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Portal:Modern Hebrew/Templates

The following is a list of templates on the main page:
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title: Portal:Modern Hebrew/Templates


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User:TowTrucker

Hi!--TowTrucker (talk) 23:30, 10 April 2012 (UTC)

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

title: User:TowTrucker


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User talk:TowTrucker

Hi--TowTrucker (talk) 23:32, 10 April 2012 (UTC)

title: User talk:TowTrucker


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User:Alberto Morales G

Perfil académico
First name: José Alberto

Second name: Morales Gómez

title: User:Alberto Morales G


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Talk:Controversies in Science/What killed the dinosaurs/A critique of The Cretaceous-Tertiary mass extinction is due to the post-impact ozone conditions from the asteroid collision in the Yucatan Peninsula in northestern Mexico (The Chicxulub crater)

The argument that there were still life after the impact of The Chicxulub is strong, but what proves that the method used to determine the existence of other animals after the impact is accurate.Lfamc (talk) 00:35, 11 April 2012 (UTC)

 * The ozone may have changed due to the impact of the asteroid, but it was not the reason for mass extinction. Studies on the impact site have shown that although the asteroid caused some change to the area, there were life forms around for many years after. These included some life forms that were very fragile. Cthom4751 (talk) 19:59, 6 April 2012 (UTC)

 * The mass extinction was not due to post-impact ozone conditions but from the after effects from the asteroid. The asteroid caused tsunamis and acid rain which first killed off all the vegetation.Acox967 (talk) 01:20, 26 September 2012 (UTC)


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User:Egm4313.s12.team11.perez.gp/R5.9

R5.9
Solved by: Gonzalo Perez

Problem Statement
Consider the L2-ODE-CC (5) p.7b-7 with $$ log(1+x) \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$  (5) p.7b-7

$$ r(x)=log(1+x) \! $$    (1) p.7c-28

and the initial conditions

$$ y(\frac{-3}{4})=1, y'(\frac{-3}{4})=0 \! $$.

Part A
Project the excitation $$ r(x) \! $$ on the polynomial basis

$$ {b_j (x) = x^j, j=0, 1, ..., n} \! $$   (1)

i.e., find $$ d_j \! $$ such that

$$ r(x) \approx r_n (x) = \sum_{j=0}^{n} d_j x^j \! $$   (2)

for x in $$ [\frac{-3}{4}, 3] \! $$, and for n = 3, 6, 9.

Plot $$ r(x) \! $$ and $$ r_n (x) \! $$ to show uniform approximation and convergence.

Note that:

$$ \left \langle x^i,r \right \rangle = \int_{a}^{b} x^i log(1+x) dx \! $$    (3)

Solution
To solve this problem, it is important to know that the scalar product is defined as the following:

$$ \left \langle b_0, b_0 \right \rangle = \int x^0\cdot x^0 dx \! $$.

Therefore, it follows that:

$$ \left \langle b_i,b_j \right \rangle = \int x^i\cdot x^j dx \! $$, where $$ b_i (x)=x^i \! $$ and $$ b_j (x) = x^j \! $$.

We know that if $$ b_1, b_2 \! $$ are linearly independent, then by theorem on p.7c-37, the matrix is solvable.

According to this and (3)p.8-14:

If $$ \Gamma \neq 0 \Rightarrow \Gamma^{-1} \! $$ exists $$ \Rightarrow c= \Gamma^{-1}d \! $$.      (3)p.8-14

Now let's define the Gram matrix $$ \Gamma \! $$ as a function of $$ b_i \! $$:

$$ \Gamma (b_i)=\begin{bmatrix} \left \langle b_0,b_0 \right \rangle & \left \langle b_0,b_1 \right \rangle & ... & \left \langle b_0,b_n \right \rangle\\ ... & ... & ... & ...\\ ... & ... & ... & ...\\ \left \langle b_n,b_0 \right \rangle & \left \langle b_n,b_1 \right \rangle & ... & \left \langle b_n,b_n \right \rangle \end{bmatrix} \! $$      (1)p.8-13

Defining the "d" matrix as was done in (3)p.8-13, we get:

$$ d = \begin{Bmatrix} \left \langle b_0,r \right \rangle\\ \left \langle b_1,r \right \rangle\\ ...\\

\left \langle b_n,r \right \rangle\end{Bmatrix} \! $$.       (3)p.8-13

And according to (1)p.8-15: $$ r_n(x) = \sum_{0}^{n}c_i x^i \! $$         (1)p.8-15

Now, we can find the values to compare $$ r_n \! $$ to $$ y \! $$.

Using Matlab, this is the code that was used to produce the results:

The Matlab code above produced the following graph:

Where $$ r_n(x) \! $$ is represented by the dashed line and the approximation,$$ y(x) \! $$, is represented by the red line. This code can work for all n values.

Part B
In a seperate series of plots, compare the approximation of the function $$ \log(x+1) \!$$ by Taylor series expansion about $$ x=0 \!$$.

Where: $$ f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(\hat x)}{n!}(x- \hat x)^n \!$$

Solution
For n=1:

$$ \log(x+1) = \frac{x}{\log(10)} \! $$

For n=2:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)} \! $$

For n=3:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)} \! $$

For n=4:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)} \!$$

For n=5:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)} \! $$

For n=6:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)} \! $$

For n=7:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)} \!$$

For n=8:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)} \! $$

For n=9:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)} \! $$

For n=10:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})} \! $$

For n=11:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})} \!$$

For n=12:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})} \! $$

For n=13:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})} \! $$

For n=14:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})} \! $$

For n=15:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})} \! $$

For n=16:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})}-\frac{x^{16}}{\log(10^{16})} \!$$

Using Matlab to plot the graph:

Part 2
Find $$ y_n (x) \! $$ such that:

$$ y_n'' + ay_n'+by_n = r_n (x) \! $$   (1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot $$ y_n (x) \! $$ for n = 3, 6, 9, for x in $$ [\frac{-3}{4}, 3] \! $$.

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution
First, we find the homogeneous solution to the ODE:

The characteristic equation is:

$$\lambda^2-3\lambda+2=0\!$$

$$(\lambda-2)(\lambda-1)=0\!$$

Then, $$\lambda=1,2\!$$

Therefore the homogeneous solution is:

$$y_h=C_1e^(2x)+c_2e^x\!$$

Now to find the particulate solution

For n=3:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0\\ 0&  2  &-6  &6  & 0\\ 0&  0&  2  &-9  &12  \\ 0&  0&  0&  2  &-12\\ 0&  0&  0& 0 &  2  \\  \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \end{bmatrix}\!$$

Then, using MATLAB and the backlash operator we can solve for these unknowns:

Therefore

$$ y_{p4}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4 \!$$

Superposing the homogeneous and particulate solution we get

$$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4+C_1e^{2x}+C_2e^{x} \!$$

Differentiating:

$$y'_{n}=3.7458+3.1486x+1.1943x^2+0.2172x^3+2C_1e^{2x}+C_2e^{x}\!$$ Evaluating at the initial conditions:

$$y(-0.75)=0.9698261719+0.231301601C_!+0.4723665527C_2=1\!$$

$$y'(-0.75)=1.9645125+0.4462603203C_1+0.4723665527C_2\!$$

We obtain: $$ C_1=-4.46\!$$ $$ C_2=0.055 \!$$ Finally we have: $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4-4.46e^{2x}+0.055e^{x} \!$$

For n=6:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)}+\frac{x^5}{5\ln(10)}-\frac{x^6}{6\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0 &0 &0 &0\\ 0&  2  &-6  &6  & 0 &0 &0 &0\\ 0&  0&  2  &-9  &12  &0 &0 &0\\ 0&  0&  0&  2  &-12 &20 &0 &0 \\ 0&  0&  0& 0 &  2 &-15 &30 &0\\ 0 &0 &0 &0 &0 &2 &-18 &42\\ 0 &0 &0 &0 &0 &0  &2 &-21\\ 0 &0 &0 &0 &0 &0 &0 &2\\  \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \end{bmatrix} \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \end{bmatrix}\!$$

Then, using MATLAB and the backlash operator we can solve for these unknowns:

Therefore

$$ y_{p7}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7 \!$$

Superposing the homogeneous and particulate solution we get

$$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+C_1e^{2x}+c_2e^x \!$$

Differentiating:

$$y'_{n}=375.3933+371.213x+181.644x^2+57.9784x^3+13.46x^4+2.1714x^5+0.214x^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions:

$$y(-0.75)=178.816+0.2231301601C_!+0.4723665527C_2=1\!$$

$$y'(-0.75)=178.413+0.4462603203C_1+0.4723665527C_2\!$$

We obtain: $$ C_1=-2.6757\!$$ $$ C_2=-375.173 \!$$ Finally $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+-2.6757e^{2x}-375.173e^x \!$$

For n=9:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)} \! $$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \\ K_{8} \\ K_{9} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \\ \frac{1}{7ln(10)} \\ \frac{1}{8ln(10)} \\ \frac{1}{9ln(10)} \end{bmatrix}\!$$

Then, using MATLAB and the backlash operator we can solve for these unknowns:

Therefore

$$ y_{p11}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11) \!$$

Superposing the homogeneous and particulate solution we get

$$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)+C_1e^{2x}+C_2e^(x) \!$$

Differentiating:

$$y'_{n}=0.2167 x^10+3.474 x^9+37.3491 x^8+323.336 x^7+2349.42 x^6+14355.1 x^5+72421.8 x^4+290980. x^3+874881. x^2\!$$ $$+1.7517x10^6 x+1.75267x10^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions:

$$y(-0.75)=828254+0.2231301601C_!+0.4723665527C_2=1\!$$

$$y'(-0.75)=828145+0.4462603203C_1+0.4723665527C_2=0\!$$

We obtain: $$ C_1=-484.022\!$$ $$ C_2=-1753750 \!$$ Finally $$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)-484.022e^{2x}-1753750e^x \!$$

Here is the graph for this problem using Matlab:

title: User:Egm4313.s12.team11.perez.gp/R5.9


text:
The RUDAS, CSDD, and the SGA: Description, analyses and recommendations for practice

Title: The Rowland Universal Dementia Assessment Scale (RUDAS), the Cornell Scale for depression in Dementia (CSDD), and the Subjective Global Assessment (SGA): Description, analyses and recommendations for practice

Abstract
Purpose

The purpose of this report is to describe how three assessment tools- the RUDAS, CSDD and SGA can be used to assess the abilities of a person with dementia. Three research articles are critically analysed on the development and use of RUDAS. Findings from this analysis are then synthesised into recommendations for practice.

Key Findings

RUDAS has both reliability and diagnostic accuracy for detecting dementia, is portable across cultures and easily used by primary health care clinicians. The assessment tool is also of benefit for its specificity in early diagnosis.

Conclusions

Assessment tools are subject to cultural, social, educational and physical factors which may influence findings. Further, while RUDAS was developed to overcome perceived biases in the MMSE it has also been found to have greater sensitivity and specificity in diagnosing dementia.

Recommendations

Users of assessment tools should be cognizant of factors which may affect results. A person-centred approach that considers the persons individual socio-cultural, educational and physical influences circumstances before selecting the most appropriate tool will produce a more reliable result. Being committed to person-centred care also requires a commitment to critical thinking, a multidimensional skill involving examination and analysis of all available ideas to inform clinical reasoning.

Dementia and assessment
John has mixed type dementia and lives in a Residential Aged Care Facility (RACF). He was admitted to the facility one year previously following the death of his wife. Over the past several months, he has been actively refusing care, lost a significant amount of weight, appears to have difficulty swallowing and is subsequently refusing diet and fluids. Three assessment tools will be discussed to provide more information and determine the best course of action in John’s case. These are the Subjective Global Assessment (SGA) for nutritional assessment, the Cornell Scale for depression in Dementia (CSDD) and the Rowland Universal Dementia Assessment Scale (RUDAS). Three research articles will be analysed on the development and use of the RUDAS scale (Appendix 1). Findings from the analysis will then be synthesised into recommendations for practice.

The SGA assessment is used to investigate John’s difficulty in swallowing (dysphagia), loss of weight and refusal of diet and fluids. The process of swallowing involves complex coordination of voluntary and involuntary psychological, sensory and motor behaviours and is at risk of compromise from neurological disorders such as dementia (Ney et al. 2008). Dysphagia is characterised by difficulty in initiation of swallowing and impaired transfer of food from the mouth to the oesophagus (RACGP 2006, p.36). There is a strong correlation between dysphagia, malnutrition and poor clinical outcome (Ney et al. 2008). Patients with eating and swallowing disturbances are often at risk of malnutrition, dehydration, depression and aspiration pneumonia (Journal of the Dietitians Association of Australia (JDAA 2009; RACGP 2006, p.35). Aspiration pneumonia is the inhalation of food, saliva and gastric content and is the most common cause of death in patients with dysphagia associated with dementia (RACGP 2006, p.36). John’s chest infection may be a clinical indication of aspiration pneumonia (RACGP 2006, p.36).

Assessment using the SGA is a team approach that examines key areas such as functional capacity, gastrointestinal systems and dietary and weight changes. Speech Pathologists assess swallowing difficulties by judging the ability to safely and efficiently transfer the food bolus through the aero-digestive track (Ney et al. 2008). A medical officer will rule on other pathological processes such as xerostomia, sarcopenia or oesophageal motility. Consultation with a dietitian is also essential to ensure that adequate hydration and nutrition can be obtained within the texture and fluid modifications that may be recommended (JDAA 2009).

CSDD is designed for the assessment of depression in older people with dementia (Royal Australian College of General Practitioners (RACGP) 2006, p.34). John is at risk of depression due to recent life changes such as loss of independence and loss of a significant other (Alexopoulos et al. 1988). Late-onset depressive disorders are also often associated with pre-existing illnesses such as dementia due to cognitive vulnerability (RACGP 2006, p.34). Depression and dementia are the most frequent psychiatric syndromes in older people and are often found to co-exist (Alexopoulos et al. 1988). Depression, a treatable condition, may accompany symptoms of dementia and exacerbate psychological and behavioural issues. Clinical features include loss of interest, irritability, disinterest in food, fatigue, psychomotor slowing and social withdrawal (RACGP 2006, p.34). For John, the three assessment tools are used specifically because dementia, depression and dysphagia are often interrelated and symptomology can overlap. Differential diagnoses that identify, treat and modify the impact of one condition may also lead to improvement in other areas and an increase in overall function and well being (RACGP 2006, p.26).

RUDAS was developed in Australia to overcome recognised cultural, educational and age-related constraints of other cognitive assessment tools. It also assesses a wider range of cognitive domains, particularly frontal lobe function, an early sign of dementia (Storey at al. 2004). The rationale for assessing John’s cognitive function with RUDAS is to form a basis for evaluating the appropriateness of care and the need for, and effectiveness of, any interventions (Farrell 2007, pp. 32-34). Cognitive assessment is crucial in short and long term management and prognosis of dementia and its symptoms. Concurrent physical and mental illness, general deconditioning, and increased anxiety may also confound the clinical picture (Pang et al. 2008). The greater specificity of RUDAS in evaluating executive function and determining meaningful deficits will provide RACF staff with ongoing determination of the degree and progression of John’s dementia (Storey et al. 2004). This enables differential diagnosis with his current presentation and identification of collaborative problems, and determines appropriate therapeutic and psychosocial interventions (Pang et al. 2008).

Article 1
Storey, J, Rowland, J, Conforti, D and Dickson, H 2004 ‘The Rowland Universal Dementia Assessment Scale (RUDAS): a multicultural cognitive assessment scale’, International Psychogeriatrics, vol.16, no.1, pp.13–31 (online Cambridge Journals).

In this article Storey et al. (2004) describe the development of RUDAS. The objective was to design and validate a simple method of detecting dementia that was portable across cultures and easily used by primary health care clinicians (Storey et al. 2004). Stage 1 involved using cultural and health advisory groups to develop culturally fair cognitive items. In Stage 2, 42 items were clinically tested. Stage 3 involved random sampling to assess the predictive accuracy of the items. The result was the RUDAS, which comprises six items: body orientation, fist palm alternation, copying of a picture of a cube, judgment in relation to crossing a busy road, animal names and recall of grocery items presented at the beginning of the test. The items are designed to assess multiple cognitive domains such as executive function, recent memory, gnosis, praxis, and category fluency (Storey et al. 2004).

The rationale for developing this tool was that the current most common tool in use – the Folstein Mini-Mental State Examination (MMSE) (comments inclusive of the amended version, the Standardised Mini-Mental State Examination (SMME) – was developed in an English-speaking population with words and concepts that were not easily translated into other cultures (Escobar et al. 1986, cited in Storey et al. 2004, p. 14). While age and education normative values have been introduced for the MMSE, they are not widely in use (Pang et al. 2008). Rossellia et al. (2006) also found significant gender and educational bias when applying both the MMSE standard ‘serial 7s’ tasks and the apparently less difficult ‘backwards spelling’ task for those with lower educational levels.

Storey et al. (2004) therefore based their design premise primarily on the Escobar finding (1986, cited in Storey et al. 2004, p. 14) that the MMSE should be revised to diminish cultural, social, educational biases. RUDAS was the culmination of this design purpose and, in 2004, Storey and co-authors found that RUDAS had both reliability and diagnostic accuracy for detecting dementia based on DISM-IV criteria. The authors recommended further validation of RUDAS by random sampling in a community-based population with broader ranges of cognitive function (Storey et al. 2004). In 2006, this validation study was achieved with federal funding and RUDAS became an accepted assessment tool, recognised not only for its cross-cultural portability but also its predictive accuracy and sensitivity (Rowland et al. 2006).

It is this latter finding that offers the student of research particular interest in how ideas are developed, explored and expanded through research and designed study to consequently emerge, in refined form, into practice. For, while the stated objective by Storey et al. (2004) from the outset was to develop a ‘simple method...that is valid across cultures’ (2004, p.13), the conversation soon changed to recognise that another emerging benefit of RUDAS was its sensitivity in detecting earlier stages of dementia due to its focus on the executive function of the cognitive domains. In effect, while the authors set out to develop a tool to solve one issue, it became evident during the study, and in later research papers, that the tool was of equal if not more value in another area. Early detection and diagnosis of dementia is important as it allows for differential diagnoses and timely therapeutic and psychosocial interventions (Storey et al. 2004).

Article 2
Lype, T, Ajitha, B, Antony, P, Ajeeth, N, Job, S & Shaji S 2006, ‘Usefulness of the Rowland Universal Dementia Assessment Scale in South India’, Journal of Neurology, Neurosurgery & Psychiatry, vol. 77, pp. 513-514 (online PubMed).

In this study Lype et al. (2006) sought to compare RUDAS with the MMSE in assessing a sample population and also to gauge how well RUDAS performed in an entirely different socio-cultural population. A total of 58 patients with mild to moderate dementia and 58 with age and sex matched controls were screened with years of formal education also correlated to the findings (Lype et al. 2006). The rationale behind the study was the concern that people without dementia but with less education, or who are illiterate and innumerate, may be screened positive for dementia (Lype et al. 2006). The study found that while RUDAS and MMSE had similar sensitivity, RUDAS had greater specificity. In other words, while both tests were similarly sensitive to detecting dementia, RUDAS was better able to identify those who did not have the condition. It was determined that the inclusion of frontal executive function tests gave RUDAS an advantage. In other areas, the authors found RUDAS to be user-friendly, easy to administer but retaining some educational bias. In conclusion, the RUDAS was of benefit in addressing the authors’ concerns of misdiagnosis of dementia due to its specificity and was usable in different socio-cultural environment. However, concerns lingered about the use of screening tools from educationally developed countries in populations with much lower levels of education (Lype et al. 2006).

Article 3
Pang, J, Yu, H, Pearson, K, Lynch, P & Fong, C 2009, ‘Comparison of the MMSE and RUDAS cognitive screening tools in an elderly inpatient population in everyday clinical use’, Internal Medicine Journal, vol. 39, no.8, pp.560-564 (online Web of Science).

Pang et al. (2008) conducted a pilot study to validate RUDAS in a clinical setting, determine patient satisfaction and gain an impression of clinicians’ views of each test’s utility. Moreover, the population setting – a convenience sample of 48 inpatients of Eastern Health in Victoria – is a heavily culturally and linguistically diverse (CALD) environment – another test of the claim made by Storey et al. (2004) that RUDAS is multicultural. In reading these objectives, there is immediate concern about lack of specificity in the research question and that the breadth may lend itself to superficiality and lack of investigative depth (Draper 2004, pp. 69-70). However, the paper does seek to answer three pertinent and very real questions for the clinician working in the pressured inpatient setting: Which test is better? Which is more user-friendly? And how do my peers regard it?

The study found that MMSE and RUDAS correlated well, took the same amount of time and had similar levels of patient satisfaction. Clinicians generally preferred the MMSE owing to lack of familiarity with RUDAS. However, RUDAS was preferred specifically in the CALD environment as there was no need to adjust normative language values as with the MMSE. The authors also recognised the specific challenges of the inpatient setting such as delirium, physical and mental illness, anxiety and environmental distractions. Where patients are bed-bound or immobile, the MMSE is easier to perform as it requires no movement.

In conclusion, this report by analysing the three articles finds that RUDAS was developed primarily to overcome perceived biases in the MMSE. It has also been found that due to its focus on the executive function of the cognitive domains, RUDAS can have a greater sensitivity and specificity in diagnosing dementia and is easily used by clinicians. However, when using any assessment tool the clinician should be mindful of the cultural, social, educational and physical factors which may influence findings and also question whether there are other tools available which may provide more accurate data. These conclusions are synthesised in the following recommendations for practice as person-centred care and critical thinking.

Person-centred care is the notion that we do not regard our patients as a collection individual of diseases but, instead, view them holistically with regard to exogenous influences such as age, education and socio-cultural differences- the human factors (Pelzang 2010; Farrell 2007, p. 25). Most important, when using assessment tools like RUDAS, is ensuring that the tool is appropriate for the patient’s situation rather than make the patient fit the parameters of the tool. A recent example from clinical placement is of a patient who scored consistently low on his Glasgow Coma Scale unless his son was present during assessment to translate into their first language, when the patient would then exhibit distinct motor responses, moving all four limbs on request. As a nurse, there is an obligation to identify when assessment tools need to be attenuated by situational awareness of cultural and other social issues (ANMC 2006). Using such tools without being holistically aware of the patient’s needs may lead to misdiagnosis.

Being committed to person-centred care also requires a commitment to critical thinking, a further recommendation for practice. Critical thinking is a multidimensional skill involving examination and analysis of all available ideas to inform clinical reasoning (Farrell 2007, p. 25). It is also about questioning whether the assessment tools being used in a particular clinical environment are providing the most accurate and valuable of data to help identify and solve the patient’s problems (ANMC 2006). Thus it may have been established that the client has dementia using the MMSE, but using the RUDAS may enable the nurse to develop a care plan specific to the functionality of the individuals’ cognitive domains and therefore better meet their needs. Collecting more cues from a wider range of informational sources enables proactive nursing that better predicts and manages a patients needs rather than reacting to a singular set of patient problems (Levett-Jones et al. 2009).

Acknowledgements
The original version of this essay was by Saul Culican and was a prize winner NSW/ACT Dementia Training Student Centre essay competition, 2011.


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User:Egm4313.s12.team11.perez.gp/R6.5

R6.5
Solved by: Gonzalo Perez

Part R4.2, p.7c-26
For each value of n=3,5,9, re-display the expressions for the 3 functions $$ y_{p,n} (x), y_{h,n} (x), y_n (x) \! $$, and plot these 3 functions separately over the interval $$ [0,20 \pi] \! $$.

R4.2 p.7-26: Consider the L2-ODE-CC (5) p.7b-7 with $$ sin x \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$              (5)p.7-7

$$ r(x) = sin x \! $$                     (1)p.9-15

and the initial conditions:

$$ y(0) = 1, y'(0) = 0. \! $$                      (3b)p.3-7

Exact solution: $$ y(x) = y_h (x) + y_p (x) \! $$                   (2)p.9-15

Re-display the expressions for $$ y_p (x), y_h (x), y(x) \! $$.

Superpose each of the above plot with that of the exact solution.

Solution
The graphs are to be separately graphed as follows.

For n = 3, the code that will generate the graph looks like this:

For n = 6:

For n = 9:

Part R4.3, p.7c-28
Understand and run the TA's code to produce a similar plot, but over a larger interval $$ [0,10] \! $$. Do zoom-in plots about the points $$ x=-0.5,0,+0.5 \! $$ and comment on the accuracy of different approximations.

R4.3 p.7-28: Consider the L2-ODE-CC (5) p.7b-7 with $$ log(1+x) \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$     (5)p.7-7

$$ r(x) = log(1+x) \! $$     (1)p.7-28

and the initial conditions:

$$ y(-\frac{3}{4})=1, y'(-\frac{3}{4})=0 \! $$    (2)p.7-28

Solution
For n=4 (from x = 0 to x = 10):

For n=4 (zoom-in plots around x = -0.5):

For n=4 (zoom-in plots around x = 0):

For n=4 (zoom-in plots around x = +0.5):

For n=7 (from x = 0 to x = 10):

For n=11 (from x = 0 to x = 10):

Where the black line represents n=11 and the blue line represents log(1+x).

The problem asks for n=4, 7, and 11. The TA had up to n=16, but the problem does not ask for this. Manipulating the code and graphs to only account for these n values, n=16 has been disregarded.

Combined plots for n=7 and n=11 (zoom-in plots around x = -0.5):

Combined plots for n=7 and n=11 (zoom-in plots around x = 0):

Combined plots for n=7 and n=11 (zoom-in plots around x = 0.5):

Part R4.4, p.7c-29
Understand and run the TA's code to produce a similar plot, but over a larger interval $$ [0.9,10] \! $$, and for n=4,7. Do zoom-in plots about $$ x=1,1.5,2,2.5 \! $$ and comments on the accuracy of the approximations.

R4.4 p.7-29: Extend the accuracy of the solution beyond $$ \hat{x}=1 \! $$.

Solution
The general code that can be used to graph the plots of n=4,7,11 is:

With the above code, we can add the following code to graph n=4.

For n = 4 (from x = 0.9 to x = 10):

For n = 7 (from x = 0.9 to x = 10):

For n=11 (from x = 0.9 to x = 10):

Repeating the same common part of the code, let's graph the other plots around x = 1, 1.5, 2, 2.5:

For n = 4, 7, 11 at x = 1:

For n = 4, 7, 11 at x = 1.5:

For n = 4, 7, 11 at x = 2:

For n = 4, 7, 11 at x = 2.5:

For the last part, the first (and unchanged TA's code) is the following:

The second part includes the change that had to be made in order to solve this problem:

title: User:Egm4313.s12.team11.perez.gp/R6.5


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User:TestPage

!@#$ This is a test.

title: User:TestPage


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User:Egm4313.s12.team8.colocar/R6.5

Problem Statement
$$ \displaystyle \text{Part 1. R4.2, p7c-26} $$ $$ \displaystyle y_n(x)=y_{h,n}(x)+y_{p,n}(x) $$ $$ \displaystyle \text{For each of n=3,5,9, redisplay the expressions for } $$ $$ \displaystyle \text{the 3 functions } y_{p,n}(x), y_{h,n}(x), y_n(x), \text{ and plot these } $$ $$ \displaystyle \text{3 functions separately over the interval } \left[ 0,20 \pi \right]. $$

$$ \displaystyle \text{Exact solution: } y(x) = y_h(x)+y_p(x) $$ $$ \displaystyle \text{Redisplay the expressions for } y_p(x), y_h(x), y(x) $$

$$ \displaystyle \text{Superpose each of the above plot with that of the}$$ $$ \displaystyle \text{exact solution.} $$

$$ \displaystyle \text{Part 2. R4.3, p7c-28} $$ $$ \displaystyle \text{Understand and run the TA's code to produce} $$ $$ \displaystyle \text{a similar plot, but over a larger interval } \left[ 0, 10 \right]. \text{ Do } $$ $$ \displaystyle \text{zoom-in plots about points } x = -0.5, 0, +0.5 \text{ and } $$ $$ \displaystyle \text{comment on the accuracy of different approximations.} $$

$$ \displaystyle \text{Part 3. R4.4, p7c-29} $$ $$ \displaystyle \text{Understand and run the TA's code to produce} $$ $$ \displaystyle \text{a similar plot, but over a larger interval } \left[ 0.9, 10 \right], \text{ and } $$ $$ \displaystyle \text{for } n=4,7. \text{ Do zoom-in plots about } x = 1, 1.5, 2, 2.5 $$ $$ \displaystyle \text{and comment on the accurace of the approximations.} $$

Solution
$$ \displaystyle \text{Part 1.}$$

$$ \displaystyle \text{n = 3} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,3}(x)=-9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4 \cdots $$ $$ \displaystyle -0.0990x^3-0.3984x^2-0.3984x-0.1992 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,3}(x)=2e^x-0.8008e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{3}(x)=2e^x-0.8008e^{2x}-9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5 \cdots $$ $$ \displaystyle -0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$

$$ \displaystyle \text{n = 5} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,5}(x)=-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^9 \cdots $$ $$ \displaystyle -4.6503*10^{-6}x^8-1.1781*10^{-4}x^7-0.0011x^6-0.0033x^5 \cdots $$ $$ \displaystyle -0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,5}(x)=2.0001e^x-0.8001e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{5}(x)=2.0001e^x-0.8001e^{2x}-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10} \cdots $$ $$ \displaystyle -1.0334*10^{-6}x^9-4.6503*10^{-6}x^8-1.1781*10^{-4}x^7-0.0011x^6 \cdots $$ $$ \displaystyle -0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$

$$ \displaystyle \text{n = 9} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,9}(x)=-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17} \cdots $$ $$ \displaystyle -8.9615*10^{-15}x^{16}-4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14} \cdots $$ $$ \displaystyle -6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11} \cdots $$ $$ \displaystyle -2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8} \cdots $$ $$ \displaystyle -1.1904*10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4 \cdots $$ $$ \displaystyle -0.1000x^3-0.4000x^2-0.4000x-0.2000 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,9}(x)=2e^x-0.8e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{9}(x)=2e^x-0.8e^{2x}-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18} \cdots $$ $$ \displaystyle -1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}-4.5405*10^{-13}x^{15} \cdots $$ $$ \displaystyle -9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12} \cdots $$ $$ \displaystyle -1.5021*10^{-8}x^{11}-2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9} \cdots $$ $$ \displaystyle -4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^7-0.0011x^6 \cdots $$ $$ \displaystyle -0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 \cdots $$

$$ \displaystyle \text{Exact Overall solution: } $$ $$ \displaystyle y(x)=1.5e^x-0.8e^{2x}+0.3 \cos x+0.1 \sin x $$

$$ \displaystyle \text{Part 2.}$$ $$ \displaystyle \text{The following plot displays the Taylor series of } log(1+x) $$ $$ \displaystyle \text{ about } \hat{x} = 0.$$

$$ \displaystyle \text{The zoom-in about } x = -0.5 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see the plot broken, showing the inaccuracy beyond this point. }$$

$$ \displaystyle \text{The zoom-in about } x = 0 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see that the plot is very exact, displaying the accuracy at x = 0.}$$

$$ \displaystyle \text{The zoom-in about } x = 0.5 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see that the plot has begun to veer off the exact solution.}$$

$$ \displaystyle \text{Part 3.}$$ $$ \displaystyle \text{The following plot displays the Taylor series of } log(1+x) $$ $$ \displaystyle \text{ about } \hat{x} = 1.$$

$$ \displaystyle \text{The zoom-in about } x = 1 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 1.5 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 2 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 2.5 \text{ is displayed below. }$$

$$ \displaystyle \text{It is noted that the plots begin to diverge from the exact}$$ $$ \displaystyle \text{solution beginning (ever so slightly) at x = 1. }$$

title: User:Egm4313.s12.team8.colocar/R6.5


text:
User:Egm4313.s12.team8.koester/R6.3

Problem
K 2011 p. 491 numbers 15 and 17

-State whether the function is even, odd, or neither

-Find it's Fourier series

-Plot the truncated Fourier series for n = 2, 4, and 8

Solution to #15
$$ \displaystyle f(x)=-x-\pi $$ on $$ \displaystyle (-\pi,-\pi/2) $$

$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi/2,\pi/2) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (\pi/2,\pi) $$

This function is an odd function.

This is proven by the fact that $$ \displaystyle f(-x)=-f(x) $$

Since this is an odd function and of period $$ \displaystyle 2\pi $$ its Fourier series reduces to the Fourier sine series shown here:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty} b_n\sin(nx) $$

Where the coefficient equals:

$$ \displaystyle b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle b_n=\frac{2}{\pi}(\int_{0}^{\pi/2}x\sin(nx)dx + \int_{\pi/2}^{\pi}(-x+\pi)\sin(nx)dx) $$

$$ \displaystyle b_n=\frac{2}{\pi}(\int_{0}^{\pi/2}x\sin(nx)dx - \int_{\pi/2}^{\pi}x\sin(nx)dx + \int_{\pi/2}^{\pi}\pi\sin(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle b_n=\frac{2}{\pi}(\frac{2\sin(\frac{\pi}{2}n)}{n^2}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty}(\frac{4\sin(\frac{\pi}{2}n)}{\pi n^2})\sin(nx) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x+\frac{1}{25}\sin5x-...) $$

If n=2

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x) $$

If n=4

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x) $$

If n=8

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x+\frac{1}{25}\sin5x-\frac{1}{49}\sin7x) $$

Solution to #17
$$ \displaystyle f(x)=x+1 $$ on $$ \displaystyle (-1,0) $$

$$ \displaystyle f(x)=-x+1 $$ on $$ \displaystyle (0,1) $$

This function is an even function.

This is proven by the fact that $$ \displaystyle f(-x)=f(x) $$

Since this is an even function and of period 2 its Fourier series reduces to the Fourier cosine series shown here:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty} a_n\cos(n\pi x) $$

Where the coefficients are equal to:

$$ \displaystyle a_0=\int_{0}^{1}f(x)dx $$

$$ \displaystyle a_n=2\int_{0}^{1}f(x)\cos(n\pi x)dx $$

$$ \displaystyle a_0=\int_{0}^{1}(1-x) $$

$$ \displaystyle a_0=\frac{1}{2} $$

$$ \displaystyle a_n=2\int_{0}^{1}(1-x)\cos(n\pi x)dx $$

$$ \displaystyle a_n=2(\int_{0}^{1}\cos(n\pi x)dx - \int_{0}^{1}x\cos(n\pi x)dx ) $$

Using integration by parts and evaluating on the interval:

$$ \displaystyle a_n=2(\frac{1-\cos(n\pi)}{n^{2} \pi^{2}}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\frac{1}{2} + \sum_{n = 1}^{\infty} \frac{2-2\cos(n\pi)}{n^{2} \pi^{2}} \cos(n\pi x) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x+ \frac{1}{25}\cos5\pi x+...) $$

If n=2

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x) $$

If n=4

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x) $$

If n=8

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x+ \frac{1}{25}\cos5\pi x+ \frac{1}{49}\cos7\pi x) $$

Plots
15. Matlab Code

>> x=linspace(-pi,pi)

>> y2=(4/pi)*sin(x)

>> y4=(4/pi)*(sin(x)-(1/9)*sin(3*x))

>> y8=(4/pi)*(sin(x)-(1/9)*sin(3*x)+(1/25)*sin(5*x)-(1/49)*sin(7*x))

>> plot(x,y2,x,y4,x,y8)

Where y2 is blue, y4 is green and y8 is red

17. Matlab Code

>> x=linspace(-1,1)

>> f2=.5+(4/pi^2)*cos(pi*x)

>> f4=.5+(4/pi^2)*(cos(pi*x)+(1/9)*cos(3*pi*x))

>> f8=.5+(4/pi^2)*(cos(pi*x)+(1/9)*cos(3*pi*x)+(1/25)*cos(5*pi*x)+(1/49)*cos(7*pi*x))

>> plot(x,f2,x,f4,x,f8)

Where f2 is blue, f4 is green and f8 is red

title: User:Egm4313.s12.team8.koester/R6.3


text:
Portal:Modern Hebrew/Department News/News Archives

The following is an archives for the Modern Hebrew Department.


text: File:Maxima test.20120411.pdf

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Template:Modern Hebrew/Staff Members

The following are staff members at the Modern Hebrew Department. If you would like to be a staff member at the departments add yourself below. Staff members are users who actively edit, contribute, and maintain the department. You should also be available for other users to ask questions.
 * Readyokaygo | Administrator and Teacher, Department of Modern Hebrew.

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TAO/Working with Volunteer Instructors/Do’s & Don’t’s

Results of a brainstorming session with Kai Grabenhorst (SeniorWeb NL), Alfons Bühlmann (Seniorweb CH) and Elvira Schmidt (Wikimedia DE):

Do's

 * to pick them up where their current interests are (e.g. hobby clubs)
 * to give seniors the impression to feel themselves at the centre of society (up to date & modern)
 * catchy, informative and clear website

Don't's

 * not to emphasize membership too much (attachment has a negative image)
 * not to raise false expectations

Do's

 * interesting newsletter
 * training
 * community management (planning and organization of concrete activities)
 * animation and guidance
 * to address them personally
 * to propose combinations and links to related activities
 * to help them overcome any inhibitions

Don't's

 * not to carry internal conflicts to the outside
 * fights vs. factual differences
 * illoyality among staff members
 * no time to get oneself accustomed
 * not to let somebody go without taking care of him/her
 * charge for services that are too expensive or too cheap


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Talk:Introduction to Computers/What is a computer

There is a subtle difference between data and information. Data is raw, unformatted, possibly even unreadable, and generally cannot be used directly. Information is data processed in some way to become readable and therefore 'useful'.


text:
Talk:Motherboard

Ports
Older serial ports (eg RS232) were used for printers before they were used for mice. Could also have included blue tooth or infra-red for local networking/communications.

Does this page have a horizontal scroll for anyone else?
Does this page have a horizontal scroll for anyone else? Maybe this page should not have a horizontal scroll, if any of you see that also.... Michael Ten (discuss • contribs) 17:23, 22 October 2022 (UTC)

 * I don't see any horizontal scrolling. I tried narrowing the width of the browser. Still no scroll. -- Dave Braunschweig (discuss • contribs) 21:19, 22 October 2022 (UTC)

title: Talk:Motherboard


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User talk:Rev RSM

A gentle middle aged man with quest for excellence in all pursuit.

 Hello Rev RSM, 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! Jtneill - Talk - c 10:42, 11 April 2012 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: User talk:Rev RSM


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TAO/Handbook

For expanding the list click "show".

You can also download a “ready for printing” version of the handbook, it always represents the current state of the work done on these pages.

title: TAO/Handbook


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TAO/Community of Practice

Third Age Online - Community of Practice

title: TAO/Community of Practice


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User talk:M'kay M'kay/myWork 2

week 2 & 3 p2 grading 1sfoerster (talk) 11:58, 11 April 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 4 p2 grading 1sfoerster (talk) 18:21, 17 April 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 2 p3 grading 1sfoerster (talk) 16:51, 1 May 2012 (UTC)
 * 10/10/50 .. very good video and link to the tools to purchase, I am putting in a purchase order

week 3 p3 grading 1sfoerster (talk) 13:07, 8 May 2012 (UTC)
 * 0/0/0 ... the riveting tool came in ... where are you documenting your work on the unicycle ... making the wheel off center?

title: User talk:M'kay M'kay/myWork 2


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User talk:Wesley Rodriguez/enes100/My Work 2.0

week 2 and 3 p2 1sfoerster (talk) 12:25, 11 April 2012 (UTC)
 * 0/0/0 .. nothing to grade

week4 p2 1sfoerster (talk) 14:28, 18 April 2012 (UTC)
 * 0/0/0 .. nothing to grade


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University of Florida/Egm4313/s12.team11.gooding/R6

Problem Statement
Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.

Solution
15.

$$f(x)=\left\{\begin{matrix} -x-\pi, -\pi<x<-\pi/2\\ x, -\pi/2<x<\pi/2\\ \pi-x, \pi/2<x<\pi \end{matrix}\right.$$

Since $$f(-x)=-f(x)$$, the function is odd.

$$a_0=(1/\pi)*[-1 \int_{-\pi}^{-\pi/2}(x+\pi)dx+ \int_{-\pi/2}^{\pi/2}(x)dx+ \int_{\pi/2}^{\pi}(\pi-x)dx]$$ After simplification, $$a_0=0$$. Using Equation (5) on page 490, the exapansion becomes; $$f(x)=8k/\pi^2[sin(\pi*x/L)-sin(3\pi*x/L)/9...)$$ With $$L=\pi,k=\pi/2$$ this becomes; $$f(x)=4/\pi(sinx-sin(3x)/9+sin(5x)/25-sin(6x)/36....)$$ for n is "odd". This means when n is even, $$f(x) =0$$

17. $$f(x)=1-|x|, -1\leq x \leq1$$

$$f(x)=\left\{\begin{matrix} 1+x-\pi, -1<x<0\\ 1-x, 0<x<1\\ \end{matrix}\right.$$

$$a_0=1/2[\int_{-1}^{0}(1+x)dx + \int_{0}^{1}(1-x)dx]$$ $$a_0=1/2[1-1/2+1-1/2]$$ $$a_0=1/2$$

$$a_n=1/1[\int_{-1}^{0}(1+x)cos(n\pi*x/1)dx+ \int_{0}^{1}(1-x)cos(n\pi*x/1)dx]$$ Simplifying results in; $$a_n=2[1/(n^2\pi^2)_(-1)^n/(n^2\pi^2)]$$ So, $$ a_n=4/(n^2\pi^2)$$ when n is odd. $$ b_n=1/1[\int_{-1}^{0}(1+x)sin(n\pi*x/1)dx+ \int_{0}^{1}(1-x)sin(n\pi*x/1)dx]$$ Simplifying results in; $$b_n=0$$ $$f(x)=1/2+ \sum_{n=0}^{\infty }4/(n^2\pi^2)*cos(n\pi*x)$$

Since $$a_n=0$$ for even n, $$f(x)$$ becomes $$f(x)=1/2$$ for even n.


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University of Florida/Egm4313/s12.team8.dupre/R6.1

Problem Statement
Given $$\displaystyle f(x)=a_{0}+\sum_{n=1}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]$$: (1) Find the (smallest) period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$. (2) Show that these functions also have period $$\displaystyle p $$. (3) Show that the constant $$\displaystyle a_{0} $$ is also a periodic function with period $$\displaystyle p $$.

Solution (1)
We know that the period of a normal $$\displaystyle sin(x) $$ or $$\displaystyle cos(x) $$ is $$\displaystyle 2\pi $$. When there are values or variables being multiplied by this $$\displaystyle x $$ variable, the period becomes $$\displaystyle 2\pi $$ divided by the values or variables. We know that: $$\displaystyle \omega=\frac{\pi }{L}=\frac{2 \pi}{p}$$    (6.1.1) Where $$\displaystyle p $$ is the period. Using this relation, along with our variable $$\displaystyle n $$, we can solve for the period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ as follows: $$\displaystyle p=\frac{2\pi }{\tfrac{n\pi }{L}}$$ $$\displaystyle p=\frac{L\pi}{n} $$    (6.1.2) Since the period will be smallest at $$\displaystyle n=1$$, plugging into equation (6.1.2) shows that the smallest period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ is: $$\displaystyle p=L\pi $$    (6.1.3)

Solution (2)
We are also given that: $$\displaystyle \omega =\frac{2\pi}{p}$$    (6.1.4) Using this relation, we can solve for the period again as follows: $$\displaystyle p=\frac{2\pi}{\frac{n2\pi}{p}}$$ $$\displaystyle p=\frac{p}{n}$$    (6.1.5) We know that the period is smallest at $$\displaystyle n=1$$, and plugging this value into (6.1.5) proves that: $$\displaystyle p=p $$    (6.1.6)

Solution (3)
We know that, starting at 0: $$\displaystyle \widetilde{a}_{0}=\frac{1}{2L}\int_{0}^{2L}f(\widetilde{x})d\widetilde{x} $$    (6.1.7) Where the period is represented by 0 to 2L. We are also given that: $$\displaystyle \frac{\pi}{L}=\frac{2\pi}{p} $$    (6.1.8) Rearranging (6.1.8) allows us to solve for L: $$\displaystyle L=\frac{p}{2}$$     (6.1.9) Multiplying (6.1.9) by 2 allows us to find the period of $$\displaystyle a_{0}$$, as follows: $$\displaystyle 2L=p$$    (6.1.10) This shows that $$\displaystyle a_{0}$$ is indeed a periodic function with a period of $$\displaystyle p$$. This also shows that at any given $$\displaystyle x$$ value or period throughout the periodic function, $$\displaystyle a_{0}$$ holds its constant value.


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TESOL/4-3-2 Exercises

A common fluency-building technique in used in teaching English as a second language is to use a 4/3/2 exercise, where a student gives a talk in four minutes, repeats it in 3 minutes to another partner, and repeats it again in 2 minutes to a third partner.

The purpose is to build fluency by reinforcing skills that the student already possesses under time pressure. Partners should be different each time so that the speaker does not feel tempted to add new information and so that there is pressure to convey a message to an audience.

Conveying a message successfully each time in the time allotted constitutes success.

A 4/3/2 exercise with small groups
This activity cannot be done with under three students without the students repeating their story to the same person. With groups of three to five students, the following patterns can be used.

In this case, the proportions of time used in 4/3/2 are used, but the amount of time for the initial telling is shortened from 4 minutes to 130 seconds. The second telling is 100 seconds, and the third is 60.

Stage 1
A and C speak. Both have 130 seconds.

Stage 2
A and B speak. A has 100 seconds. B has 130 seconds.

Stage 3
B and C speak. Both have 100 seconds.

Stage 4
A and C speak. Both have 60 seconds.

Stage 5
It's B's lucky day, because they get to speak to everyone. B has 60 seconds.

4 students
4 students is an ideal number for a 4/3/2 exercise in small groups because every student gets to talk to every other student, the times per student at each stage are uniform, and the teacher can focus on the class as a whole at each step.

Stage 1
A and C speak. Both have 130 seconds.

Stage 2
B and D speak. Both have 130 seconds.

Stage 3
A and B speak. Both have 100 seconds.

Stage 4
C and D speak. Both have 100 seconds.

Stage 5
A and D speak. Both have 60 seconds.

Stage 6
B and C speak. Both have 60 seconds.

Stage 1
A, C, and E speak. Each have 130 seconds.

Stage 2
A, B, and D speak. B and D have 130 seconds but A only has 100.

Stage 3
B, C, and E speak. Each has 100 seconds.

Stage 4
A, B, and D speak. D has 100 seconds but A and B only have 60 seconds.

Stage 5
C, D, and E speak. Each has 60 seconds.

title: TESOL/4-3-2 Exercises


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User:Thx1147

helllo world!

title: User:Thx1147


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User:Egm4313.s12.team8.kyle/R6.4

Problem Statement
Consider the L2-ODE-CC with the window function from R6.2, $$f(x)$$, as excitation:
 * $$y'' - 3y' + 2y = r(x)$$
 * $$r(x) = f(x)$$

And the initial conditions:
 * $$y(0) = 1,\;y'(0) = 0$$

Part 1
Find $$y_n(x)$$ such that:
 * $$y_n'' + a y_n' + b y_n = r_n(x)$$

with the same initial conditions.

Plot $$y_n(x)$$ for $$n=2,\,4,\,8$$, for $$x \in \left[ 0, \, 10 \right]$$.

Part 2
Use the MATLAB command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Part 1
From R6.2, we know that our excitation functions are given by:
 * $$r_0(x) = f_0(x) = \frac{A}{2}$$
 * $$r_1(x) = f_1(x)= \frac{A}{2} + \frac{A}{\pi}\sin (\frac{\pi (x-.25)}{2})$$

Case $$n = 0$$
Setting $$n=0$$ yields the following:

$$r_0(x) = \frac12 A$$

So the ODE takes the form:

$$y''(x)-3y'(x)+2y(x)=A/2$$

Inputting into Wolfram Alpha with the initial conditions:

$$y(0)=1$$ and $$y'(0)=0$$

Yields the solution:

$$ y_{0}(x) = \frac14 (A (e^x-1)^2-4 e^x (e^x-2))$$

Case $$n = 1$$
Setting $$n=1$$ yields the following:

$$r_1(x) = \frac12 A + \frac{A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2}$$

So the ODE takes the form:

$$y''(x)-3y'(x)+2y(x)=\frac12 A + \frac{A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2}$$

Inputting into Wolfram Alpha with the initial conditions:

$$y(0)=1$$ and $$y'(0)=0$$

Yields the solution:

$$ y_{1}(x) = -0.598093 A e^x+0.283756 A e^{2x}+0.019468 A sin(1.5708 x)+0.0643367 A cos(1.5708 x)+0.25 A+2e^x-e^{2 x}$$

Part 2
Using the following Matlab function and inputs allows us to compare the analytical and numerical solutions:

 * [[Image:Egm4313.s12.team8.r6.4.y0.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y1.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y+y0+y1.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y+y0+y1.zoomed.png]]

The zoomed plot shows, as expected, an increase in accuracy as n increases from 0 to 1.

title: User:Egm4313.s12.team8.kyle/R6.4