text:
University of Florida/Egm4313/s12.team1/R6

Statement
Find the (smallest) period of $$ \cos n \omega x $$ and $$ \sin n \omega x $$. Show that these functions also have period $$ p $$. Show that the constant $$ a_0 $$ is also a periodic function with period $$ p $$.

Solution
First, we identify the property of periodic functions, as shown on p9-2 of the notes, equation (1)

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

$$\displaystyle{f(x+np)=f(x)}$$ (1.0)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

So for $$ \cos n \omega x $$, the following determines the smallest period.

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

$$\displaystyle{\cos (n\omega x})$$ (1.1) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{\cos (n\omega x + 2 \pi)}$$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\cos[n\omega ( x + \frac{2 \pi}{n \omega})]}$$ (1.3) This means the smallest period is $$ \frac{2 \pi}{n \omega} $$.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For $$ \sin n \omega x $$, the following determines the smallest period.

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

$$\displaystyle{\sin (n\omega x)}$$ (1.4) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{\sin (n\omega x + 2 \pi)}$$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\sin[n\omega ( x + \frac{2 \pi}{n \omega})]}$$ (1.6) This means the smallest period is $$ \frac{2 \pi}{n \omega} $$, just as for its cosine equivalent.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From the notes (page 9-5, (1)), it can be stated that $$ \omega = \frac{2 \pi}{p} $$ where $$ p $$ is the period. Both of these functions have $$ \frac{2 \pi}{n \omega} $$ as their period. If we put in 1 for n, and $$ \frac{2 \pi}{p} $$ for $$\omega $$, the period simplifies to simply $$ \displaystyle p$$.

As for the constant $$ a_0 $$, a constant is, almost by definition, a periodic function. The property of periodic functions as referenced in (1.0) supports this.

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

$$\displaystyle{f(x+np)=f(x)}$$ (1.0) For a function of f that is a constant, x is not a factor effecting its value. This means, no matter what is added to x, the function will return the same constant number. This added value can just as well be $$ p $$, and thus, $$ a_o $$ is a periodic function with period $$\displaystyle p $$.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Author
Solved and Typed By -Egm4313.s12.team1.silvestri (talk) 02:19, 7 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.rosenberg (talk) 01:19, 11 April 2012 (UTC)

Statement
Kreyszig pg 491 pbs. 11,12 Is the given function odd or even or neither even nor odd? Show details of your work. -11)$$ f(x) = x^2 (-1 < x <1), p=2 $$ -12)$$ f(x) = 1- \frac{x^2}{4} (-2 < x <2), p=4 $$ -

Continued)Find the Fourier series expansion for $$ f(x) $$ on p.9-8 as follows: 1. Develop the Fourier series expansion of $$f(\bar{x})$$ Plot $$f(\bar{x})$$ and the truncated Fourier series $$f_n(\bar{x})$$

dsfawf

for n=0,1. Observe the values of $$f(\bar{x})$$ at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$ f(x) $$. Hint: $$f(\bar{x})$$ is an even function; what would become of the coefficients $$\bar{b}_k$$? 2. Do the same as above, but using $$f(\tilde{x})$$ to obtain the Fourier series expansion of $$ f(x) $$; compare to the result obtained above. Hint: Transform $$f(\tilde{x})$$ into an odd function, and expand this odd function into Fourier series; observe the coefficients of this series.

Solution
'''Part 1: K 2011 p.491 pb. 11'''
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=x^2, (-1<x<1), p=2 $$     (2.0) The preceding function is an even function. Even functions are functions where $$f(-x)=f(x)\!$$. Odd functions on the other hand are functions where $$f(-x)=-f(x)\!$$. The Fourier series of even and odd functions can be found more easily than the Fourier series of a function that is neither even nor odd. Normal Fourier series contain both sine and cosine terms. Because of the nature of even functions, the coefficient of all sine terms in their series is zero. Likewise, all cosine terms in odd functions have a coefficient of zero. Therefore its Fourier series expansion will be of the form:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=a_0+\sum_{n=1}^{\infty}a_n \cos \frac{n\pi}{L}x $$     (2.1) Where:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_0=\frac{1}{L} \int_0^Lf(x)dx $$     (2.2) And
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_n=\frac{2}{L} \int_0^Lf(x) \cos \frac{n\pi x}{L}dx $$     (2.3) To get the constant term of the expansion, equation (2.0) will be substituted into equation (2.2) with L=1:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_0=\frac{1}{1} \int_0^1x^2dx=\left [ \frac{x^3}{3} \right ]^1_0=\frac{1}{3} $$     (2.4) To get the coefficients of the rest of the terms in the expansion, equation (2.0) will be substituted into equation (2.3) with L=1:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_n=\frac{2}{1} \int_0^1x^2 \cos \frac{n\pi x}{1}dx=2\left [ \frac{(\pi^2n^2x^2-2)\sin (\pi n x)+2\pi n x \cos(\pi nx)}{\pi^3n^3} \right ]^1_0 $$     (2.5) Simplifying, this yields:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_n=\frac{4}{n^2 \pi^2} \cos n\pi $$     (2.6) Therefore, the Fourier expansion of f(x) is:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="1"

$$  \displaystyle x^2=\frac{1}{3}+\sum_{n=1}^{\infty}\left (\frac{4}{n^2 \pi^2} \cos n\pi \right ) \cos n\pi x $$ (2.7) '''Part 2: K 2011 p.491 pb. 12'''
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=1-\frac{x^2}{4}, (-2<x<2), p=4 $$     (2.8) The preceding function is an even function. Therefore, equations (2.1)-(2.3) are applicable. To get the constant term of the expansion, equation (2.8) will be substituted into equation (2.2) with L=1:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle a_0=\frac{1}{2} \int_0^21-\frac{x^2}{4}dx= \frac{1}{2}\left[ x-\frac{x^3}{12} \right ]^2_0=\frac{4}{3} $$     (2.9) To get the coefficients of the rest of the terms in the expansion, equation (2.0) will be substituted into equation (2.3) with L=1:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle a_n=\frac{2}{2} \int_0^2\left (1-\frac{x^2}{4} \right ) \cos \frac{n\pi x}{2}dx=\left [ \frac{(8-\pi^2n^2(x^2-4))\sin\frac{n\pi x}{2}-4n\pi x\cos\frac{n\pi x}{2}}{2\pi^3n^3} \right ]^2_0 $$     (2.10) Simplifying, this yields:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle a_n=\frac{-4\cos(n\pi)}{n^2 \pi^2} $$     (2.11) Therefore, the Fourier expansion of f(x) is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle 1-\frac{x^2}{4}=\frac{4}{3}+\sum_{n=1}^{\infty}\left (\frac{-4\cos(n\pi)}{n^2 \pi^2} \right ) \cos \frac{n\pi x}{2} $$     (2.12) Part 3 The Fourier series expansion of the function f(x) in Figure 1 can be found in one of two ways. The first way involves transforming x in the following way:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \bar x=x-\frac54 $$     (2.13) The resulting function is now an even function that alternates between the constant value A and zero and has period 4. Using the same methods used above, the Fourier series of the function can be found. The constant coefficient of the series is found by:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \bar a_0=\frac12 \int_0^2f(x)dx=\frac12 \int_0^1Adx+\frac12 \int_1^20dx=1/2\left [ Ax \right ]_0^1=\frac A2 $$ (2.14) Note that because the function is discontinuous at the midpoint, the integral must be split into two different integrals. The coefficients of the cosine terms of the series can be found by:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \bar a_n=\frac22 \int_0^2f(x)\cos \frac{n\pi x}{2}dx= \int_0^1A\cos \frac{n\pi x}{2}dxdx+\frac12 \int_1^20dx=\left [ \frac{2A}{n\pi}\sin\frac{n\pi x}{2} \right ]_0^1=\frac{2A}{n\pi}\sin \frac{n\pi}{2} $$     (2.15) Therefore, the Fourier series for the transformed function is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle f(\bar x)=\frac A2 + \sum_{n=0}^{\infty} \left ( \frac{2A}{n\pi}\sin\frac{n\pi}{2} \right )\left ( \cos \frac{n\pi\bar x}{2} \right ) $$     (2.16) Through a change in variables, the Fourier series of the original function, f(x), can be found:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle f(x)=\frac A2 + \sum_{n=0}^{\infty} \left ( \frac{2A}{n\pi}\sin\frac{n\pi}{2} \right )\left ( \cos \frac{n\pi(x-1.25)}{2} \right ) $$     (2.17) The following figure shows a plot of $$f(\bar x)\!$$ with its truncated Fourier series for n=0,1.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Figure 2 Figure 3 shows the same plot with more values of n:

Note that all plots assume that A=1. It should be noted that the accuracy of the expansion increases with an increase in the number of terms in the expansion. Also, there is strong fluctuation at points of discontinuity. These fluctuations are apparent even at high values of n, but they decrease in size and move closer to the points of discontinuity. This is a property of Fourier series known as the Gibbs phenomenon. The other method of developing the Fourier series is to transform the function into an odd function. This can be done by transforming both the independent variable and overall function as follows:
 * {| style="width:100%" border="0"

$$  \displaystyle \tilde x=x-\frac14 $$     (2.18)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle f(\hat x)=f(\tilde x)-\frac A2 $$ (2.19) An odd function has a Fourier series expansion of the form:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle f(x)=\sum_{n=0}^{\infty}b_n\sin \frac{n\pi x}{L} $$     (2.20) Where:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_n=\frac2L\int_{0}^{L}f(x)\sin \frac{n\pi x}{L}dx $$     (2.21) Applying this to the transformed function, with L=2:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \hat b_n=\frac22\int_{0}^{2}f(x)\sin \frac{n\pi \hat x}{2}d\hat x=\int_{0}^{2}\frac A2\sin \frac{n\pi \hat x}{2}d\hat x=\left [ -\frac{A}{n\pi}\cos\frac{n\pi\hat x}{2} \right ]^2_0=\frac{A}{n\pi}(1-\cos n\pi) $$     (2.22) The Fourier series is therefore:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle f(\hat x)=\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi\hat x}{2} \right ) $$     (2.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle f(\tilde x)=\frac A2+\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi\tilde x}{2} \right ) $$     (2.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle f(x)=\frac A2+\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi(x-0.25)}{2} \right ) $$     (2.25) The following figure shows a plot of $$f(\tilde x)\!$$ with its truncated Fourier series for n=0,1.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Figure 4 Note that all plots assume A=1. The accuracy of the expansion as shown on the plot is expected given the low number of terms used in the expansion. As Figure 3 shows, if the number of terms used to form the plot in Figure 4 is increased, the accuracy of the expansion will greatly increase (but more slowly at points of discontinuity because of the Gibbs phenomenon). This level of accuracy is shown in Figure 5, which contains the expansion to 100 and 200 terms.

Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 07:10, 11 April 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri (talk) 18:30, 11 April 2012 (UTC)

Statement
Kreyszig pg 491 pbs 15,17 Is the given function even or odd or neither even not odd? Find its Fourier series. Show details of your work.

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

Solution
By definition, a function, $$ f(x)\!$$ is even if $$ f(-x) = f(x)\!$$, and its Fourier series reduces to a Fourier cosine series:
 * {| style="width:100%" border="0"

$$\displaystyle{f(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}cos{\frac{n\pi}{L}}x}$$ (3.0)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

A function, $$ f(x)\!$$ is odd if $$ f(-x) = -f(x)\!$$, and its Fourier series reduces to a Fourier sine series:
 * {| style="width:100%" border="0"

$$\displaystyle{f(x) = \sum_{n=1}^{\infty}b_{n}sin{\frac{n\pi}{L}}x}$$ (3.1) -15) By inspection, it is evident that the given function in problem 15 is odd because for every $$ x\!$$ value that is negative, $$ f(x)\!$$ is negative. Therefore, $$ f(-x) = -f(x)\!$$.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following are the defining functions for the given graph:
 * {| style="width:100%" border="0"

$$\displaystyle{f(x) = \begin{cases} -\pi - x & \text{ if } x= -\pi < x < -\frac{\pi}{2}\\ x & \text{ if } x= -\frac{\pi}{2} < x < \frac{\pi}{2}\\ \pi - x & \text{ if } x= \frac{\pi}{2} < x < \pi \end{cases}}$$ (3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Because the function is odd, we use equation (3.1) to develop the Fourier series, where the period is $$2\pi\!$$ and $$ L\!$$ equals $$\pi\!$$:
 * {| style="width:100%" border="0"

$$\displaystyle{f(x) = \sum_{n=1}^{\infty}b_{n}sin{\frac{n\pi}{L}}x = \sum_{n=1}^{\infty}b_{n}sin(nx)}$$
 * style="width:95%" |
 * style="width:95%" |
 *

(3.3)
 * }

Now we need to find $$ b_{n}\!$$, which is of the following form for odd functions:
 * {| style="width:100%" border="0"

$$\displaystyle{b_{n} = \frac{2}{L}\int_{0}^{L}f(x)\sin{\frac{n{\pi}x}{L}}dx}$$ (3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{b_{n} = \frac{2}{\pi}\int_{0}^{\pi}sin(nx)dx = \frac{2}{\pi}[\int_{0}^{\pi/2}xsin(nx)dx + \int_{\pi/2}^{\pi}(\pi-x)sin(nx)dx]}$$ (3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{b_{n} = \frac{2}{\pi}[[\frac{sin(nx)-nxcos(nx)}{n^2}]_{0}^{\pi/2}+[\frac{-sin(nx)+n(\pi-x)cos(nx)}{n^2}]_{\pi/2}^{\pi}]}$$     (3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating and simplifying the definite integral gives the following:

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

$$\displaystyle{b_{n} = \frac{2}{n^2\pi}[2sin(\frac{n\pi}{2})-sin(n\pi)]}$$ (3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating equation (3.7) at $$ n = 1, 2, 3, 4, 5, 6, 7, 8\!$$

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

$$\displaystyle{b_{1} = \frac{4}{\pi}, b_{2} = 0, b_{3} = -\frac{4}{9\pi}, b_{4} = 0, b_{5} = \frac{4}{25\pi}, b_{6} = 0, b_{7} = -\frac{4}{49\pi}, b_{8} = 0}$$ (3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus:

 * {| style="width:100%" border="1"

$$ f(x) = \frac{4}{\pi}(sinx - \frac{1}{9}sin3x + \frac{1}{25}sin5x - \frac{1}{49}sin7x +...) $$ (3.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plots of truncated Fourier series: n = 2: n = 4: n = 8:

-17) By inspection, it is evident that the given function in problem 17 is even because for every $$ x\!$$ value that is negative, $$ f(x)\!$$ is positive. Therefore, $$ f(-x) = f(x)\!$$.

The following are the defining functions for the given graph:
 * {| style="width:100%" border="0"

$$\displaystyle{f(x)=\begin{cases}x+1 & \text{ if } -1<x<0\\ 1-x & \text{ if } 0<x<1 \end{cases}}$$ (3.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Because the function is even, we use equation (3.0) to develop the Fourier series, where the period is $$2\!$$ and $$ L = 1\!$$:

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

$$\displaystyle{f(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}cos{\frac{n\pi}{L}x} = a_{0} + \sum_{n=1}^{\infty}a_{n}cos(n\pi x)}$$ (3.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we need to find $$a_{0}\!$$ using Euler's Formula:
 * {| style="width:100%" border="0"

$$\displaystyle{a_{0} = \frac{1}{2L}\int_{-L}^{L}f(x)dx}$$ (3.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{a_{0} = \frac{1}{2}\int_{-1}^{1}f(x)dx = \frac{1}{2}[\int_{-1}^{0}(x+1)dx + \int_{0}^{1}(1-x)dx]}$$ (3.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{a_{0} = \frac{1}{2}[[\frac{x^2}{2}+x]_{-1}^{0}+[x-\frac{x^2}{2}]_{0}^{1}]}$$     (3.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating the definite integral in equation (3.14) yields $$ a_{0}\!$$:

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

$$\displaystyle{a_{0} = \frac{1}{2}}$$ (3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we find $$ a_{n}\!$$ using Euler's Formula:

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

$$\displaystyle{a_{n} = \frac{1}{L}\int_{-L}^{L}f(x)cos{\frac{n{\pi}x}{L}}dx = \int_{-1}^{1}f(x)cos(n\pi x)dx}$$ (3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{a_{n} = \int_{-1}^{0}(x+1)cos(n\pi x)dx + \int_{0}^{1}(1-x)cos(n\pi x)dx}$$ (3.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{a_{n} = [\frac{\pi n(x+1)sin(\pi nx) + cos(\pi nx)}{n^2\pi^2}]_{-1}^{0} + [-\frac{\pi n(x-1)sin(\pi nx)+cos(\pi nx)}{n^2\pi^2}]_{0}^{1}}$$ (3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating the definite integral in equation (3.18) yields the following equation for $$a_n\!$$:

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

$$\displaystyle{a_{n} = \frac{2(1-cos(\pi))}{n^2\pi^2}}$$ (3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating equation (3.19) at $$ n = 1, 2, 3, 4, 5, 6, 7, 8\!$$ yields:

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

$$\displaystyle{a_{1} = \frac{4}{\pi^2}, a_{2} = 0, a_{3} = \frac{4}{9\pi^2}, a_{4} = 0, a_{5} = \frac{4}{25\pi^2}, a_{6} = 0, a_{7} = \frac{4}{49\pi^2}, a_{8} = 0}$$ (3.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus, the Fourier series is:

 * {| style="width:100%" border="1"

$$ f(x) = \frac{1}{2} + \frac{4}{\pi^2}(cos(\pi x) + \frac{1}{9}cos(3\pi x) + \frac{1}{25}cos(5\pi x) + \frac{1}{49}cos(7\pi x) + ...) $$ (3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plots of truncated Fourier series: n = 2: n = 4: n = 8:

Author
Solved and Typed By ---Egm4313.s12.team1.wyattling (talk) 06:19, 11 April 2012 (UTC) Reviewed By -Egm4313.s12.team1.armanious (talk) 15:18, 11 April 2012 (UTC)

Statement
Consider the L2-ODE-CC (5) p7b-7 with the window function f(x) p.9-8 as excitation:
 * {| style="width:100%" border="0"

$$\displaystyle{y''-3y'+2y=r(x)}$$ $$r(x)=f(x)$$ (4.0) and the initial conditions
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{y(0)=1, y'(0)=0}$$ (4.1) 1. Find $$ y_n(x) $$ such that:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle{y''_n +ay'_n + by_n=r_n(x)}$$ (4.2) with the same initial conditions (4.1). Plot $$ y_n(x)$$ for n=2,4,8, for x in [0,10]. 2.Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution. Level 1:n=0,1
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Solution
First, we shift the excitation f(x) to the left by introducing a new independent variable, t. This allows the period to start at zero.
 * {| style="width:100%" border="0"

$$\displaystyle t=x-\frac{1}{4}$$ (4.3) The piecewise representation of the window function is now (in terms of t) as follows:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle f(t)=A$$ for $$\displaystyle t=\left[ 0,2 \right]$$ $$\displaystyle f(t)=0$$ for $$\displaystyle t=\left[ 2,4 \right]$$ (4.4) To find the Fourier transform, the period of oscillation is determined:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle p=2L=4$$ (4.5) And the frequency of oscillation:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle \omega =\frac{2\pi }{p}=\frac{\pi }{2}$$ (4.6) The general form of a Fourier transform is
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle f(t)={{a}_{0}}+\sum\limits_{1}^{n}{\left[ {{a}_{n}}\cos (n\omega t)+{{b}_{n}}\sin (n\omega t) \right]}$$ (4.7) The following equations are given values of the constants in (4.7), evaluated for the function in (4.4):
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{a}_{0}}=\frac{1}{2L}\int\limits_{0}^{2L}{f(t)dt}=\frac{1}{4}\int\limits_{0}^{2}{Adt}=\frac{A}{4}(2-0)=\frac{A}{2}$$ (4.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{a}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\cos (n\omega t)dt} =\frac{1}{2}\int\limits_{0}^{2}{A\cos (\frac{n\pi }{2}t)dt} =\frac{A}{n\pi }(\sin n\pi -\sin 0)$$ (4.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{b}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\sin (n\omega t)dt}=\frac{1}{2}\int\limits_{0}^{2}{A\sin (\frac{n\pi }{2}t)dt}=-\frac{A}{n\pi }(\cos n\pi -\cos 0)=\frac{A}{n\pi }(1-\cos n\pi )$$ (4.10) In their simplest forms:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{a}_{n}}=0$$ (4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{b}_{n}}=\left\{ 0,\frac{A}{\pi },0,\frac{A}{2\pi },0,\frac{A}{3\pi },0,\frac{A}{4\pi },...,0,\frac{2A}{n\pi } \right\}$$ (4.12) Plugging in r(x)=f(t) in (4.0),
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y''-3y'+2y=\frac{1}{2}+\frac{2}{n\pi }\sin (\frac{n\pi }{2}t)$$ (4.13) From the general form, a particular solution to (4.13) and its derivatives are as follows:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{y}_{n}}={{A}_{n}}\cos (\frac{n\pi }{2}t)+{{B}_{n}}\sin (\frac{n\pi }{2}t)$$ (4.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{y}_{n}}'=-{{A}_{n}}\frac{n\pi }{2}\sin (\frac{n\pi }{2}t)+{{B}_{n}}\frac{n\pi }{2}\cos (\frac{n\pi }{2}t)$$ (4.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{y}_{n}}''=-{{A}_{n}}\frac{4}\cos (\frac{n\pi }{2}t)-{{B}_{n}}\frac{4}\sin (\frac{n\pi }{2}t)$$ (4.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging in (4.14), (4.15), and (4.16) into (4.13), we find An and Bn in terms of another constant, Cn:

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

$$\displaystyle {{A}_{n}}=6n\pi {{C}_{n}}$$ (4.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{B}_{n}}=(8-{{n}^{2}}{{\pi }^{2}}){{C}_{n}}$$ (4.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{C}_{n}}={{\left[ \tfrac{1}{8}n\pi ({{(8-{{n}^{2}}{{\pi }^{2}})}^{2}}+{{(6n\pi )}^{2}}) \right]}^{-1}}$$ (4.19) After substituting t with (4.3), the solutions are shown for n=2,4,8.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{y}_{2}}={{y}_{1}}+{{y}_{2}}$$ (4.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{y}_{4}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}$$ (4.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {{y}_{8}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{y}_{5}}+{{y}_{6}}+{{y}_{7}}+{{y}_{8}}$$ (4.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Solution
The excitation for n=0 and n=1 are the same, because $$a_1=b_1=0$$.

The homogeneous solution to (4.0) is
 * {| style="width:100%" border="0"

$$\displaystyle {{y}_{h}}={{C}_{1}}{{\operatorname{e}}^{2t}}+{{C}_{2}}{{\operatorname{e}}^{t}}$$ (4.23) For $$r(x)=a_0$$, the particular solution is
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{y}_{p}}={{C}_{3}}$$ (4.24) Evaluating, we find that
 * style="width:95%" |
 * style="width:95%" |
 *

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

$${{C}_{1}}=-\tfrac{3}{4}$$ $${{C}_{2}}=\tfrac{3}{2}$$ $${{C}_{3}}=\tfrac{1}{4}$$ (4.25) Thus, the complete solution $$y_0$$ is
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ {{y}_{0}}=\tfrac{1}{4}-\tfrac{3}{4}{{\operatorname{e}}^{2(x-\tfrac{1}{4})}}+\tfrac{3}{2}{{\operatorname{e}}^{x-\tfrac{1}{4}}}$$ (4.26) Integrating $$y_0$$ using MATLAB's ode45 command, the following plot is obtained for y:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 00:00, 11 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.wyattling (talk) 14:02, 11 April 2012 (UTC)

Statement
Continuation of R4.2, R4.3, R4.4. R4.2,p.7c-26:$$y_n(x)=y_{h,n}(x) + y_{p,n}(x)$$ For each value of n=3,5,9, redisplay 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 $$ [0,20 \pi] $$. Exact solution: $$ y(x) = y_h(x) + y_p(x)$$ Redisplay the expressions for $$y_p(x),y_h(x), y(x)$$ Superimpose each of the above plots with that of the exact solution. R4.3,p.7c-28: Level 1: Understand and run the TA's code to produce a similar plot, but over a larger interval [0,10]. Do zoom-in plots about points $$ x=-0.5,0,+0.5 $$ and comment on the accuracy of different approximations. R4.4,p.7c-29: Level 1: 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.

R4.2,p.7c-26
This first part of the problem is to redo 4.2:

In order to find the overall solution for the L2-ODE-CC corresponding to the Taylor series expansion of the sine function, both the homogenous and particular solutions must be found. The homogenous equation can be found through this method:
 * {| style="width:100%" border="0"

$$  \displaystyle y''-3y'+2y=0 $$     (5.0)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \lambda ^2-3\lambda+2=0 \rightarrow \lambda = 1,2 $$     (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{h,n}(x)=C_1e^x+C_2e^{2x} $$     (5.2) Next, the excitation must be expanded to the desired n.  The following shows the excitation expanded to n=3, 5, and 9:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle r_3(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} $$     (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle r_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!} $$     (5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle r_9(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}-\frac{x^{15}}{15!}+\frac{x^{17}}{17!}-\frac{x^{19}}{19!} $$     (5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Next, the particular solution must be found. The particular solution will be of the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p,n}=\sum_{i=0}^{n}k_ix^i $$     (5.6) Using the derivation discussed in R4.1, a matrix equation in the form Ak = d can be found, where k is the matrix containing the coefficients of the particular solution and d is the matrix containing the coefficients of each power of x in the expansion of the excitation. The general formula for A is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \begin{bmatrix} b& a & 2 & &  & &\\ & b & 2a & 6 & & &\\ & & b & 3a & 12 & &\\ & &  & ... & ... & ... &\\ &  &  &  & b & a(n-1) & n(n-1)\\ & &  &  &  &b &an\\ & &  &  &  & &b \end{bmatrix} $$     (5.7) n=3 The matrix A is found as:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle A = \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\ & 2 & -6 & 6 &  &  &  & \\ &  & 2 & -9 & 12 &  & &  \\ &  &  & 2 &  -12& 20 &  & \\ &  &  &  & 2 & -15 & 30 & \\ &  &  &  &  & 2 & -18 & 42\\ &  &  &  &  &  & 2 &-21 \\ &  &  &  &  &  &  & 2 \end{bmatrix} $$     (5.8) Therefore, the matrix equation is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\ & 2 & -6 & 6 &  &  &  & \\ &  & 2 & -9 & 12 &  & &  \\ &  &  & 2 &  -12& 20 &  & \\ &  &  &  & 2 & -15 & 30 & \\ &  &  &  &  & 2 & -18 & 42\\ &  &  &  &  &  & 2 &-21 \\ &  &  &  &  &  &  & 2 \end{bmatrix} \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{1}{6}\\ 0\\ \frac{1}{120}\\ 0\\ -\frac{1}{5040} \end{bmatrix} $$     (5.9)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} -0.1992\\ -0.3984\\ -0.3984\\ -0.0990\\ -0.0078\\ -0.0031\\ -0.0010\\ -9.9206\times 10^{-5} \end{bmatrix} $$     (5.10) Therefore the particular solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,3}(x)=-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.11) And the overall solution for n=3 is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{3}(x)=C_1e^x+C_2e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.12) Using the initial conditions:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 1=C_1+C_2-0.1992$$ &  $$ 0=C_1+2C_2-0.3984 $$     (5.13) Solving yields $$C_1=2 $$  &  $$ C_2=-0.8008 \!$$. Therefore the overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle y_{3}(x)=2e^x-0.8008e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.14) n=5 The matrix equation is: Therefore the particular solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,5}(x)=-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-$$ $$0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (5.15) And the overall solution for n=5 is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{5}(x)=C_1e^x+C_2e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8$$ $$-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (5.16) Using the initial conditions:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving yields $$C_1=2.0001 \& C_2=-0.8001 \!$$. Therefore the overall solution is:
 * {| style="width:100%" border="1"

$$  \displaystyle y_{5}(x)=2.0001e^x-0.8001e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8$$ $$-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2 $$     (5.18) n=9 The matrix equation is: Therefore the particular solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,9}(x)=-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}$$ $$-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8$$ $$-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.19) And the overall solution for n=9 is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{9}(x)=C_1e^x+C_2e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}$$ $$-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9$$ $$-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.20) Using the initial conditions:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 1=C_1+C_2-0.2 \& 0=C_1+2C_2-0.4 $$     (5.21) Solving yields $$C_1=2 \& C_2=-0.8 \!$$. Therefore the overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle y_{9}(x)=2e^x-0.8e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-$$ $$8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}$$ $$-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9$$ $$-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.22) The following plot shows the overall solution of the ODE at n=3,5,9 over the domain [0,4π]: Figure 4.2-2
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The near perfect overlap between all three graphs shows that the approximations converge quickly for very low values of n.

Next, the exact ODE will be calculated to find the accuracy of the above approximations. The homogeneous solution is the same as the homogeneous solution above, only with different coefficients. The particular solution will be in the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p(x)=K\cos x + M\sin x $$ (5.23)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p'(x)=-K\sin x + M\cos x $$ (5.24)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p''(x)=-K\cos x - M\sin x $$ (5.25) Plugging these values into the ODE:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -K\cos x - M\sin x + 3K\sin x - 3M\cos x + 2K\cos x + 2M\sin x =\sin x $$ (5.26) Separating sine and cosine terms yield two linear equations which can be used to solve for the unknown coefficients:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle K-3M=0 $$ &  $$  3K+M=1 $$     (5.27) Solving these equations yield K = 0.3 and M = 0.1.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_p(x)=0.3\cos x + 0.1\sin x $$ (5.28) Therefore the exact overall solution is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y(x)=C_1e^x + C_2e^{2x}+0.3\cos x + 0.1\sin x $$ (5.29)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y'(x)=C_1e^x + 2C_2e^{2x}-0.3\sin x + 0.1\cos x $$ (5.30) Using the initial conditions yield the equations:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 1=C_1 + C_2 + 0.3 $$  &  $$  0=C_1 + 2C_2 + 0.1 $$     (5.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving these two equations yield $$C_1=1.5 \& C_2=-0.8 \!$$. Therefore the exact overall solution is:
 * {| style="width:100%" border="1"

$$  \displaystyle y(x)=1.5e^x - 0.8e^{2x}+0.3\cos x + 0.1\sin x $$ (5.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following figure shows the plot of the exact solution over the plot in Fig. 4.2-2: Figure 4.2-3

R4.3,p.7c-28
In this problem we are given the TA's Matlab code but must change the interval on which we are plotting. This give us the following code:

x=-1:0.01:1.5; y=log(1+x); x1=-1.5:0.01:1.5; y1=zeros(1,301); for i=1:4 for j=1:301 y1(j)=y1(j)-((-x1(j))^i)/i; end end y2=zeros(1,301); for i=1:7 for j=1:301 y2(j)=y2(j)-((-x1(j))^i)/i; end end y3=zeros(1,301); for i=1:11 for j=1:301 y3(j)=y3(j)-((-x1(j))^i)/i; end end y4=zeros(1,301); for i=1:16 for j=1:301 y4(j)=y4(j)-((-x1(j))^i)/i; end end h=plot(x,y); orange = [1 0.5 0.2]; set(h,'Color',orange); hold on; plot(x1,y1,'r'); plot(x1,y2,'g'); plot(x1,y3,'b'); plot(x1,y4,'c'); axis([0 10 -10 10]); legend('log(1+x)','T_4','T_7','T_1_1','T_1_6'); grid on; Zooming at -0.5 This graph shows fairly accurate results with slight deviation, with largest deviation occurring at the T4 term. Zooming about point 0 In this graph, only T16 shows, thus meaning that it is the best approximation. Zooming at 0.5 All lines appear to be fairly good approximations with the exception of T4 which deviates from the rest.

R4.4,p.7c-29
Again we must adjust the TA's Matlab code:

syms x f=log(x+1); fT1=taylor(f,5,1); fT2=taylor(f,8,1); fT3=taylor(f,12,1); X=.9:0.1:10; Y(:,1)=subs(fT1,'x',X); Y(:,2)=subs(fT2,'x',X); Y(:,3)=subs(fT3,'x',X); Y(:,4)=log(1+X); figure plot(X,Y); axis([.9,10,-10,10]);

Zooming at 1 Since the three lines are all stacked, both n=4 and n=7 are good approximations at x=1. Zooming at 1.5 Again, since the three lines are all stacked, both n=4 and n=7 are good approximations at x=1.5. Zooming at 2 Here, both can again be said to be good approximations, however, you can see some very slight deviation in n=4. Zooming at 2.5 In this plot, n=7 is still a pretty strong approximation, while n=4 is beginning to deviate at a fairly rapid rate and thus will soon be a poor approximation.

Author
Solved and Typed By - --Egm4313.s12.team1.rosenberg (talk) 01:19, 11 April 2012 (UTC) Reviewed By ---Egm4313.s12.team1.durrance (talk) 19:12, 11 April 2012 (UTC)

Statement
Verify (2) p.10-7 as follows: Given the ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}=2e^{-2x}cos(3x) $$     (6.1) 1)Simplify the 1st term $$y''_p$$ on the lhs of (2) p.10-4. 2)Simplify the 2nd term $$4y'_p $$, and combine with the simplified 1st term. 3)Finally, add the 3rd term $$13y_p$$ 4)Find the final expression for $$y_p(x)$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec10-1.djvu Lecture 10 Pg. 9]

To find the particular solution of an ODE we can use the method of undetermined coefficients. In report 3 we used the table 2.1 below to find the standard form of the particular solution given the form of the excitation $$r(x)\!$$. Table 2.1 So the general form of the particular solution is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x] $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Task 1 asks to simplify they first term of the given ODE which is $$y_{p}''(x)\!$$. This is done by taking the second derivative of $$y_{p}(x)\!$$ and plugging it into equation 6.1
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

$$y_{p}''(x)\!$$ simplifies to:
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we must do the same for the second term $$4y_{p}'(x)\!$$.
 * {| style="width:100%" border="0"

$$  \displaystyle 4y'_{p}(x)=\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)] $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Add the second term to the first to continue building the original ODE (equation 6.1).
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}+\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]$$ $$+e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally add the third term $$13y_{p}(x)\!$$ to the other two to have the lhs of the ODE 6.1
 * {| style="width:100%" border="0"

$$  \displaystyle 13y_{p}(x)=13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)+13y_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])+4e^{-2x}[sin(3x)(-3Mx-2Nx+N)$$ $$+cos(3x)(-2Mx+M+3Nx)]+13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x]=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This simplifies to the particular solution below:
 * {| style="width:100%" border="1"

$$  \displaystyle y_{p}(x)=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and Typed By ---Egm4313.s12.team1.stewart (talk) 23:50, 10 April 2012 (UTC) Reviewed By - Egm4313.s12.team1.essenwein (talk) 18:12, 11 April 2012 (UTC)

Statement
Find the separated ODEs for the heat equation:
 * {| style="width:100%" border="0"

$$ \frac{\partial u}{\partial t}=\kappa \frac{\partial^2 u}{\partial x^2} $$     (7.0) Where $$\kappa$$ is the coefficient of heat conductivity.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ u(x,t)=F(x) \cdot G(t) $$     (7.1) So, by take partial derivatives of Eq. (7.1) with respect to the individual variables:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \frac{\partial u(x,t)}{\partial t}=F(x) \cdot \dot G(t) $$     (7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \frac{\partial^2 u(x,t)}{\partial x^2}=F''(x) \cdot G(t) $$     (7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By substituting the partial derivatives back into Eq. (7.0):
 * {| style="width:100%" border="0"

$$ F(x) \cdot \dot G(t)=\kappa F''(x) \cdot G(t) $$     (7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Separating the equations to similar variables yields:
 * {| style="width:100%" border="0"

$$ \frac{\dot G(t)}{G(t)}=\kappa \frac{F''(x)}{F(x)}=C \text{ (constant)} $$     (7.4a)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore the separated ODEs are:
 * {| style="width:100%" border="1"

$$ \dot G(t)-CG(t)=0$$ $$ F''(x)-\frac{C}{\kappa}F(x)=0 $$     (7.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and Typed By - --Egm4313.s12.team1.durrance (talk) 02:13, 9 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 23:48, 10 April 2012 (UTC)


text:
User:Raity

Welcome to my page :)

ar:مستخدم:Raity cs:Uživatel:Raity de:Benutzer:Raity el:Χρήστης:Raity es:Usuario:Raity fi:Käyttäjä:Raity fr:Utilisateur:Raity it:Utente:Raity ja:利用者:Raity pt:Utilizador:Raity ru:Участник:Raity sl:Uporabnik:Raity sv:Användare:Raity

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

title: User:Raity


text:
User:Slugger0822

http://en.wikiversity.org/wiki/User:Yourusername/Calculus1

title: User:Slugger0822


text:
User talk:Bronows

hi,

I am newly arrived user. I am interested in Wikipedia's high school physics pages.--Bronows (talk) 04:21, 12 April 2012 (UTC)

title: User talk:Bronows


text:
Multi Entity Bayesian Logic

MULTI ENTITY BAYESIAN NETWORKS (MEBN)
MEBN is  breakthrough Bayesian reasoning system in which complex probabilistics models are constructed from modular components that can be replicated and combined in an infinite variety of ways.MEBN allows models to capture important and subtle aspects of objects and their inter-relationships that would be impossible to model using existing technologies.

MODELLING WITH MEBN
MEBN logic represent the world as comprised of entities that have attributes and are related to other entities.Random variables represent features of entities and relationships among entities.MEBN logic expresses knowledge about attributes and relationships as a collection of MEBN fragments (MFrags) organized into MEBN Theories (MTheories).

An MFrag represents a conditional probabilty distribution for instances of its resident random variables given their parents in the fragment graph and the context node.

An MTheory is a set of MFrags that collectively satisfies consistency constraints ensuring the existence of a unique joint probability distribution over instances of the random variables represented in each of the MFRAGS within the set.

title: Multi Entity Bayesian Logic


text:
CIVICS/Zaqatala Case Study

Geography
Zaqatala district borders Georgia in the south, Dagestan] in the north and [[w:Balaken:Balaken and Gakh districts in the west and east. Zaqatala town is located 445 km away from Baku, the capital of Azerbaijan. It sits on the bank of the Tala river and at the Yevlakh-Baku highway. It is 25-30 km away from the Eskibazar, that was one of the biggest trade centres of the East at the Great Silk Way. Gala plain is at the 25-35 metres above the town.

Zaqatala district is composed of 1 town, 60 villages and two additional 2 settlements. It has an area of 1,350 km² and a population of 120,300 people (sources?). From Zagatala to Baku City, it is about 450 km.

 * Landscape being existed in the north and northeast sides is mountain, and in the south and southwest is flatted. The highest point is Guton mountain (3648 m). Climate at the plain is warm, in mountains – cold. The mid temperature of January on the plain - 1 C, in mountains - 10 C, mid temperature of July accordingly +24 C and +5 C. Level of precipitation constitutes 600-1600 mm a year. Half of the territory of the area is occupied with woods, basically, in mountains. It grows rare the most valuable types of trees: chestnuts, hazelnuts, nuts, filbert, oak, beech, hornbeam, garagaj and others. Forests cover about 49% of the territory of the district. At the plain, the greenery is privilege like bushes. Noble deer, it is habituated pheasant, snowbird, partridge, turaj, mountain eagle, hawk, nightingale and others.

 * Having founded in 1929 Zagatala State Nature Reserve occupies part of the area territory. Square of the reserve constitutes 23844 Hectares. In the National park, it is forbidden to carry out any economic activity, but there are conditions available for ecological tourism. Having making walk tour or on horses upon tourist routes one may watch animals and birds habituated in reserves, make camera shooting and photo. At the aforesaid routes there are places where tourists might have relaxation term and to stay at night. It is presumably hunting on pheasant and water floating bird and running to such animals as wild bore, hare. In outskirts of Gas village, you may have skating in winter. Zagatala – international region. Herein in conjunction with Azerbaijanis, there live representatives of more than 20 nations: Avars, Tsakhurs, Russian, Lezghins, Tatars, Ingiloys and others.
 * Population
 * Population

 * As of the 1st January 2011, population of Zagatala district was 120,266 people. Average density of the population was 89 people per 1 sq. km. Population mainly inhabits foothills and plains. Zagatala town, Aliabad, Mukhakh, Upper Tala, Lower Tala, Danachi villages are large settlements. Zagatala is a multinational district. Here there are representatives of up to 20 nations (avars, sakhurs, lezgies, tatars, udins, geogians, jews and etc.).
 * Tourism
 * Tourism

 * Picturesque nature, attractive climate and historical monuments of the district boosted the development of tourism. There are unique natural areas good for recreation (Zagatala waterfall, Khalakhi lake, Gebizdere, Honzogor mountain, Moorchay, Malarasa pass and etc.). Zagatala State Reserve (area – 23.8 thsd. ha) is the first reserve established in the Caucasus (1929) region with very rich flora and fauna is very important for the development of ecotourism.
 * 2 fortresses, 38 mosques, 3 minarets and 2 churches are preserved as historical monuments. They include fortress in Zagatala town ( 1830), mosque in Aliabad village (XIX century), Chingizgala in Jar village (XIV century), «Sheytangala» ( XVII century), fortress in Mukhakh village, tomb and mosque in Lower Tala village, tower in Upper Chardaglar village (XIV century) and etc.
 * Economy
 * Economy

 * Main areas of the economy are – tobacco growing, silkworm breeding, nut growing, grain growing (corn for forage). Zagatala district has beautiful and rich nature, minerals and fertile lands. Zagatala is one of the main districts of the country specialising in tobacco growing, silkworm breeding and nut growing. Moreover, grain growing, cattle breeding and processing of agricultural products play an important role in the economy. Mainly nuts and seedy fruits (apple, pear) are grown in the territory of the district.
 * Furthermore, grapes, tea, plants containing ether are also grown. There is a fruit cannery, nut processing factory, tobacco fermentation, butter-cheese plants, brickworks, poultry incubator in the district. Economic reforms pursued in recent years created a favourable condition for the reconstruction of the economy and development of the agrarian sector of Zagatala district.

title: CIVICS/Zaqatala Case Study


text:
Zagatala State National Park

Zagatala State Nature Reserve

 * Year of foundation: 1929
 * Area (hectare): 47349 ha
 * Location: Within the territory of Zagatala and Balakan administrative districts, on the southern micro slope in the middle part of the Major Caucasus Ridge.

Description: Established in administrative areas of Zagatala and Balakan regions in 1929. Its area totals to 47349 ha. In During the period of its existence, the reserve has belonged to different organization and its area and borders have been changed. The reserve is situated at a height of 650 – 3,646 m above sea level. The reserve territory has a complicated relief due to spurs of the major ridge extending to the south and south-east: Agkemal, Katslar, Rochigel, Pichgel, Khalagel, Ruchug, Mrovdag and others, which are separated from each other by deep river valleys (canyons). Slopes with a steepness of 40-80 m and more occupy an area of more than 450 hectares; slopes with a steepness of 25-40 m prevail. The relief reflects the activity of glacier and other forms of erosion.

The asymmetry of the river basin is typical: the mountain slopes, deep canyons and valleys are sharply shaped. The Zaqatala reserve was organized with the purpose of protecting and studying the fauna and flora of the southern slopes of the Major Caucasus. The reserve territory is referred to by botanists as the Iberian area of the Caucasus flora province. At the end of the Tertiary period, the forests of this area had a different composition, with a considerable touch of elements of the Hirkan forests, and were much richer than the present ones. Contemporary flora of the reserve has more than a thousand species.

Such representatives of ancient plants as rhododendron yellow, Laurocerasus officinalis, Caucasian bilberry-bush, Taxus baccata, maple, Polypodiophyta and others are preserved on this territory. The main forest – forming species of the reserve – Fagus orientalis, as well as Quercus iberica and Corylus colurna are also referred to as ancient plants. The representatives of rare plants: Taxus baccata, apple-tree, ash-tree, birch-tree, cherry-tree, pear-tree and others are observed as well. The fauna of the reserve is rich in species composition. They are: Dagestan aurochs, chamois, red deer, roe, brown bear, fox, badger, Mustela nivalis, pine marten and stone marten, lynx, squirrel and others.

There are 104 species of birds, including some birds of prey: long-eared owl, golden eagle (Aquilla chrysaetos), Cerchneis tinnunculus, Neophron percnopterus, griffon (Gyps fulvus), bearded vulture (Cypaetus barbatus), black vulture and others. There are some rare, specially protected species of birds: bearded vulture (Gypaetus barbatus), golden eagle (Aquilla chrysaetus), peregrine (Falco peregrinus), Tetraogallus, Accipiter badius, which are registered in the Red Book.

title: Zagatala State National Park


text: Zagatala State Nature Reserve

title: Zagatala State Nature Reserve


text:
Talk:Exercise and metabolic disease/prostate

REFLECTION ON PROSTATE ADT

1.	What parts of the resource were most useful in working with the "client" and why?

The exercise training and prescription pages would be most useful; it is clear and concise in the exercise prescription page regarding the correct exercise modality and the prescribed reps and sets and length of time. The diagnosis page was also useful in gaining an understanding of the diagnosis phase. There was an important point made regarding when prostate testing should take place; “Men over 40 with a family history are advised to get tested annually and so are all men over the age of 50. In the early stages of prostate cancer there can be no signs or symptoms in some cases, so getting tested is usually the only way to detect it. Early detection is vital as far as whether or not the cancer can be cured” the reference is useful, but it could have been linked to the actual statement for easier navigation, this is an important point that should stand out well in the wiki.

'''2.	Can you suggest improvements to the practical use of the resource? Rationalise any improvements and provide examples where possible.'''

The prevalence of prostate cancer was reported, but not against other diseases. Due to the amount of deaths from prostate cancer in relation to other diseases, it is an important point to make. Some information to alert EP’s on the size and importance of this disease and its lack of diagnosis and education could be made clearer in the wiki. The references below may be useful to back this statement up;

•	AIHW 2010. Cancer in Australia 2010: an overview. Cancer series no. 60. Cat. no. CAN 56. Canberra: AIHW. •	AIHW, CA (Cancer Australia) & AACR (Australasian Association of Cancer Registries) 2008. Cancer survival and prevalence in Australia: cancers diagnosed from 1982 to 2004. Cancer series no. 42. Cat. No. CAN 38. Canberra: AIHW.

A prevention page would also be helpful, as an EP we get clients who come to see us with a family history of prostate cancer and want to know about ways in which to prevent the likelihood of them getting it. There is quite a bit of information published on the prevention of prostate cancer, links to this would be very useful in this type of situation. Some examples below;

•	Brasky, T. M., Till, C., White, E., Neuhouser, M. L., Song, X., Goodman, P., et al. (2011). Serum phospholipid fatty acids and prostate cancer risk: Results from the prostate cancer prevention trial. American Journal of Epidemiology, 173(12), 1429-1439. •	Rota, M., Scotti, L., Turati, F., Tramacere, I., Islami, F., Bellocco, R., et al. (2011). Alcohol consumption and prostate cancer risk: A meta-analysis of the dose-risk relation. European Journal of Cancer Prevention, •	Schenk, J. M., Kristal, A. R., Arnold, K. B., Tangen, C. M., Neuhouser, M. L., Lin, D. W., et al. (2011). Association of symptomatic benign prostatic hyperplasia and prostate cancer: Results from the prostate cancer prevention trial. American Journal of Epidemiology, 173(12), 1419-1428.

A prognosis page would be beneficial for prostate cancer, most chronic disease wikis would benefit from this type of page. Clients will talk about this with their EP, having the information or links to prognosis would be beneficial and helpful in understanding the disease and its progression. I realise that prognosis is individualised, but even the latest generic information would be great. Some links below for prognosis;

•	Cima, I., Schiess, R., Wild, P., Kaelin, M., Schüffler, P., Lange, V., et al. (2011). Cancer genetics-guided discovery of serum biomarker signatures for diagnosis and prognosis of prostate cancer. Proceedings of the National Academy of Sciences, 108(8), 3342-3347. •	Donovan, M. J., Costa, J., & Cordon-Cardo, C. (2011). Personalized approach to prostate cancer prognosis. Archivos Espanoles De Urologia, 64(8), 783-791. •	Martin, N. E., Mucci, L. A., Loda, M., & DePinho, R. A. (2011). Prognostic determinants in prostate cancer. The Cancer Journal, 17(6), 429. •	Oleksowicz, L., Liu, Y., Bracken, R. B., Gaitonde, K., Burke, B., Succop, P., et al. (2011). Secretory phospholipase A2‐IIa is a target gene of the HER/HER2‐elicited pathway and a potential plasma biomarker for poor prognosis of prostate cancer. The Prostate,

'''3.	Was the resource factually accurate? Use references to support your response.''' Some references were not very recent for example in the prevalence page the following reference from 2002 was used;

•	Quinn, M, and Babb, P. Patterns and trends in prostate cancer incidence, survival, prevalence and mortality. Part I: international comparisons. British Journal of Urology 90(2): 162-173, 2002

More recent information regarding prevalence can be found by searching WHO or Australian government websites, for example;

•	http://www.who.int/mediacentre/news/releases/2003/pr27/en/ •	AIHW 2010. Cancer in Australia 2010: an overview. Cancer series no. 60. Cat. no. CAN 56. Canberra: AIHW. •	AIHW, CA (Cancer Australia) & AACR (Australasian Association of Cancer Registries) 2008. Cancer survival and prevalence in Australia: cancers diagnosed from 1982 to 2004. Cancer series no. 42. Cat. No. CAN 38. Canberra: AIHW.

'''4. What are your thoughts on the layout of the resource? Describe what changes you might make to improve ease of understanding of the resource. If you are aware of anything that may improve the resource (e.g. a video, reference to further reading etc) include them here.'''

The wiki was difficult to navigate and therefore read, some good information was reported though. Improvements could have been made by adding in subtitles, pictures and some links to other pages. Links to other pages means that you do not have to do the search yourself to find the information you need; it saves time if all the important points are linked to the actual page. For example right at the start of the wiki “prostate cancer” could have been linked to the prostate cancer page on Wikipedia, a link could also have been made to ADT for further information.

Below are some excellent links that could be included to add more information to the wiki. Wikipedia have some fantastic images of prostate cancer that are open access and can be used. See below; (to be able to see the prostate in images, would be a great benefit to an EP). http://en.wikipedia.org/wiki/Prostate_cancer http://en.wikipedia.org/wiki/File:Prostatelead.jpg

There is quite a lot of information on dietary links to prostate cancer; this was not included in the wiki. I thought that it would be an important add. Knowing the correct advice to give a client on dietary intake, including referral options would be relevant for the overall treatment of an individual with prostate cancer. Some references about dietary intake and prostate cancer are below;

•	Venkateswaran, V; Klotz, LH (2010 Aug). "Diet and prostate cancer: mechanisms of action and implications for chemoprevention". Nature reviews. Urology 7 (8): 442–53. •	Hébert, J. R., Hurley, T. G., Harmon, B. E., Heiney, S., Hebert, C. J., & Steck, S. E. (2011). A diet, physical activity, and stress reduction intervention in men with rising prostate-specific antigen after treatment for prostate cancer. Cancer Epidemiology, •	Hori, S., Butler, E., & McLoughlin, J. (2011). Prostate cancer and diet: Food for thought? BJU International, 107(9), 1348-1359. •	Marshall, J. R. (2012). Diet and prostate cancer prevention. World Journal of Urology,, 1-9.

5. Describe any personal experiences you have had, or anticipate, in professional life and how such a resource might be used in practice.

Personally, as a health professional I know that time is limited and to correctly treat a client you need to have the right information that is up to date and very relevant. This ensures the best outcome for a client. This resource is a quick and effective tool to use to navigate and build knowledge and understanding of a type of condition or disease state and the correct exercises to prescribe. It would be a great tool if it is continually updated (maybe each year) to include new and relevant information, statistics and research. The time needed to search for up to date research can be limited, it takes a great deal of time to look in all of the places needed (text books, on-line, scholar, Google, finding the correct search terms) to get the information required to stay up to date. To do a great job as an EP you need to spend the majority of your time with a client, so this type of resource could potentially cut down time, which may lead to more time for and with clients! This wikiversity resource, if developed further with more conditions, could also ensure that EP’s and other health professionals are all uniformed in the way in which they train people. This could potentially stop or cut down on any unnecessary injuries or incorrect training methods for these types of conditions. As suggested previously, to be most effective it would need to be kept up to date, made very easy to read and navigate and be accessible to everyone. Overall, the wiki is an excellent tool that I personally would use.


title: File:CORDIC.cpp.20120412.pdf


title: File:CORDIC.octave.20120412.pdf


text: File:Screen Shot 2012-04-12 at 10.00.19 AM.png


title: File:Pea2.s12.sec48.l.djvu


text: File:2012-04-11 07.03.21.jpg

title: File:2012-04-11 07.03.21.jpg


text: File:2012-04-11 07.03.59.jpg

title: File:2012-04-11 07.03.59.jpg


text: File:2012-04-11 07.04.29.jpg

title: File:2012-04-11 07.04.29.jpg


text:
User talk:Silver seren

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

To get started, you may

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

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

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! Cirt (talk) 17:27, 12 April 2012 (UTC)

-- Cirt (talk) 17:27, 12 April 2012 (UTC)

Regarding some falsehoods
Silver seren, on the English Wikipedia, you have twice published falsehoods about me that I would like to see you refactor. What is the best way to reach you outside of Wikipedia? I have tried your Facebook account, but my message received no response. You don't seem to have "Email this user" set up on any Wikimedia projects. Please send your contact info to me -- I think you know various ways of reaching me. I'm amenable to e-mail, postal mail, or telephone during the day and evening. -- Thekohser (discuss • contribs) 15:36, 14 February 2013 (UTC)

title: User talk:Silver seren


text:
University of Florida/Egm4313/s12.team5.R7

Question
Verify (4)-(5) p.19-9. (4): $$ <\Phi _i,\Phi _j> = 0, i\neq j $$ (5): $$ <\Phi _i,\Phi _j> = \frac{L}{2}, i=j $$

Solution
We know that: $$ \Phi _j(x) := sin\omega _jx, j=1,2, ... $$ We also know by definition that: $$ \langle \bar f, \bar g \rangle := \int_0^L \bar f(x) \, \bar g(x) \, dx $$ Therefore, we start with Eq. (4) and use the above definitions to obtain the following results: $$ <\Phi _i,\Phi _j> = \int_{0}^{L} sin(\omega _ix)sin(\omega _jx)dx $$ For simplicity's sake, we set L equal to $$ \pi $$. This yields the following result: $$ \int_{0}^{ \pi} sin(\omega _ix)sin(\omega _jx)dx = \frac{(\omega_i-\omega_j)sin((\omega_i+\omega_j)x)+(-\omega_i-\omega_j)sin((\omega_i-\omega_j)x)}{2\omega_i^2-2\omega_j^2} $$ This last expression is evaluated from $$ \pi $$ to zero. We know as a geometric property that the following two things are true: $$ sin(C\pi) = 0 $$ $$ sin(0) = 0 $$ Where "C" is any positive integer. We also know that since L = $$ \pi $$, p = 2L = $$ 2 \pi $$, and $$ \omega = \frac{2 \pi}{p} = \frac{2 \pi}{2 \pi} = 1 $$, so therefore every sine term in the definite integral is 0 when both $$ \pi $$ and 0 are evaluated. Therefore, we know that: $$ <\Phi _i,\Phi _j> = 0, i\neq j $$ Now, for Eq. (5), we evaluate is as follows: $$ <\Phi _j,\Phi _j> = \int_{0}^{L} sin(\omega _jx)^2 dx $$ Again, we set L equal to $$ \pi $$ to make the expression easier to evaluate. This yields: $$ \int_{0}^{ \pi} sin(\omega _jx)^2 dx = -(\frac{sin(2\omega_jx)-2\omega_jx}{4\omega_j}) $$ This integral is evaluated from $$ \pi $$ to 0. Again, we know that the sine terms will be zero no matter which boundary we plug again, since $$ \omega = 1 $$. Therefore, the evaluation simplifies to: $$ -(-\frac{2\omega_j\pi}{4\omega_j}-0) = \frac{\pi}{2} $$ Since we set L equal to $$ \pi $$, we have thus proven Eq. (5).

Author
This problem was solved and uploaded by: William Knapper

Question

 * Plot the truncated series $$\displaystyle u(x,t)=\sum_{j=1}^{n}a_{j}cos(cw_{j}t)sin(\omega _{j}x)$$
 * with n=5 and for $$\displaystyle t=\alpha \frac{2L}{c} $$
 * where $$\displaystyle \alpha = 0.5,1,1.5,2 $$

Solution
Matlab Code:

% Report 7 Problem 2 % Plot truncated series for n=5 % Given values L = 2; c = 3; % Calculate alphas for use in t alpha = (0.5:.5:2); % four alpha terms t = ((alpha.*2*L)/c); % four "t" terms corresponding to the four alpha terms % Calculate omegas (w) j = (1:1:5); % series terms 1-5 w = ((j*pi)/L); % omega correspoding to series terms 1-5 % Calculate "a" coefficients % Check if "a" is odd or even, then assign value a = j; for k=1:5; if mod(k,2) == 0 % even term a(k) = 0; else % odd term a(k) = -4/(pi^3*k^3); end end % Calculate truncated series (note that a2=a4=0, so those terms in the % series have no contribution x=(0:0.01:2); % For t=0.5 y1=zeros(size(x)); for k=1:5   y1 = y1 + a(k)*cos(c*w(k)*t(1))*sin(w(k)*x); end % For t=1.0 y2=zeros(size(x)); for k=1:5    y2 = y2 + a(k)*cos(c*w(k)*t(2))*sin(w(k)*x); end % For t=1.5 y3=zeros(size(x)); for k=1:5   y3 = y3 + a(k)*cos(c*w(k)*t(3))*sin(w(k)*x); end % For t=2.0 y4=zeros(size(x)); for k=1:5    y4 = y4 + a(k)*cos(c*w(k)*t(4))*sin(w(k)*x); end % Compare to original function yorg = x.*(x-2); plot(x,y1,x,y2,x,y3,x,y4,x,yorg) legend('t=0.5','t=1.0','t=1.5','t=2.0','org function')

Plot of 4 different t's and original function:

Author
This problem was solved and uploaded by: Josh House

Question
Find the a) scalar product, b) the magnitude, and c) the angle between:

$$ \displaystyle 1. f(x) = \cos(x), g(x) = x\ $$ for $$ \displaystyle -2\leq x\leq 10\ $$

and

$$ \displaystyle 2. f(x) = \frac{1}{2} (3x^2-1), g(x) = \frac{1}{2} (5x^3-3x)\ $$ for $$ \displaystyle -1\leq x\leq 1\ $$

Solution
The scalar product between 2 functions on the interval [a,b] is defined by:

$$ \displaystyle < f,g> = \int_{a}^{b} f(x) g(x)dx\ $$

The magnitude of a function is defined by:

$$ \displaystyle \left \| f \right \| = < f,f> ^\frac{1}{2} = \left [ \int_{a}^{b} f^2(x)dx\right ]^\frac{1}{2}\ $$

The angle between 2 functions is defined by:

$$ \displaystyle \cos(\theta ) = \frac{< f,g> }{\left \| f \right \|\left \| g \right \|}\ $$

Problem 1:

$$ \displaystyle  = \int_{-2}^{10} x \cos(x) dx\ $$

Using integration by parts: $$ \displaystyle u = x, du = dx, dV = \cos(x) dx, V = \sin(x)\ $$

The integral becomes: $$ \displaystyle x\sin(x) - \int \sin(x) dx = x\sin(x) + \cos(x)\mid_{-2}^{10} = -7.68\ $$

$$ \displaystyle \left \| f \right \| = [\int_{-2}^{10}\cos^2(x) dx] ^ \frac{1}{2} = [\int_{-2}^{10} \frac{1+\cos(2x)}{2} dx] ^\frac{1}{2} = [\frac{x}{2} + \frac{\sin(2x)}{4}\mid _{-2}^{10} ] ^ \frac{1}{2} = 2.46\ $$

$$ \displaystyle \left \| g \right \| = [\int_{-2}^{10}x^2 dx]^\frac{1}{2} = [\frac{x^3}{3}\mid _{-2}^{10}] ^\frac{1}{2} = 18.33\ $$

$$ \displaystyle \cos(\theta ) = \frac{-7.68}{18.33*2.46} \Rightarrow \theta = 1.742\ $$ rads

Problem 2:

$$ \displaystyle  = \int_{-1}^{1} \frac{1}{2} (3x^2-1) * \frac{1}{2} (5x^3 - 3x) dx = \int_{-1}^{1} (\frac{15}{4} x^5 - \frac{7}{2} x^3 + \frac{3}{4}x) dx\ $$

The integral becomes: $$ \displaystyle \frac{5}{8} x^6 - \frac{7}{8} x^4 + \frac{3}{8} x^2 \mid_{-1}^{1} = 0\ $$

$$ \displaystyle \left \| f \right \| = [\int_{-1}^{1} [\frac{1}{2}(3x^2-1)]^2 dx] ^\frac{1}{2} = [\int_{-1}^{1} (\frac {9}{4}x^4 - \frac{3}{2}x^2 + \frac{1}{4}) dx ] ^\frac{1}{2} = [\frac{9}{20}x^5-\frac{1}{2}x^3 + \frac{1}{4}x \mid_{-1}^{1} ] ^\frac{1}{2} = 0.632\ $$

$$ \displaystyle \left \| g \right \| = [\int_{-1}^{1} [\frac{1}{2} (5x^3-3x)]^2 dx]^\frac{1}{2} = [\int_{-1}^{1} (\frac{25}{4}x^6 - \frac{15}{2}x^4 + \frac{9}{4} x^2) dx] ^\frac{1}{2} = [\frac{25}{28}x^7 - \frac{3}{2}x^5 +\frac{3}{4}x^3 \mid _{-1}^{1} ] ^\frac{1}{2} = 0.535\ $$

$$ \displaystyle \cos(\theta) = 0 \Rightarrow \theta = \pi / 2\ $$ rads.

Author
This problem was solved and uploaded by: David Herrick

Question
K. 2011 p482 pb. 6,9,12,13:

problems 6,9: Sketch or graph f(x) for $$-\pi < x < \pi $$.

problems 12,13: Find the fourier series for the f(x) in problems 6 and 9 respectively up to n = 5.

Problem 6: f(x) = |x| for $$ -\pi < x < \pi $$

Problem 9: f(x) = x for $$ -\pi < x < 0 $$

and f(x) = $$ \pi - x $$ if $$ 0 < x < \pi $$

Solution
Problem 6: f(x) = |x|

MATLAB code used:

Problem 9: $$ f(x) = x $$ if $$ -\pi < x < 0 $$ $$ f(x) = \pi - x $$ if $$ 0 < x < \pi $$

MATLAB code used:

Problem 12: f(x) = |x| breaks down into: $$ f(x) = -x $$ if $$ -\pi < x < 0 $$ and $$ f(x) = x $$ if $$ 0 < x < \pi $$ The formula for the fourier series is: $$ f(x) = a_0 + \sum_{n=1}^\infty [a_n \cos(nx) + b_n \sin(nx)] $$ The euler formulas are also: $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) dx $$ $$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) dx $$ $$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) dx $$ Therefore: $$ a_0 = \frac{1}{2\pi} [\int_{-\pi}^0 -xdx + \int_0^\pi xdx] $$ $$ a_n = \frac{1}{\pi} [ \int_{-\pi}^0 -x\cos(nx)dx + \int_0^\pi x\cos(nx)dx] $$ $$ b_n = \frac{1}{\pi} [ \int_{-\pi}^0 -x\sin(nx)dx + \int_0^\pi x\sin(nx)dx] $$ After integration for $$ a_0 $$ and integration by parts for $$ a_n $$ and $$ b_n $$: $$ a_0 = \frac{\pi}{2} $$ $$ a_n = \frac{2}{\pi n^2} (\cos(-n\pi) - 1) $$ $$ b_n = 0 $$ Similarly, for n = even, $$ a_n = 0 $$ and for n = odd, $$ a_n = \frac{-4}{n^2 \pi} $$ Therefore, the fourier series out to n = 5 is: $$ f_5 (x) = \frac{\pi}{2} - \frac{4}{\pi} \cos(x) - \frac{4}{9 \pi} \cos(3x) - \frac{4}{25 \pi} \cos(5x) $$ Problem 13: $$ f(x) = x $$ if $$ -\pi < x < 0 $$ $$ f(x) = \pi - x $$ if $$ 0 < x < \pi $$ The euler formulas are the same as in problem 12. Plugging in problem 13's f(x): $$ a_0 = \frac{1}{2 \pi} [\int_{-\pi}^0 xdx + \int_0^\pi (\pi-x)dx] $$ $$ a_n = \frac{1}{\pi} [ \int_{-\pi}^0 x\cos(nx)dx + \int_0^\pi (\pi- x)\cos(nx)dx] $$ $$ b_n = \frac{1}{\pi} [ \int_{-\pi}^0 x\sin(nx)dx + \int_0^\pi (\pi - x)\sin(nx)dx] $$ Therefore, after using integration for $$ a_0 $$ and integration by parts for $$ a_n $$ and $$ b_n $$: $$ a_0 = 0 $$ $$ a_n = \frac{2}{n^2 \pi} (1-\cos(n \pi)) $$ $$ b_n = \frac{1}{n}(1-\cos(n \pi)) $$ Therefore, when n is odd, $$ a_n = \frac{4}{n^2 \pi} $$ and $$ b_n = \frac{2}{n} $$. Similarly, when n is even, $$ a_n = 0 = b_n $$. Therefore, the fourier series out to n = 5 is: $$ f_5(x) = \frac{4}{\pi} [\cos(x) + \frac{1}{9} \cos(3x) + \frac{1}{25} \cos(5x)] + 2 [\sin(x) + \frac{1}{3} \sin(3x) + \frac{1}{5} \sin(5x)] $$

Author
The problem was solved and uploaded by John North.

Question
Consider (1)p.12-4: $$ <\Phi_{2j-1} \Phi_{2k-1}> = \int_0^p \sin j \omega x \cdot \sin k \omega x dx $$ where $$ p=2 \pi, j=2, k=3. $$

1. Find the exact integration of (1)p.12-4 with the given data. 2. Confirm the result with matlab's trapz command for the trapezoidal rule as explained in (4)p.11-2.

Part 1
Using (3)p19-10 we get: $$ \int_0^{2 \pi} \sin j \omega x \cdot \sin k \omega x dx = \int_0^{2 \pi} \frac12 \left [cos(j \omega - k \omega)x - cos(j \omega + k \omega)x \right] dx $$

Integrating we get: $$ \frac{sin((j\omega-k\omega)x)}{2(j\omega-k\omega)}-\frac{sin((j\omega+k\omega)x)}{2(j\omega+k\omega)} $$

Plugging in the known values we have: $$ \frac{sin((2\omega-3\omega)x)}{2(2\omega-3\omega)}-\frac{sin((2\omega+3\omega)x)}{2(2\omega+3\omega)} = \frac{sin(-1\omega x)}{-2\omega}-\frac{sin(5\omega x)}{10\omega}$$

Solving from 0 to $$ 2 \pi $$: $$ \left [ \frac{sin(-2\omega \pi)}{-2\omega}-\frac{sin(10\omega \pi)}{10\omega} \right ] - \left [\frac{sin(0)}{-2\omega}-\frac{sin(0)}{10\omega} \right ] $$

We know that: $$ sin(C\pi)=0 $$ if C is any integer $$ sin(0)=0 $$

Thus, if $$ \omega $$ is an an integer, the solution to the equation becomes: $$ 0-0=0 $$

Therefore: $$ <\Phi_{3},\Phi_{5}> = \int_0^{2 \pi} \sin 2 \omega x \cdot \sin 3 \omega x dx = 0 $$

Part 2
Using MATLAB's TRAPZ method, a variable X was created from 0 to $$ 2 \pi $$ and the integral of

$$ \sin j \omega x \cdot \sin k \omega x dx $$

was found from 0 to $$ 2 \pi $$. The assumption of $$ \omega $$ being an integer and in this case = 1 was made.

The following was the outcome in MATLAB:

-3.8317e-007 is used as an equivalent to 0.

Therefore through MATLAB's TRAPZ method,

$$ \int_0^{2 \pi} \sin 2 \omega x \cdot \sin 3 \omega x dx = 0 $$

Author
Part 1 of this problem was solved and uploaded by Radina Dikova.

Part 2 of this problem was solved and uploaded by Derik Bell.

Contribution Summary
Problem 1 was solved by: William Knapper

Problem 2 was solved by: Joshua House

Problem 3 was solved by: David Herrick

Problem 4 was solved by: John North

Problem 5 part 1 was solved by: Radina Dikova

Problem 5 part 2 was solved by: Derik Bell


text: File:2012-04-10 17.04.11.jpg

title: File:2012-04-10 17.04.11.jpg


text: File:2012-04-10 17.13.01.jpg

title: File:2012-04-10 17.13.01.jpg


text: File:2012-04-10 17.13.07.jpg

title: File:2012-04-10 17.13.07.jpg


text: File:2012-04-10 17.36.12.jpg

title: File:2012-04-10 17.36.12.jpg


text: File:2012-04-10 17.36.22.jpg

title: File:2012-04-10 17.36.22.jpg


text: File:2012-04-10 17.43.10.jpg

title: File:2012-04-10 17.43.10.jpg


text: File:2012-04-10 17.45.40.jpg

title: File:2012-04-10 17.45.40.jpg


text: File:2012-04-10 17.47.37.jpg

title: File:2012-04-10 17.47.37.jpg


text: File:2012-04-10 17.55.30.jpg

title: File:2012-04-10 17.55.30.jpg


text: File:2012-04-10 17.58.14.jpg

title: File:2012-04-10 17.58.14.jpg


text: File:2012-04-10 17.58.19.jpg

title: File:2012-04-10 17.58.19.jpg


text: File:2012-04-10 18.24.16.jpg

title: File:2012-04-10 18.24.16.jpg


text: File:2012-04-10 18.36.22.jpg

title: File:2012-04-10 18.36.22.jpg


text: File:2012-04-10 18.58.43.jpg

title: File:2012-04-10 18.58.43.jpg


text: File:Standard Zonation.jpg

title: File:Standard Zonation.jpg


text: File:PEAS12HW13.5LANGxiaI n.jpg

title: File:PEAS12HW13.5LANGxiaI n.jpg


text:
User:Egm6322.s12.team2.Xia/RP13.5

Given the following function with the weight e^{-x^{2}}

I(1)=\int_{-\infty}^{\infty}1\cdot e^{-x^{2}}dx

find

1. Analytically: Square up the intergral (1) to convert the line integral in to an integral over the plane:

I^{2}=\left(\int_{-\infty}^{\infty}1\cdot e^{-x^{2}}dx\right)\left(\int_{-\infty}^{\infty}1\cdot e^{-y^{2}}dy\right)

=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy

Convert the area intergral to polar coordinates to integrate, and show that

I^{2}=\pi\Rightarrow I=\sqrt{\pi}

2. Verify with Wolfram Alpha

integrate e^{-x^2} for x in [-\infity, +\infity]

3.Numerically: Use the Gauss-Hermite quadrature with the table for \{x_{i}\} and \{H_{i}\} in p.48-36 at DLMF: Gauss-Hermite formula. Plot the convergence of the numerical value of

I_{n}(1)=\sum_{i=1}^{n}H_{i}f(x_{i})=\sum_{i=1}^{n}H_{i}

as a function of n.

x=r\cos\theta,\quad y=r\sin\theta

and

dx=dr\cos\theta-r\sin\theta d\theta

dy=dr\sin\theta+r\cos\theta d\theta

thus

J=\frac{\partial(x,y)}{\partial(r,\theta)}=|\begin{array}{cc} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array}|=r^{2}

dxdy=Jdrd\theta

\begin{array}{l} H_{7}(x)=128x^{7}-1344x^{5}+3360x^{3}-1680x\,\\ H_{8}(x)=256x^{8}-3584x^{6}+13440x^{4}-13440x^{2}+1680\,\\ H_{9}(x)=512x^{9}-9216x^{7}+48384x^{5}-80640x^{3}+30240x\,\\ H_{10}(x)=1024x^{10}-23040x^{8}+161280x^{6}-403200x^{4}+302400x^{2}-30240\, \end{array}

title: User:Egm6322.s12.team2.Xia/RP13.5


text: File:Y0 pos.png

title: File:Y0 pos.png


text: File:Y1 neg.png

title: File:Y1 neg.png


text: File:Y1 pos.png

title: File:Y1 pos.png


text: File:Y2 neg.png

title: File:Y2 neg.png


text: File:Y2 pos.png

title: File:Y2 pos.png


text: File:Y4 neg.png

title: File:Y4 neg.png


text: File:Y4 pos.png

title: File:Y4 pos.png


text: File:Y8 neg.png

title: File:Y8 neg.png


text: File:Y8 pos.png

title: File:Y8 pos.png


text: File:Y, y2, y4, y8 ode45 zoom1.png

title: File:Y, y2, y4, y8 ode45 zoom1.png


text: File:Y, y2, y4, y8 ode45.png

title: File:Y, y2, y4, y8 ode45.png


text:
User talk:Mananshah15

 Hello Mananshah15, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

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

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

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! -- dark lama  21:12, 12 April 2012 (UTC)

title: User talk:Mananshah15


text: File:Y, y2, y4, y8 neg NDSolve.png

title: File:Y, y2, y4, y8 neg NDSolve.png


text: File:Y, y2, y4, y8 neg NDSolve zoom.png

title: File:Y, y2, y4, y8 neg NDSolve zoom.png


text:
Draft talk:Reef geomorphology

This article is based loosely on one I recently published in the Encylopedia of Modern Coral Reefs (Blanchon 2011; see my academia.edu site for the paper). However, this open version is more complete, and includes additional images, text, references and analyses. In other words it is the upgrade version. --Blanchon (talk) 21:31, 12 April 2012 (UTC)

If you would like to contribute, I have organized it in the following way: 1. what is presently known about the geomorphology of coral reefs (ie objective description with no analysis of origin) 2. What is presently known about the formative processes from ecologic, oceanographic or geologic data 3. What is presently known about their development through time largely from geologic data.--Blanchon (talk) 21:39, 12 April 2012 (UTC)

title: Draft talk:Reef geomorphology


title: File:Fe1.s11.elastic.bar.svg


text:
P3550W

Problem Statement
Develop an Arduino based system for a RC Hovercraft.

Team Members

 * Wendell Phipps

Summary
Initially we agreed to construct a basic hovercraft from scratch. However, after a working toy hovercraft was located, the project changed directions to trying to program the hovercraft to react to sensor inputs.

Story
The project began with the group deciding to design/build a basic working hovercraft. Initially work on this hovercraft commenced with us breaking up the work between group members. The project was split into 4 parts; design, hovercraft movement coding, ESC programming, and hovercraft construction. However, between weeks 1 and 2, we were able to locate a working toy hovercraft. This pushed the project forward alot. All the group members came together to decide upon new jobs and a new direction for the project.

With a working hovercraft already available, our group decided to work on powering/controlling the hovercraft via an Arduino and various sensors in place of the RC controller's potentiometers.

New jobs were decided upon; power control, RC Hack, ESC programming, and infrared sensor programming. However after some slow progress, a few jobs were swapped, and the final project jobs were divided as follows; power control, RC Hack, infrared sensor programming, and touch sensor programming.

We got each individual job finished, but we have not gotten to make them all work with eachother yet.

Decision List
List all formal decisions made with links to their documentation such as a decision tree or decision matrix.

Material List
Materials used
 * Toy Hovercraft
 * Arduino
 * Motor Shield
 * Infrared Sensor
 * Touch Sensor
 * Soldering Iron
 * Solder
 * Camera

Software List
We used the Arduino software for tutorials for use on the RC Toy Hack, as well as creating new code to allow for additional potentiometer.

We also used 123D for preliminary design of a hovercraft.

Time
30 Hours

Tutorials
All projects create new tutorials of technical details future participants are going to want to know. They are going to be separate pages that are linked to here.

Next Steps
The next project should included fully hacking the RC Controller to control the hovercraft via the arduino. After RC Control is integrated, work should be done to replace the RC Controller input with a sensor input(infrared/touch/sonar).

title: P3550W


text:
Engineering Projects/Illuminate Clothing/Howard Community College/spring2012/I2550INC

Electronic Sections Expected

 * 1.Reverse engineer the EL wire driver hardware.
 * 2.Troubleshoot the EL wire drivers that are not working.
 * 3.Design circuit to enable the arduino to control the EL wire.
 * 4.Tear apart old lillypad diode circuit and sew another one.
 * 5.Design sewing system so lillypad conductive thread does not short out.
 * 6.Create a new script or improve current lillypad arduino LED program.
 * 7.Find a inventor for control the led from adrinuo
 * 8.match the adrinuo with new inventor

Team Members

 * Navid Jahanpour
 * Chris Nunez

Summary
on project 2 we try to design the inventor to control the led with adrinuo but we had the problem with transformer so we decide to find the inventor.because the transformer was ac and we supply it with dc voltage. son we buy the one inventor that it working with 3 volt battery and its produce the 3 Khz sin AD pulse.

Poster
3 KHZ-180VOLT inventor

Story
such as i say before we try to find the inventor from internet.our inventor EL inverters allow us  to drive EL (electroluminescent) wires. This particular EL inverter accepts a 12 VDC wall adapter. The output is terminated with a JST PH connector, so it plugs directly into the input on the EL Escudo or EL Sequencer board. You can drive EL wires with them directly, but you may need to re-terminate them. There is also a small switch on the side which allows you to switch between 'on', 'off', and 'blink' settings.so after this we try to control the more than one el wire with the EL Sequencer.

here is the schematic of EL Sequencer

after this i contact the inventor to the power port on the el board and contact the battery and then contact the el wire two the el Sequencer then we have to program the adrinu. in the end we control the el wire with adrinu.

Decision List
for first week we try to did the project with our circuit that we design in project two. second week we try to find the inventor from internet.

for third week we test the plans and data sheet for new inventor.

and for forth week we contact the inventor and control the led with adrinu.

Material List
1.3KHZ 180 volt inventor that we buy from internet 2.led wire 3.EL Sequencer 4.adrinuo software 5.adrinuo board 6.two 1.5 volt AA battery

Software List
adrinuo soft ware

Time
48 hour

Tutorials
http://www.ladyada.net/learn/el-wire/#inverter_driver_details

Next Steps
you have to hard time with control the led wire without the inventor i tried to did it but i needed the some stuffs.

1. high frequency (3khz)3to 180 volt dc transformer. 2.dc to ac inventor. 3.and RC filter to fix the frequency.


text:
Engineering Projects/HoverCraft/Howard Community College/Spring2012/p3550W

Please delete everything in italics after filling it out.

''Project documentation focuses on how the team organized itself to tackle the unknown. This is a creative process. The details learned should be captured in tutorials. Project documentation should focus on how the team organized itself to step out into the unknown. Failure is common. The single most effective way to increase engineering respect is to document failures in a creative way that stimulates, inspires and helps others continue the project. Projects are never done. There are always loose ends. There are always decisions that might have been made more clearly. There are always other alternatives that might have been considered. None of this stuff appears in a tutorial. It does appear in the team documentation.''

Problem Statement
To control a hovercraft using the Arduino and sensor input instead of RC controller.

Team Members

 * Wendell Phipps
 * Josh Bers

Summary
We have a new and working RC hovercraft. Our goal is to integrate an Arduino and use it to control the hovercraft. To achieve this we are going to be using a sensor input in order to control the hovercraft. Specifically, we will be using Digital QRE1113 IR Sensors as the input devices. We will be using the IR sensors follow a line placed on the floor. By using contrasting colors(black stripe on white floor) we will program the hovercraft to follow the black stripe based on the varying values returned from the IR Sensors.

Poster
''Put a graphic in wikimedia, include the graphic here or link to it here. The graphic should be suitable for creating a traditional project poster.''

Story
''Tell a detailed story of the project. Describe how split up, what the obstacles were, what testing was done, what informal decisions were made, what assumptions were made, what the results were.'' The first week's work was the same for both of us. We were both on our own trying to get feedback from our newly acquired IR Sensors. This was achieved and using the basic code located in the tutorial linked on the product page for the Digital QRE1113 IR Sensor. This basic tutorial simply activated the sensor and allowed it to return a value visible in serial monitor in the Arduino software. We were successful at getting values from the sensor, however encountered obstacles getting consistent readings. This has to do with the distance the sensor was being held from whatever surface it was supposed to be sensing. We have decided that it will need to be mounted on the hovercraft with and adjustable arm that will allow us to set its position relative to the ground/guide stripe.

During the second week we worked on getting the sensors to control small fans as well as started to work on basic mounting solutions for the arduino, sensors, and motorshield. Therefore the work was split between hovercraft design and arduino coding. By combining the basic code again from the Bildr Tutorial for the IR Sensor, and utilizing code from this Instructables Tutorial we were able to create a hybrid code that allowed first for control of a single fan(code), then eventually for a second fan(2FanCode). We also figured out that the best mounting method for the Arduino and MotorShield would be to remove the hovercraft's battery cover and install a small cushion/piece to level the boards as well as small elastic straps to hold the entire assembly in place. For basic test purposes, the control leads were run through the induction fans intake, however for final assembly, the body of the hovercraft will be modified so that the control leads will not pass near the induction fan.

For the third week of this project we had to further refine the design of the hovercraft as well as incorporating the fan control coding and setup with the hovercraft as a whole. Initially we attempted to control the hovercraft fans directly off of the Arduino. However, two problems were encountered. First, the Arduino board itself did not have power itself to control the hovercraft's DC fans. The second problem was that the hovercraft fans would need to rotate in both direction to allow the hovercraft to turn and reverse direction. Both of these problems were solved with the addition of the MotorShield. The Motorshield contains a chip called an H-Bridge that allows for motors to spin in both directions. Using the basic Arduino Example: MotorParty we were able to achieve basic fan operation, in both directions, with the Arduino/Motorshield. However, this example only allows for a single fan to operate, and did not have enough power(with just the USB cable) to operate at full speed. By adding the power from the DC wall outlet, the operation improved, but was still not quite up to full power.

For the final week of the project the coding for the Arduino/Motorshield was expanded to allow both fans to operate at the same time(code). The hovercraft battery was also integrated into the hovercraft setup to allow for both fans to run at full power. Video of this latest modification to the hovercraft coding can be seen HERE.

Decision List
List all formal decisions made with links to their documentation such as a decision tree or decision matrix. 1. During the first week of this project both team members had the same focus. We individually experimented with the QRE1113 Digital IR Sensor to get feedback via the Arduino/Serial Monitor.

2. With the IR Sensor now functioning the group split our focus. While Josh worked on developing a code that would allow for the hovercraft fans to operate via IR Sensor input. Concurrently, Wendell began work on mounting the Arduino/IR Sensor Hardware.

3. The next aim was to scale the coding for basic fan operation up, in order to operate the higher powered fans of the hovercraft. We decided that we would then need to redesign the hovercraft to allow for additional hardware mounting, as well as a modified code to operate the new hardware.

4. Discovered that hovercraft fans switch rotation by alternating which lead is power and which lead is ground. In order to operate these fans properly, we will need to integrate the Motorshield with the Arduino. This will also allow for operation of higher power motors.

Material List

 * 1 Arduino Uno
 * 1 Toy Hovercraft
 * Motor Shield
 * Infrared Sensor
 * Touch Sensor
 * Soldering Iron
 * Solder
 * Camera
 * Scanning Head from a Printer

Software List
Arduino

Time
36 Hours

Tutorials
''All projects create new tutorials of technical details future participants are going to want to know. They are going to be separate pages that are linked to here.''

Applying Shrink Wrap to Electrical Wire

Next Steps
List specific details, advice, describe the current problems that the next team faces associated with the project here. The next step of this project will be to: 1. finish mounting Arduino/MotorShield/IR Sensor hardware, 2. modify coding to spin individual fans or spin both fans in opposing directions, 3. add coding lines to operate induction fan, 4. add coding for multiple sensors.


text:
User:Christopher.nunez/enes100/illuminationProject2

illuminationTeamPage

Write problem/project Goal
Create a light show with a lilypad ardiuno. To include LED's and EL wire.

My First Task
Figure out how to get EL wire working with Lilypad

Summary of actual work over first weekend
I've changed my apporach on the project. This time I want to use the EL Sequencer instead of using the Lilypad Arduino. This is because the EL sequencer makes it easier for Several EL wire to be used and it is programmed the same way as a lilypad arduino.

Week1 Narrative
I research the EL sequencer arduino and found several tuturial sites. I also found that sparkfun website will sell everything I need. My goal is to show this information in a presentation to the professor so he can purchase what I need to complete my project. Here are the sites I will use.br />

Sites:

Getting started with your sparkfun EL Sequencer

Programming EL wire with lilypad

Videos:

EL Wire Tutorial

Eventually I want to get the Lilypad project in sync with the EL wire Project

My Second Task
Hopefully the Materials arrive this week. So I can start working on the project. If not I will Continue learning scripting for the new arduino.

Summary of actual work over third weekend
The material Finally came in. I started moving the project forward. Along with the materials arriving I also researched more. Tested the 12V inverter and new EL Wire. Below i will provide pictures of the testing and a video. Also started to create a code.

Week3 Narrative
The Equipment arrived and I started working on it. Came to some problems I learned as I went through that you can not just place the USB FDTI connector on the EL Sequencer in any way. You must match the the connectors. If not you will not be able to upload code. I found several example codes that worked great. Now I must Create my own code. Here is part of the:

EL Code

Here is a Video of my success:

EL Wire Project

Here some Pictures of my progress:

My Fourth task
On my fourth week I will Try to come up with a code that will work great. i will also try to Get the battery working so the EL sequencer can run off the battery and not the USB. I will also ask the professor to buy a 3.3v inverter from spark fun so we can make this process mobile.

Summary of actual work over fourth weekend
How is what you did different than what you planned?

Week4 Narrative
Tell a detailed story describing what you did for your team over the weekend.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).