifically where brain lesions lie and the tasks that are impaired because of them. VLSM's ability to identify white matter regions which can play causal roles in certain cognitive domains allows for professionals closely identify these problem regions. A 'voxel' is the three dimensional analog of a pixel and represents a volume of 1 cubic millimeter; The image produced displays 3 dimensional picture of the human brain as depicted in the picture to the right. Studies using VSLM have suggested that language functions are not as localized as the Geschwind model posits it to be.

Forms of Aphasia
Aphasia can be divided into three category types depending upon the quality of the deficit acquired. Fluent, non-fluent, and pure:

 * Fluent
 * Fluent aphasias are caused by impairment to language reception. These aphasic individuals have little issue with fluent verbal output but have difficulty in the language that they are speaking, which often seems like nonsense. Some fluent aphasias include: Wernicke’s Aphasia, Conduction Aphasia, and Anomic Aphasia.

 * Non-fluent
 * Non-fluent aphasias are impaired in their ability to articulate. Unlike some fluent aphasias, non-fluent aphasics retain a relatively intake auditory verbal comprehension. Some examples of non-fluent aphasias include: Broca’s aphasia, Transcortical motor aphasia, and global aphasia.

 * Pure-Aphasia
 * Pure aphasias display themselves as selective deficits to writing, reading or the recognition of words altogether. Pure aphasias include: Pure alexia, Pure word deafness, and Agraphia.

Broca's Aphasia

 * Broca's aphasia, also referred to as 'expressive aphasia, occurs as a result of damage to the frontal lobe of the brain. As a result of the deletion of function words (i.e. 'and', 'the'), speech by a Broca's aphasic appears telegraphically. As opposed to wernicke's aphasia, speech is nonfluent, and highly laboured depending upon the severity of the aphasia.

Global Aphasia

 * This type of aphasia is a result of extensive damage to the language areas of the brain with the impairments appearing globally across all areas of speech processing and production.

Transcortical Motor Aphasia (TMA)

 * Patients with TMA, also referred to as adynamic aphasia, display nonfluent speech as well as phonemic and global paraphasias and the omission of function words. This aphasia results from damage, typically caused by a stroke, of the anterior superior frontal lobe.

Wernicke's Aphasia

 * Wernicke’s Aphasia occurs as a result of damage to the temporal lobe of the brain. Patients exhibit fluent speech and paraphasias and in some types are unable to comprehend speech as well as produce comprehensible speech.

Conduction Aphasia

 * Conduction aphasia, also referred to as 'associative aphasia', is a fluent disorder caused by damage to the left hemisphere of the brain above and below the posterior sylvian fissure. Auditory comprehension remains fairly intact in these individuals, with speech production being mostly affected. Speech repetition is poor, and spontaneous speech production is laboured, with frequent substitution of words or transposing of sounds. As opposed to wernicke's aphasia, patients with conduction aphasia are often aware of their mistakes and make efforts at correcting their errors.

Anomic Aphasia

 * Anomic aphasia occurs as a result of damage to the language areas outside the perisylvian circle. Characteristic of this aphasia is difficulty in word retrieval, while fluent and well-articulated speech remains intact.

Transcortical Sensory Aphasia (TSA)

 * TSA patients display damage to the temporal occipital parietal junction located behind Wernicke's area.

TSA differs from Wernicke's aphasia in that, although they are both fluent disorders, TSA sufferers have fluent and usually comprehensible speech, save for semantic paraphasia in which similar words are substituted for an item. Auditory comprehension is often severely impaired.

 * The following table separates the different types of aphasias, identifies the area of the brain affected and then names the deficits incurred by each.

(From Biological Psychology by Klein, Stephen B., © 2000.)

Signs and Symptoms
Individuals may experience one or many of the following symptoms of an acquired aphasia due to stroke or brain damage:

 * Disturbances in Naming 'Paraphasia' :

Paraphasia refers to the inability to use correct words in speech through substitution of other words in a way that makes speech incomprehensible. In normal human speech, the ability to quickly retrieve words from a mental lexicon makes it easy for speech to sound fluid, well- structured and effortless. This process of selecting a word, termed ‘word finding’, is automatic for most of the population of normal speakers, but for aphasics displaying paraphasia, this process is greatly impaired resulting in the substitution of random words for the intended word. It is called ‘global’ aphasia when an aphasic substitutes an entire word and ‘semantic’ aphasia when a word belonging to the same semantic field is substituted.

 * Disturbance of Fluency:

Aphasic patients are grouped into two categories: fluent, and non-fluent. Fluent aphasics retain the ability to speak in continuous strains of words, with the meaning of the uttered words being the point of issue. Non-fluent aphasics, such as Broca’s aphasics, suffer from low speech rate, short sentence length, with the production of sentences, and even single words being laboured.

 * Disturbance of Repetition:

In some aphasia patients, the ability to repeat words may be lost. More specifically, impairment of the ability to repeat indicates that the damage to the brain is located in the perisylvian region of the dominant hemisphere. Conversely, patients suffering from transcortical aphasia repeat too frequently, and engage in a echolalia, meaning they repeat what is being said to them without knowledge that they are doing so.

 * Disturbance of Grammatical Processing 'Agrammatism' :

Agrammatism refer to an aphasic’s inability to speak correctly in terms of grammatical morphemes. As a result of this, free grammatical morphemes, as well as inflectional affixes which indicate tense or aspect, are not present in speech. Sentences are oversimplified with the omission of these function words, resulting in a telegraphic speech. Also, these patients have difficulty with questions or complex sentence, such as passive sentences. The severity of the aphasia predicts the amount of errors an aphasic makes in this area.

 * Disturbances of Reading and Writing:

Certain Aphasias can affect reading and/or writing as well as speech production and processing. Reading and writing are not necessarily affected together or even equally when the pairing does occur, however, in some aphasias, all of the above are affected together, only in varying degrees of intensity.

 * Apraxia of Speech:

This type of Apraxia, also referred to as dyspraxia, occurs when the brain is damaged in a way that disrupts voluntary movement involved in speech production. When patients with Apraxia are asked to perform a physical command, they are unable to do so even though the command is understood and the speech muscles are not impaired due to paralysis. Two types of Apraxia exist: Acquired Apraxia of Speech, and Developmental Apraxia of Speech. The focus, as it related to Aphasia, is in acquired apraxia of speech as it is a result of sustaining injury to the central nervous system.

Treatments

 * Aphasia Rehabilitation

Learning Exercise
1. The following youtube clips show different situations in which people suffer from aphasia. During the first segment, you encounter councilman Ken Albrecht. Given the way he speaks during therapy, what type of Aphasia does he likely have? As well, referring back to the table on the types of aphasia, what type of Aphasia does the man in the second clip suffer from? http://www.youtube.com/watch?v=Bk13HLma2CI

http://www.youtube.com/watch?v=ZDbSh5DoVJA&feature=related

2.Refer back to the readings of this chapter. What is the primary issue with the Geschwind model?

3.Watch the following video: http://www.youtube.com/watch?v=NUTpel04Nkc

 * What qualities of Broca’s aphasia does Charles display?
 * Why is it that he has such an issue with the simple sentence concerning the leopard and the lion?
 * What types of symptoms would you be witnessing if Charles was suffering from Conduction Aphasia?

4. What does the study of Aphasias tell us about the brain and language?

5. What recent advancements have made it possible to study the brain in detail?

6. As a doctor you encounter a patient who begins to describe experiencing what you suspect to be a stroke. Upon studying their behaviour, you come to find that their fluency of speech is normal, but their ability to name things is greatly impaired and proves to be laboured and frustrating for the patient. What type of aphasia has this patient experienced? What part of the brain was likely damaged due to the stroke?

7. If the temporoparietal region of the brain is damaged, what type of aphasia is a patient suffering from and what would their symptoms be?

Autism and Language
Autism Spectrum Disorder

Autism Spectrum Disorder encompasses five neurobiological disorders in which there is a marked impairment in social and communication abilities. Included in this grouping are: Autistic Disorder, Asperger’s Disorder, Pervasive Developmental Disorder Not Otherwise Specified (PDD-NOS), Childhood Disintegrative Disorder and Rett’s Disorder. Each of these disorders begin in childhood and results in an atypical development which impairs several areas of the child’s functioning. Each case of ASD is different from the next, with degrees varying from mild to severely impaired. Language abilities are affected by varying degree of ASD, with the exception of those suffering from Aspergers. Individuals with the higher functioning ASD retain their linguistic as well as proper cognitive development. Although the research poured into understanding ASD has been substantial, our understanding of the disorder and how it truly functions largely remain a mystery.

Autism

The word Autism means “autos” and “self” in Greek, which refers to how individuals with autism seem to exist in a world of their own, often times unable to communicate and interact with others. Autism is highly variable in nature. The disorder first appears in childhood, with a diagnosis usually occurring by two or three years of age. A triad of symptoms must be present for the child to be classified as autistic, and they include; impairments in social interaction, restricted interests/repetitive behaviour, and lastly, impairments in communication. While not all individuals with autism suffer from language impairments, the majority do, with varying degrees of severity existing among the autistic population. As well, many aspects of language abilities are affected by varying degree of autism, as well as non-verbal communication methods such as gesturing, normal facial expression and eye contact with other individuals.

Autism and Language

Language impairments in individuals with Autism often present itself early in life with delayed onset of babbling and diminished responsiveness; however, not every child with autism will have issues concerning language. Typically, as the child grows older, communication becomes further impaired as the individual with autism are less likely to communicate with their caregivers and other children. Some autistic individuals are incapable of speech entirely, while others are able to speak rather normally. Often, autistic communication presents itself in echolalia in which the autistic individual repeats another person’s words, or reverses their pronouns. Two forms of echolalia exist: Immediate echolalia refers to the immediate repetition of a word, whereas delayed echolalia results in the repetition of a phrase or word after a delay. Research also suggests that autistic children may have difficulty in developing symbols in language, further impairing their language abilities. The autistic’s difficulty in relating to others impairs their language in speaking with others. Meaning, rhythm, bodily language and the nuances of vocal tones are all area of speech in which autistics may have varying levels of difficulty.

Patterns of Language

Repetitive or rigid language: Individuals with autism will often speak words or phrases that have little contextual meaning to the conversation, or will repeat words heard by other people during conversation, called echolalia. Echolalia may also occur after watching television programs in which the child repeats portions of the dialogue. Some autistic individuals may also speak with a high pitched voice.

Intense focus on a single subject: Some individuals with autism may hold intense interest about a single subject or event in which they prefer to revisit through conversation with others Difficulty with non-verbal language: Because of the nature of autism, people sufferings from it are often unable to properly use gestures which add to conversational speech. They lack eye contact, and the normal interactive social nuances which often time leads the autistic individual to act in frustration with inappropriate behaviour or unusual vocal outbursts.

Fragmented language development: Autistic individuals who are fortunate to develop language abilities often develop them at an abnormal rate. While one area may be highly developed, another may be lacking. For example, some children may be able to read by the age of 5, but may have little comprehension about what they have just read.

Therapies and Tools

The primary form of therapy to aid in language development is through a Speech Language Pathologist. An SLP begins by evaluating the child’s speech and language comprehension, then from this evaluation develops a specialized program in order to improve one or many of the language areas affected. Early and intensive intervention is believed to be best by many professionals.

Picture Exchange Communication System (PECS)

 * The PECS is used as a means of alternative communication for autistic individuals. Developed in 1985, PECS was formed to teach functional communication behaviour to children with autism. It acts as a way for those with autism to initiate conversation through the indication of a picture associated with the desired action or object. Small thumbnail pictures depict the actions and/or objects that the individual uses or does in everday life. The picture to the right contains functional pictures such as "boots", "jacket", and the action of "washing hands" that a person with autism may select and hand to their caregiver in order to indicate that they would like their boots, jacket or want to wash their hands.

Speech Generating Devices (SGD)
 * Speech Generating Devices are systems that enable nonverbal individuals to communicate via software installed on PCs, PDAs or other electronic devices. The device can have a speech output in which the words entered are said electronically, allowing for these nonverbal individuals to actively participate in society. One study suggested that 86% of studies concerning the positive effects of SGD were evident, giving credibility to SGD as a viable option for communicating with individuals with autism.

 * Every so often, an underestimated autistic individual surfaces and changes the way in which the world views the mysterious world of autism. One such person is Carly Fleischmann who has astounded professionals upon learning to use a computer to communicate with the outside world- exposing a hidden intellect. To read more about Carly's breakthrough story, visit http://carlysvoice.com/

References

1. Butz, J. (2009). Speech therapy. In E. A. Boutot, M. Tincani, E. A. Boutot & M. Tincani (Eds.), Autism encyclopedia: The complete guide to autism spectrum disorders. (pp. 65-69). Waco, TX US: Prufrock Press.

2. Charlop-Christy, M., Carpenter, M., Le, L., LeBlanc, L. A., & Kellet, K. (2002). Using the picture exchange communication system (PECS) with children with autism: Assessment of PECS acquisition, speech, social-communicative behavior, and problem behavior. Journal of Applied Behavior Analysis, 35(3), 213-231.

3. Jurgens, A., Anderson, A., & Moore, D. W. (2009). The effect of teaching PECS to a child with autism on verbal behaviour, play, and social functioning. Behaviour Change, 26(1), 66-81.

4. Sloman, G. M. (2011). Training paraprofessionals to implement the picture exchange communication system (PECS). ProQuest Information & Learning). Dissertation Abstracts International Section A: Humanities and Social Sciences, 71 (8-)

5. van, d. M., & Rispoli, M. (2010). Communication interventions involving speech-generating devices for children with autism: A review of the literature. Developmental Neurorehabilitation, 13(4), 294-306.

6. http://www.nidcd.nih.gov/health/voice/autism.asp

title: Psycholinguistics/Aphasia


title: File:Hw1aplot.jpg


text:
User talk:Eml5526.s11.team2.reiss

 Hello Eml5526.s11.team2.reiss, 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.

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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) 03:37, 24 January 2011 (UTC)

title: User talk:Eml5526.s11.team2.reiss


text:
User:Eml5526.s11.team4.rhee/HW1

Problem Statement
Derive the dynamic case (1D PDE)

The PDE is

Solution
For regular static case, summing the forces in the x-direction gives

where

 * $$N(x)$$ || is the normal force at $$x$$
 * $$f(x)$$ || is the body force normal force (in the units: force/length)
 * }
 * $$f(x)$$ || is the body force normal force (in the units: force/length)
 * }

Dividing Eq(2) by dx, we obtain

If we take the limit as $$dx->0 $$, it becomes

Now, let's define several terms. The stress is

The strain is

Taking the limit as $$dx->0$$

In our case, $$u=u(x,t)$$, so it is

And we know Hook's law:

where

 * $$E$$ || is Young's modulus.
 * }
 * }

Substituting Eq(8) into Eq(9)

Substituting in (5)

Substituting Eq(11) into Eq(4)

Eq(12) is same with the inertia force according to balance of forces. Therefore, it can be written as

where

 * $$\mathbf \rho $$ || is the mass density.
 * }
 * }

Rewriting the above equation

where

 * $$ m(x) $$ || is the linear mass density. (in the units: mass/unit length)
 * }
 * }

Finally, we got the dynamic response of an elastic bar of variable cross section. From Eq(12) and Eq(14), we obtain

title: User:Eml5526.s11.team4.rhee/HW1


text:
User talk:Egm6341.s11.team4.BM/HW1

Looks great so far. I will post problem 4 when I can make my own personal account.

-William

Egm6341.s11.team4 04:37, 24 January 2011 (UTC)

title: User talk:Egm6341.s11.team4.BM/HW1


text:
User:Eml5526.s11.team4.rhee/Help

To writing mathematical equations, you need to use Latex editor. http://www.codecogs.com/latex/eqneditor.php
 * Useful websites for Latex:

http://www.numberempire.com/texequationeditor/equationeditor.php

Copy from the Latex editor and Paste!

Don't Forget $$and$$ options!

$$\int_{a}^{b}f(x)dx,{x}^{a},{x}_{a},\sum_{a}^{b}f(x),\frac{a}{b}$$

title: User:Eml5526.s11.team4.rhee/Help


text:
User talk:Paulmcdonald

Please feel free to leave a message here, but be sure to notify me at my Wikipedia talk page because I check there much more often. THANK YOU!!--Paulmcdonald 05:18, 24 January 2011 (UTC)

title: User talk:Paulmcdonald


text:
User:Paulmcdonald

Want to learn about me, read my page at Wikipeida. Want to leave me a message, that's the best place too. Or if you prefer, leave a message on my talk page here and notify me on my Wikipedia page to truck over here to read it. Cheers!

title: User:Paulmcdonald


text: Survey research and design in psychology/Header


text:
User:Margit51

I' am hungarian wikipedista. :) My hungarian page

title: User:Margit51


text:
User talk:Margit51

Hi!

title: User talk:Margit51


text:
User:Kethryvis

The Disclaimer
I am a Community Associate for the Wikimedia Foundation, though my work on this username is in my capacity as a volunteer. My work with the Foundation began officially on October 15, 2010. Unless otherwise stated, any edit to this wiki by me is an act of a regular member of the community, not a legal or official action of any part of the Wikimedia Foundation. Official acts by me in the line of my work will be done from my office account, User:Christine_(WMF).

You can find me on IRC under the nick ChristineM.

The Other Stuff
I don't edit a lot, I mainly edit on English Wikipedia mainly around The Muppets and other Muppet Related Things. I'd like to get more involved with editing as time allows between work and grad school. They'll mostly be small edits and hopefully nothing to get excited over but hey... ya gotta start somewhere, right?

title: User:Kethryvis


text:
File:FUN WITH ENGLISH 8A Lesson 7 Version 0.42.pdf

Summary
Charles Jeffrey Danoff 08:07, 24 January 2011 (UTC)


text:
User:Marbeeno

Just another wiki-student...

title: User:Marbeeno


text:
TESOL/English preposition combinations

Remembering the correct preposition combinations is a difficult task second-language English learners. Usually a verb phrase takes only one or two prepositions and other combinations are incorrect. Learning them is a matter of memorization.

Quiz
{Accuse +of -for -on -at
 * type=""}

{be good -for -on +at -in
 * type=""}

{apologize - somebody -for -at +to -by
 * type=""}

{believe - God -of -to -by +in
 * type=""}

{insist -of -for +on -by
 * type=""}

title: TESOL/English preposition combinations


text:
Survey research and design in psychology/Readings/Textbooks/Pallant/2010

Pallant, J. (2010). SPSS survival manual (4th ed.). London: Open University Press. (Download 1st ed. online - no cost)


text:
User talk:Marbeeno

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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! Jtneill - Talk - c 11:51, 24 January 2011 (UTC)

 Woof! My name is Jack Russell. I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

title: User talk:Marbeeno


title: File:NM1.team1.HW1.figure1.png


text:
User talk:Nandhy

 Hello Nandhy, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Nandhy


text:
User talk:Bahram afshar

 Hello Bahram afshar, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Bahram afshar


text:
User talk:Bhargavk04

 Hello Bhargavk04, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Bhargavk04


text:
User talk:RobotEducativo

 Hello RobotEducativo, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:RobotEducativo


text:
User talk:Kethryvis

 Hello Kethryvis, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Kethryvis


text:
User talk:Eml5526.s11.team6.tupsakhare

 Hello Eml5526.s11.team6.tupsakhare, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Eml5526.s11.team6.tupsakhare


text:
User talk:Eml6341.s11.team4

 Hello Eml6341.s11.team4, 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.

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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) 14:58, 24 January 2011 (UTC)

title: User talk:Eml6341.s11.team4


text:
User talk:Courcelles

 Hello Courcelles, 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.

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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) 14:59, 24 January 2011 (UTC)

title: User talk:Courcelles


text:
User:Eml5526.s11.Team3.risher

Team Page
http://en.wikiversity.org/wiki/User:Eml5526.s11.team3

User:Eml5526.s11.team3

Homework
User:EML5526.S11.Team3.risher/homework2

title: User:Eml5526.s11.Team3.risher