text:
User:Egm6321.f10.team03/Hwk7

= Problem 1 - Plot Legendre functions=

Given
Legendre polynomials (functions /first homogeneous solutions)

 * $$\begin{align}

& {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ & {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1) \\ & {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x) \\ & {{P}_{4}}(x)=\frac{1}{8}(35{{x}^{4}}-30{{x}^{2}}+3) \\ \end{align}$$

Legendre functions (second homogeneous solutions)

 * $$\begin{align}

& {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{Q}_{1}}(x)=\frac{1}{2}x\log \left( \frac{1+x}{1-x} \right)-1 \\ & {{Q}_{2}}(x)=\frac{1}{4}(3{{x}^{2}}-1)\log \left( \frac{1+x}{1-x} \right)-\frac{3}{2}x \\ & {{Q}_{3}}(x)=\frac{1}{4}(5{{x}^{3}}-3x)\log \left( \frac{1+x}{1-x} \right)-\frac{5}{2}{{x}^{2}}+\frac{2}{3} \\ & {{Q}_{4}}(x)=\frac{1}{16}(35{{x}^{4}}-30{{x}^{2}}+3)\log \left( \frac{1+x}{1-x} \right)-\frac{35}{8}{{x}^{3}}+\frac{55}{24}x \end{align}$$

Find

 * (1) Plot $$\displaystyle \left\{ {{P}_{0}}(x),{{P}_{1}}(x),{{P}_{2}}(x),{{P}_{3}}(x) \right\}$$ and observe $$\displaystyle {{P}_{n}}(x)$$ as $$\displaystyle x\to \pm 1$$

 * (2) Plot $$\displaystyle\left\{ {{Q}_{0}}(x),{{Q}_{1}}(x),{{Q}_{2}}(x),{{Q}_{3}}(x) \right\}$$ and observe $$\displaystyle {{Q}_{n}}(x)$$ as $$\displaystyle x\to \pm 1$$

 * (3) Plot $$\displaystyle \left\{ {{P}_{0}}(x),{{Q}_{0}}(x) \right\}$$, $$\displaystyle \left\{ {{P}_{1}}(x),{{Q}_{1}}(x) \right\}$$, $$\displaystyle \left\{ {{P}_{2}}(x),{{Q}_{2}}(x) \right\}$$, $$\displaystyle \left\{ {{P}_{3}}(x),{{Q}_{3}}(x) \right\}$$, $$\displaystyle \left\{ {{P}_{4}}(x),{{Q}_{4}}(x) \right\}$$

 * (4) Observe even-ness and odd-ness of $$\displaystyle \left\{ {{P}_{i}}(x),{{Q}_{i}}(x) \right\}$$, $$\displaystyle i=0,1,2,3,4$$ and guess

 * $$\displaystyle \int\limits_{-1}^{+1}{{{P}_{i}}(x){{Q}_{i}}(x)dx=}?$$

(1) Plot $$\displaystyle \left\{ {{P}_{0}}(x),{{P}_{1}}(x),{{P}_{2}}(x),{{P}_{3}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig1.jpg|center|650px]]

 * As $$\displaystyle x\to \pm 1$$, $$\displaystyle {{P}_{n}}(x)\to \pm 1$$. When n is even (0,2..), $$\displaystyle {{P}_{n}}(\pm 1)=1$$ and when n is odd (1,3,...), $$\displaystyle {{P}_{n}}(\pm 1)=\pm 1$$. It can also be seen that $$\displaystyle {{P}_{n}}(x)$$ have steepest slope as $$\displaystyle x\to \pm 1$$.

(2) Plot $$\displaystyle\left\{ {{Q}_{0}}(x),{{Q}_{1}}(x),{{Q}_{2}}(x),{{Q}_{3}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig2.jpg|center|650px]]

 * As $$\displaystyle x\to \pm 1$$, $$\displaystyle {{Q}_{n}}(x)$$ also have steepest slope like $$\displaystyle {{P}_{n}}(x)$$.

(3a) Plot $$\displaystyle \left\{ {{P}_{0}}(x),{{Q}_{0}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig3.jpg|center|650px]]

(3b) Plot $$\displaystyle \left\{ {{P}_{1}}(x),{{Q}_{1}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig4.jpg|center|650px]]

(3c) Plot $$\displaystyle \left\{ {{P}_{2}}(x),{{Q}_{2}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig5.jpg|center|650px]]

(3d) Plot $$\displaystyle \left\{ {{P}_{3}}(x),{{Q}_{3}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig6.jpg|center|650px]]

(3e) Plot $$\displaystyle \left\{ {{P}_{4}}(x),{{Q}_{4}}(x) \right\}$$

 * [[Image:Sudheesh HW7 1 fig7.jpg|center|650px]]

 * When $$\displaystyle n$$ is 'even' (0,2,4), $$\displaystyle {{P}_{n}}(x)$$ is 'even' and $$\displaystyle {{Q}_{n}}(x)$$ is 'odd'

 * When $$\displaystyle n$$ is 'odd' (1,3), $$\displaystyle {{P}_{n}}(x)$$ is 'odd' and $$\displaystyle {{Q}_{n}}(x)$$ is 'even'

 * As a result, in each $$\displaystyle \left\{ {{P}_{n}}(x),{{Q}_{n}}(x) \right\}$$ combination, one is odd and the other is even.

 * So Guess:

 * $$\displaystyle \int\limits_{-1}^{+1}{{{P}_{i}}(x){{Q}_{i}}(x)dx=}0$$

Contributing members

 * Solved and posted by Egm6321.f10.team3.Sudheesh 21:25, 5 December 2010 (UTC)
 * proof by Egm6321.f10.team3.Hong SJ 22:45, 5 December 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 01:00, 7 December 2010 (UTC)

=Problem 2: Summation of functions =

Given
Given is the definition of a function, which is the sum of two or more functions:
 * {| style="width:100%" border="0"

$$f(x)=\sum\limits_{i}{{{g}_{i}}\left( x \right)}$$
 * $$\,(Eq. 2.1) $$
 * style= |
 * }

Find
It has to be shown that if each gi(x) is an even function also f(x) is an even function and that if gi(x) is odd also f(x) is an odd function.

Solution
As first we need to look at how even and odd functions are defined.

Even functions: A function gi(x) is even if:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.2) $$
 * style= |
 * }

Odd functions: A function gi(x) is odd if:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={-{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.3) $$
 * style= |
 * }

Now let's split the function g(x) into 3 parts: The part for all x<0, the part for x=0 and the part for all x>0, denoted by g-(x), g0(x) and g+(x) respectively. So that the function g(x) can now be written as:
 * {| style="width:100%" border="0"

$$g\left( x \right)={{g}_{-}}\left( x \right)+{{g}_{0}}\left( x \right)+{{g}_{+}}\left( x \right)$$
 * $$\,(Eq. 2.4) $$
 * style= |
 * }

The properties of the single functions are:
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_{-}}\left( x \right)=\left\{ _{0\quad for\quad x\ge 0}^{\mathbb{Z}\quad for\quad x<0} \right. \\ & {{g}_{0}}\left( x \right)=\left\{ _{0\quad for\quad x\ne 0}^{\mathbb{Z}\quad for\quad x=0} \right. \\ & {{g}_{+}}\left( x \right)=\left\{ _{0\quad for\quad x\le 0}^{\mathbb{Z}\quad for\quad x>0} \right. \\ \end{align}$$ THE EVEN CASE:
 * $$\,(Eqs. 2.5) $$
 * style= |
 * }

Now, accordingly to Eq. 2.2 and with the properties of Eqs. 2.5 for an even function gi(x) we can set g-(x) = g+(x) so that the function gi(x) can be rewritten as:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}_{+}\left( x \right)+{{g}_{_{i}0}}\left( x \right)+{{g}_{i}}_{+}\left( x \right)$$ Summing n even functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right)} \\ & {{f}_{+}}\left( x \right)+{{f}_{0}}\left( x \right)+{{f}_{+}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.6 which we found to be even, therefore also the sum of all gi(x), which is f(x), has to be even!'''
 * $$\,(Eqs. 2.7) $$
 * style= |
 * }

THE ODD CASE:

Accordingly to Eq. 2.3 and with the properties of Eqs. 2.5 for an odd function gi(x) we can set g-(x) = -g-(x) so that the function gi(x) can be rewritten as:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}_{-}\left( x \right)+{{g}_{_{i}0}}\left( x \right)-{{g}_{i}}_{-}\left( x \right)$$ Summing n odd functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right)} \\ & {{f}_{-}}\left( x \right)+{{f}_{0}}\left( x \right)-{{f}_{-}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.8 which we found to be odd, therefore also the sum of all gi(x), which is f(x), has to be odd!'''
 * $$\,(Eqs. 2.9) $$
 * style= |
 * }

Contribution

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

The solution above was derived from scratch by the author without consulting last years Homework!
 * style="width:45%; padding:1px; border:2px solid #CCFF33" |
 * style="width:45%; padding:1px; border:2px solid #CCFF33" |
 *
 * style= |
 * }

Solved and posted by Egm6321.f10.team03.sigillo 21:02, 5 December 2010 (UTC)

= Problem 3: Even and Odd Legendre Polynomials =

Given
The Legendre Polynomials can be found using the following equation
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3.1.1)
 * }
 * }

Find
Show that,

1. $$\displaystyle P_{2k}(x)$$ is even

2. $$\displaystyle P_{2k+1}(x)$$ is odd

given $$\displaystyle k = 0,1,2,3,...$$.

Solution
1. An even function is $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ into Eq.3.1.1.
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3.1.2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3.1.3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3.1.4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \rightarrow P_{2k}\left(x\right) = P_{2k}\left(-x\right) $$ $$ 2. An odd function is $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ into Eq.3.1.1
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 3.1.5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3.1.6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3.1.7)
 * }
 * }

Evaluating Eq.3.1.7 at $$\displaystyle -x$$ gives
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$ Comparing Eqs.3.1.7 and 3.1.8 shows that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3.1.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \rightarrow P_{2k+1}\left(x\right) = -P_{2k+1}\left(-x\right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 3.1.5)
 * }
 * }

Given
For $$q(x)=\sum_{i=0}^4c_ix^i$$ where $$c_0=2,\ c_1=-5,\ c_2=-3,\ c_3=11,\ c_4=7,\ c_5=6$$

Find
Find $$\displaystyle a_i$$ such that $$q=\sum_{i=0}^4a_iP_i$$.

Plot $$\displaystyle q=\sum_ic_ix^i$$ and $$\displaystyle q=\sum_ia_iP_i$$.

Solution
$$q(x)=\sum_{i=0}^5c_ix^i$$ can be shown with components as column matrix

$$=\left[ c \right]\left[ \begin{matrix} 1 \\   x  \\ {{x}^{2}} \\ {{x}^{3}} \\ {{x}^{4}} \\ {{x}^{5}} \\ \end{matrix} \right]$$

Next, evaluate $$q(x)=\sum_{i=0}^5a_iP_i$$ and show its components

$$q(x)=\sum\limits_{i=0}^{5}{{P}_{i}}=\left[ a \right]\left[ P \right]$$

Now $$\displaystyle \left[ P \right]$$ can be written as

$$\left[ P \right]=\left[ \begin{matrix} 1 \\   x  \\ \frac{1}{2}(3{{x}^{2}}-1) \\ \frac{1}{2}(5{{x}^{3}}-3x) \\ \begin{align} & \frac{1}{8}(35{{x}^{4}}-30{{x}^{2}}+3) \\ & \frac{1}{8}(63{{x}^{5}}-70{{x}^{3}}+15x) \\ \end{align} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 & 0 & 0  \\   -1/2 & 0 & 3/2 & 0 & 0 & 0  \\   0 & -3/2 & 0 & 5/2 & 0 & 0  \\   3/8 & 0 & -15/4 & 0 & 35/8 & 0  \\   0 & 15/8 & 0 & -35/4 & 0 & 63/8  \\ \end{matrix} \right]}_{\left[ \alpha  \right]}\left[ \begin{matrix} 1 \\   x  \\ {{x}^{2}} \\ {{x}^{3}} \\ {{x}^{4}} \\ {{x}^{5}} \\ \end{matrix} \right]$$

Setting $$q(x)=\sum_{i=0}^5a_iP_i(x)=\sum_{i=0}^5c_ix^i $$ yields

$$\left[ c \right]=\left[ a \right]\left[ \alpha \right]$$

$$\Rightarrow \left[ a \right]=\left[ c \right]{{\left[ \alpha \right]}^{-1}}$$

Using the above equation $$\displaystyle[a]$$ can be evaluated as

$$[a]=\left[ \begin{matrix} 2.40 \\   4.17  \\   2.00  \\   7.07  \\   \begin{align} & 1.60 \\ & 0.76 \\ \end{align}  \\ \end{matrix} \right]$$

The plot of $$\displaystyle q=\sum_ic_ix^i=\sum_ia_iP_i$$ is shown below.

Reference

 Author solved the problem referring the procedure[| Hw6,Prob#11-Team1, 2009].

Contributing Members

 * Solved and posted by Egm6321.f10.team3.franklin 16:57, 5 December 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 01:02, 7 December 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 03:47, 7 December 2010 (UTC)

= Problem 4 Find Coefficcients =

Given
Consider the boundary condition
 * {| style="width:100%" border="0" align="left"

a)f(\theta) = T_0 \cos^6 \theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.1.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

b)f(\theta) = T_0 \exp{\left(-\left(\frac{2\theta}{\pi}\right)^2\right)} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.1.b)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a)f(\mu) = T_0 \left(1 - \mu^2\right)^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.2.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

b)f(\mu) = T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} $$ $$ is even, and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.2.b)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mu = \sin \theta. $$ $$ $$\displaystyle A_n $$ is
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

A_n = \frac{2n + 1}{2} <f, P_n> $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.4)
 * }
 * }

1.
Without calculation, find property of $$\displaystyle A_n$$, i.e.
 * {| style="width:100%" border="0" align="left"

A_{2k} \stackrel{?}{=} 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.5.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

A_{2k+1} \stackrel{?}{=} 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4.5.b)
 * }
 * }

2.
Compute 3 non-zero coefficients of $$\displaystyle A_n .$$

Solution for 1.
The scalar product of $$\displaystyle f $$ and $$\displaystyle P_n $$ is
 * {| style="width:100%" border="0" align="left"

<f, P_n> = \int_{-1}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ We can split the integral up in two intervals
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_n> = \int_{-1}^0 f(\mu) P_n(\mu) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ When $$\displaystyle n$$ is odd, so is $$\displaystyle P_n$$, and $$\displaystyle f(\mu)$$ is $$\displaystyle even$$. Hence, when $$\displaystyle n$$ is odd, we have that $$\displaystyle n = 2k + 1$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = \int_{-1}^0 f(-\mu) (-P_{2k+1}(-\mu)) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$ We can substitute $$\displaystyle \xi = -\mu$$ in the first integral so that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = \int_1^0 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = - \int_0^1 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu = 0 $$ $$ Hence
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle A_{2k+1} = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 4a.11)
 * }
 * }

Part a.
The expression for $$\displaystyle A_n $$ is given by,

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

A_n = \frac{2n+1}{2} \int_{-1}^{1} f(\mu)P_n(\mu)d\mu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.12)
 * }
 * }

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

f\left(\mu\right) = \left(1-\mu^2\right)^3 , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.13)
 * }
 * }

Knowing that $$\displaystyle P_i\left(\mu\right)$$ for $$\displaystyle i = 0, 2, 4, 6 $$ is

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

\begin{align} &P_0\left(\mu\right)=1 ,\\ &P_2\left(\mu\right)=\frac{1}{2}\left(3\mu^2-1\right) ,\\ &P_4\left(\mu\right)=\frac{1}{8}\left(35\mu^4-30\mu^2+3\right) ,\\ &P_6\left(\mu\right)=\frac{1}{16}\left(231\mu^6-315\mu^4+105\mu^2-5\right) .\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 4a.14)
 * }
 * }

Plugging in Eqs 4a.14 and 4a.13 into Eq 4a.12 gives

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

A_n = \frac{2(n)+1}{2} \int_{-1}^{1}T_0 \left(1-\mu^2\right)^3 P_n\left(\mu\right) d\mu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.15)
 * }
 * }

Using Matlab to evaluate the definite integral above, we get
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{align} & A_0=T_0\frac{16}{35} \\ & A_2=-T_0\frac{16}{21} \\ & A_4=T_0\frac{144}{385} \\ & A_6=-T_0\frac{16}{231} \\ \end{align} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style= |
 *
$$\displaystyle (Eq. 4b.15)
 * }
 * }

Part b.
The expression for $$\displaystyle A_n $$ is given by,

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

A_n = \frac{2n+1}{2} \int_{ \frac{-\pi}{2}}^{\frac{\pi}{2}} f(\theta)P_n(sin\theta)d\theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4b.12)
 * }
 * }

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

f(x)={{T}_{0}}\exp \left( \left(-\frac{2x}{\pi}\right)^2 \right) , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4b.13)
 * }
 * }

Knowing that $$\displaystyle P_i\left(\mu\right)$$ for $$\displaystyle i = 0, 2, 4, 6 $$ is

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

\begin{align} &P_0\left(sin\theta\right)=1 ,\\ &P_2\left(sin\theta\right)=\frac{1}{2}\left(3\sin\theta^2-1\right) ,\\ &P_4\left(sin\theta\right)=\frac{1}{8}\left(35\sin\theta^4-30\sin\theta^2+3\right) ,\\ &P_6\left(sin\theta\right)=\frac{1}{16}\left(231\sin\theta^6-315\sin\theta^4+105\sin\theta^2-5\right) .\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 4b.14)
 * }
 * }

Plugging in Eqs 4b.14 and 4b.13 into Eq 4b.12 gives

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

$$
 * $$\displaystyle A_n = \frac{2(n)+1}{2} \int_{ \frac{-\pi}{2}}^{\frac{\pi}{2}}T_0 \exp \left( \left(-\frac{2x}{\pi}\right)^2 \right) P_n\left(\sin\theta\right) d\theta $$
 *
$$\displaystyle (Eq. 4b.15)
 *
$$\displaystyle (Eq. 4b.15)
 * }
 * }

Using Wolfram Alpha to evaluate the definite integral above, we get
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{align} & {{A}_{0}}=T_0 1.173108 \\ & {{A}_{2}}=T_0 0.639393 \\ & {{A}_{4}}=T_0 0.714901 \\ & {{A}_{6}}=T_0 0.726179 \\ \end{align} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style= |
 *
$$\displaystyle (Eq. 4b.15)
 * }
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team3.franklin 16:57, 5 December 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 03:51, 7 December 2010 (UTC)

= Problem 5 - Relations Between Inverse Hyperbolic Tangent Function And Legendre Functions $$\displaystyle {{Q}_{0}}(x),{{Q}_{1}}(x)$$=

Given

 * Legendre functions:

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

 * $${{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
 *
$$\displaystyle (Eq. 5.1)$$
 * }
 * }
 * }

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

 * $${{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1$$
 *
$$\displaystyle (Eq. 5.2)$$
 * }
 * }
 * }

Find

 * Show that:

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

 * $$\displaystyle{{Q}_{0}}(x)={{\tanh }^{-1}}(x)$$
 * }
 * }
 * }

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

 * $$\displaystyle{{Q}_{1}}(x)=x\cdot {{\tanh }^{-1}}(x)-1$$
 * }
 * }
 * }

Solution

 * We know that:

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

 * $$\tanh (\alpha )=\frac{{{e}^{\alpha }}-{{e}^{-\alpha }}}{{{e}^{\alpha }}+{{e}^{-\alpha }}}$$
 *
$$\displaystyle (Eq. 5.3)$$
 * }
 * }
 * }

 * Substitute $$\displaystyle \alpha =\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$ in $$\displaystyle Eq. 5.3$$ implies:

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

 * $$\tanh \left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)=\tanh \left( \log \sqrt{\frac{1+x}{1-x}} \right)=\frac{{{e}^{\log \sqrt{\frac{1+x}{1-x}}}}-{{e}^{-\log \sqrt{\frac{1+x}{1-x}}}}}{{{e}^{\log \sqrt{\frac{1+x}{1-x}}}}-{{e}^{-\log \sqrt{\frac{1+x}{1-x}}}}}$$
 * }
 * }
 * }

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

 * $$=\frac{{{e}^{\log \sqrt{\frac{1+x}{1-x}}}}-{{e}^{\log \sqrt{\frac{1-x}{1+x}}}}}{{{e}^{\log \sqrt{\frac{1+x}{1-x}}}}-{{e}^{\log \sqrt{\frac{1-x}{1+x}}}}}=\frac{\sqrt{\frac{1+x}{1-x}}-\sqrt{\frac{1-x}{1+x}}}{\sqrt{\frac{1+x}{1-x}}+\sqrt{\frac{1-x}{1+x}}}=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}=\frac{2x}{2}$$
 * }
 * }
 * }

 * That is,

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

 * $$\tanh \left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)=x$$
 * }
 * }
 * }

 * Therefore,

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

$${{\tanh }^{-1}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{Q}_{0}}(x)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.4)$$
 * }
 * }

 * Now substitute $$\displaystyle Eq. 5.4$$ in $$\displaystyle (Eq. 5.2)$$ yields

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

$${{Q}_{1}}(x)=x\cdot {{\tanh }^{-1}}(x)-1$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.5)$$
 * }
 * }

.

 Author solved this problem without referring to the $$\displaystyle F\phi 9$$

Contributing Members

 * Posted by Egm6321.f10.team3.Sudheesh 22:45, 5 December 2010 (UTC)
 * Solved by Egm6321.f10.team3.Schulze
 * Proofread by Egm6321.f10.team03.sigillo 01:04, 7 December 2010 (UTC)

= Problem 6 : Odd and Even Solutions of Legendre Polynomials=

Given

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

\begin{align}
 * $$\displaystyle
 * $$\displaystyle

Q_{n}(x) = P_{n}(x) \cdot tanh^{-1}(x) - 2 \sum_{i=1,3,5}^{J} \left[ \frac {2n-2J-1}{(2n-j+1)J} \cdot P_{n-j}(x) \right]

\end{align} $$ $$
 *
$$\displaystyle Eqt (7.1)
 * }
 * }

Find
Show that Qn(x) = EVEN or ODD function depending on "n"

Solution
"P_{2k}(x) = even function & P_{2k+1}(x) = odd" is shown before.

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

\begin{align} (1) P_{2k}(x) = [even] \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle Eqt (7.2)
 * }
 * }

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

\begin{align} (2) P_{2k+1}(x) = [odd] \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle Eqt (7.3)
 * }
 * }

Thus, Qn(x) is determined by n.

From Eqt(7.2),

When n = even = 2k, where k = 1,2,3...

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

\begin{align}
 * $$\displaystyle
 * $$\displaystyle

Q_{2k}(x) &= \underbrace{P_{2k}(x)}_{[even]} \cdot \underbrace{tanh^{-1}(x)}_{[odd]} - \underbrace{2 \sum_{i=1,3,5}^{J} \left[ \frac {4k-2J+1}{(4k-j+1)J} \cdot P_{2k-j}(x) \right]}_{[even] (\because 2 \times all values = even)}\\ &= [even] \cdot [odd] - [even]\\ &= [odd] - [even]\\ &= [odd] \end{align} $$ $$
 *
$$\displaystyle
 * }
 * }

From Eqt(7.3),

When n = odd = 2k+1, where k = 1,2,3...

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

\begin{align}
 * $$\displaystyle
 * $$\displaystyle

Q_{2k+1}(x) &= \underbrace{P_{2k+1}(x)}_{[odd]} \cdot \underbrace{tanh^{-1}(x)}_{[odd]} - \underbrace{2 \sum_{i=1,3,5}^{J} \left[ \frac {4k-2J+3}{(4k-j+3)J} \cdot P_{2k+1-j}(x) \right]}_{[even] (\because 2 \times all values = even)}\\ &= [odd] \cdot [odd] - [even]\\ &= [even] - [even]\\ &= [even] \end{align} $$ $$
 *
$$\displaystyle
 * }
 * }

As a result, Qn(x) can be odd or even

Given

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

\begin{align}
 * $$\displaystyle
 * $$\displaystyle

<f(x), g(x)> = \int_{-1}^{+1} f(x) \cdot g(x) dx

\end{align} $$ $$
 *
$$\displaystyle
 * }
 * }

Find
Show that <Pn(x), Qn(x)> = 0, where Pn(x)& Qn(x)> are Legendre Polynimials.

Solution
As problem 7-6, part 1 shows that,

When Pn(x) = odd function, Qn(x) = even function.

When Pn(x) = even function, Qn(x) = odd function.

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

\begin{align}
 * $$\displaystyle
 * $$\displaystyle

<P_{n}, Q_{n}> = \int_{-1}^{+1} P_{n}(x) \cdot Q_{n}(x) dx = \int_{-1}^{+1} \underbrace {[odd]}_{all the time} dx

\end{align} $$ $$
 *
$$\displaystyle
 * }
 * }

Reference

 Author solved the problem referring the procedure[Hw7,Prob#3-Team3, 2009]

Contributing Members
Solved and posted by Egm6321.f10.team03.Hong SJ 16:52, 03 December 2010 (UTC)

= Problem 7: Show the length of vector $${r}_{PQ}$$=

Given
Given the definition of the vector from the lecture notes:
 * {| style="width:100%" border="0"

$$r_{PQ}^{2}=\sum\limits_{i=1}^{3}{\left( x_{Q}^{i}-x_{P}^{i} \right)^{2}}$$
 *
$$\displaystyle (Eq. 7.1) $$
 * }
 * }

Find
Show that 7.1 is similar to
 * {| style="width:100%" border="0"

$$\left( r_{PQ} \right)^{2}=r_{Q}^{2}\left( \left( \frac{r_{P}}{r_{Q}} \right)^{2}+1-2\left( \frac{r_{P}}{r_{Q}} \right)\cos \gamma, \right)$$
 *
$$\displaystyle (Eq. 7.2) $$
 * }
 * }

Solution
Given in lecture that

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

$$\cos \gamma =\cos \left( \theta _{Q} \right)\cos \left( \theta _{P} \right)\cos \left( \phi _{Q}-\phi _{P} \right)+\sin \left( \theta _{Q} \right)\sin \left( \theta _{P} \right)$$ Multiplying out the square in equation 7.1
 *
$$\displaystyle (Eq. 7.3) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\left( r_{PQ} \right)^{2}=\sum\limits_{i=1}^{3}{\left( x_{Q}^{i} \right)^{2}-2x_{Q}^{i}x_{P}^{i}+\left( x_{P}^{i} \right)^{2}}$$ Expanding and converting to spherical coordinates as follows:
 *
$$\displaystyle (Eq. 7.4) $$
 * }
 * }
 * $$x_{P}^{1}=r_{P}\cos \left( \theta _{P} \right)\cos \left( \phi _{P} \right)$$
 * $$x_{P}^{2}=r_{P}\cos \left( \theta _{P} \right)\sin \left( \phi _{P} \right)$$
 * $$x_{P}^{3}=r_{P}\sin \left( \theta _{P} \right)$$
 * $$x_{Q}^{1}=r_{Q}\cos \left( \theta _{Q} \right)\cos \left( \varphi _{Q} \right)$$
 * $$x_{Q}^{2}=r_{Q}\cos \left( \theta _{Q} \right)\sin \left( \varphi _{Q} \right)$$
 * $$x_{Q}^{3}=r_{Q}\sin \left( \theta _{Q} \right)$$

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

 * $$\left( r_{PQ} \right)^{2}=r_{Q}^{2}\cos ^{2}\left( \theta _{Q} \right)\cos ^{2}\left( \phi _{Q} \right)-2r_{Q}\cos \left( \theta _{Q} \right)\cos \left( \phi _{Q} \right)r_{P}\cos \left( \theta _{P} \right)\cos \left( \phi _{P} \right)+r_{P}^{2}\cos ^{2}\left( \theta _{P} \right)\cos ^{2}\left( \phi _{P} \right)$$

 * $$+r_{Q}^{2}\cos ^{2}\left( \theta _{Q} \right)\sin ^{2}\left( \phi _{Q} \right)-2r_{Q}\cos \left( \theta _{Q} \right)\sin \left( \phi _{Q} \right)r_{P}\cos \left( \theta _{P} \right)\sin \left( \phi _{P} \right)+r_{P}^{2}\cos ^{2}\left( \theta _{P} \right)\sin ^{2}\left( \phi _{P} \right)$$

 * $$+r_{Q}^{2}\sin ^{2}\left( \theta _{Q} \right)-2r_{Q}\sin \left( \theta _{Q} \right)r_{P}\sin \left( \theta _{P} \right)\cos \left( \phi _{P} \right)+r_{P}^{2}\sin ^{2}\left( \theta _{P} \right)$$

Re-arranging and grouping similar terms to use trigometric identities, equation 7.5 becomes
 *
$$\displaystyle (Eq. 7.5) $$
 * }
 * }

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

 * $$r_{Q}^{2}\cos ^{2}\left( \theta _{Q} \right)\left( \cos ^{2}\left( \phi _{Q} \right)+\sin ^{2}\left( \phi _{Q} \right) \right)-2r_{Q}r_{P}\cos \left( \theta _{Q} \right)\cos \left( \theta _{P} \right)\left( \cos \left( \phi _{Q} \right)\cos \left( \phi _{P} \right)+\sin \left( \phi _{Q} \right)\sin \left( \phi _{P} \right) \right)$$

 * $$r_{P}^{2}\cos ^{2}\left( \theta _{P} \right)\left( \cos ^{2}\left( \phi _{P} \right)+\sin ^{2}\left( \phi _{P} \right) \right)+r_{Q}^{2}\sin ^{2}\left( \theta _{Q} \right)-2r_{Q}r_{P}\sin \left( \theta _{Q} \right)\sin \left( \theta _{P} \right)+r_{P}^{2}\sin ^{2}\left( \theta _{P} \right)$$

 *
$$\displaystyle (Eq. 7.6) $$
 * }
 * }

Simplifying and re-arranging again
 * {| style="width:100%" border="0"

$$r_{Q}^{2}\left( \cos ^{2}\left( \theta _{Q} \right)+\sin ^{2}\left( \theta _{Q} \right) \right)-2r_{Q}r_{P}\left[ \cos \left( \phi _{Q}-\phi _{P} \right)\cos \left( \theta _{P} \right)\cos \left( \theta _{Q} \right)+\sin \left( \theta _{P} \right)\sin \left( \theta _{Q} \right) \right]+r_{P}^{2}\left( \cos ^{2}\left( \theta _{P} \right)+\sin ^{2}\left( \theta _{P} \right) \right)$$ Using equation 7.3 and trigometric identities, equation 7.7 can be further simplified
 *
$$\displaystyle (Eq. 7.7) $$
 * }
 * }

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

$$\left( r_{PQ} \right)^{2}=r_{Q}^{2}-2r_{Q}r_{P}\cos \gamma +r_{P}^{2}$$
 *
$$\displaystyle (Eq. 7.8) $$
 * }
 * }

With some re-arranging, we arrive at our solution
 * {| style="width:100%" border="0"

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left( r_{PQ} \right)^{2}=r_{Q}^{2}\left( \left( \frac{r_{P}}{r_{Q}} \right)^{2}+1-2\left( \frac{r_{P}}{r_{Q}} \right)\cos \gamma, \right)$$

 *
$$\,(Eq. 7.9) $$
 * style= |
 * }

Contributing Members
Solved and Posted by James Roark 04:14, 7 December 2010 (UTC), with help with trigometric idenities from F09 Team 4

=Problem 8: Find the Binomial expansion of $$ { \left(1-x \right) }^{- \frac{1}{2}} $$ =

Given
From [|Mtg 40-4] of the lecture notes we have the Binomial Theorem generalized for real numbers.

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

$$ { \left(x+y\right) }^{r}= \sum_{k=0}^{\infty} \binom{r}{k}{x}^{r-k}{y}^{k}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 8.1)
 * }

Where the Binomial coefficient

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

$$ \binom{r}{k}= \frac{r \left( r-1 \right) \left( r-2 \right).... \left( r-k+1 \right)}{k!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 8.2)
 * }

Note: By the [| Multiplicitve Formula] $$\binom{r}{k} = \frac{{2}^{r}}{k!}$$. This allows the Binomial coefficient to be generalized to include real and complex numbers.

Find
Use the General Binomial Theorem to obtain (6)&(7) from [|Mtg 40-3] of the lecture notes:

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

{\left(1-x\right) }^{- \frac{1}{2}}= \sum_{i=0}^{\infty} {\alpha }_{i}{x}^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8.3)
 * style= |
 * }

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

{ \alpha }_{i} = \frac{1*3*5..... \left(2i-1 \right)}{2*4*6.... \left( 2i\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8.4)
 * style= |
 * }

Solution
Apply the General Binomial Theorem for the parameters $$\displaystyle r= -\frac{1}{2} ; y = 1 ; x = -x$$

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

{ \left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{- \frac{1}{2}-k} {1}^{k}$$ $$ Using the hint provided in the lecture notes we will assume that $$ \left| x\right|  < 1 $$ to obtain
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * style= |
 * }

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

{ \left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{k} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8.5)
 * style= |
 * }

Now we evaluate the binomial coefficient using $$\displaystyle (Eq. 8.2)$$ we have

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

\binom{ -\frac{1}{2}}{k}= \frac{-\frac{1}{2} \left( -\frac{1}{2}-1 \right) \left( -\frac{1}{2}-2 \right).... \left( -\frac{1}{2}-k+1 \right)}{k!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * style= |
 * }

Now we pull out the common factor of $$ \frac{1}{2} $$ and distribute through the negative we obtain

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

\binom{ -\frac{1}{2}}{k}= \frac{1}{2}\frac{1 \left( 1+2 \right) \left( 1+4 \right).... \left( 1+2k - 2 \right)}{k!}=\frac{\left( 1 \right) \left( 3\right) \left( 5  \right).... \left( 2k - 1 \right)}{2k!} = \frac{\left( 1 \right) \left( 3\right) \left( 5  \right).... \left( 2k - 1 \right)}{\left( 2 \right) \left( 4  \right) \left( 6 \right).... \left( 2k \right) } $$ $$ We can see that $$ k=i$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8.6)
 * style= |
 * }
 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

\binom{ -\frac{1}{2}}{k}=  { \alpha }_{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * style= |
 * }

Changing the index from k to i we have:
 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{\left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{k} = \sum_{i=0}^{\infty} {\alpha }_{i}{x}^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * style= |
 * }

Thus (6) & (7) can be obtained from the General Binomial Theorem.

Contributing Members
Solved and Posted by --Egm6321.f10.team3.cook 13:28, 7 December 2010 (UTC)

= Problem 9 - Generate Legendre Polynomials Using Second Reccurence Relation=

Given
Second reccurence relation of Legendre polynomials:

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

 * $$\left( n+1 \right){{P}_{n+1}}-\left( 2n+1 \right)x{{P}_{n}}+n{{P}_{n-1}}=0$$
 *
$$\displaystyle (Eq. 9.1)$$
 * }
 * }
 * }

Legendre polynomials for n=0,1:

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

 * $$\displaystyle {{P}_{0}}(x)=1$$
 * }
 * }
 * }

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

 * $$\displaystyle {{P}_{1}}(x)=x$$
 * }
 * }
 * }

Find
Generate $$\displaystyle \left\{ {{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}},{{P}_{6}} \right\}$$ using second reccurence relation starting from $$\displaystyle {{P}_{0}}$$ and $$\displaystyle {{P}_{1}}$$ and compare with [[media:2010_11_09_15_00_14.djvu|Eq.4-6 in page-2 Mtg-36]]

Solution

 * $$\displaystyle Eq. 9.1$$ can be written as:

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

 * $${{P}_{n+1}}=\frac{\left( 2n+1 \right)x{{P}_{n}}-n{{P}_{n-1}}}{\left( n+1 \right)}$$
 *
$$\displaystyle (Eq. 9.2)$$
 * }
 * }
 * }

Generate $$\displaystyle {{P}_{2}}$$

 * Substitute $$\displaystyle n=1$$ in $$\displaystyle Eq. 9.2$$ implies:

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

 * $${{P}_{2}}=\frac{\left( 2\times 1+1 \right)x{{P}_{1}}-1\times {{P}_{0}}}{\left( 1+1 \right)}$$
 * }
 * }
 * }

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

 * $${{P}_{2}}=\frac{3x\cdot x-1\times 1}{2}$$
 * }
 * }
 * }

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

$${{P}_{2}}=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.3)$$
 * }
 * }

Generate $$\displaystyle {{P}_{3}}$$

 * Plugg in $$\displaystyle n=2$$ in $$\displaystyle Eq. 9.2$$ implies:

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

 * $${{P}_{3}}=\frac{\left( 2\times 2+1 \right)x{{P}_{2}}-2{{P}_{1}}}{\left( 2+1 \right)}$$
 * }
 * }
 * }

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

 * $${{P}_{3}}=\frac{5x\cdot \frac{1}{2}\left( 3{{x}^{2}}-1 \right)-2x}{3}$$
 * }
 * }
 * }

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

$${{P}_{3}}=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.4)$$
 * }
 * }

Generate $$\displaystyle {{P}_{4}}$$

 * Put $$\displaystyle n=3$$ in $$\displaystyle Eq. 9.2$$ implies:

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

 * $${{P}_{4}}=\frac{\left( 2\times 3+1 \right)x{{P}_{3}}-3{{P}_{2}}}{\left( 3+1 \right)}$$
 * }
 * }
 * }

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

 * $${{P}_{4}}=\frac{7x\cdot \frac{1}{2}\left( 5{{x}^{3}}-3x \right)-3\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)}{4}$$
 * }
 * }
 * }

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

 * $${{P}_{4}}=\frac{35{{x}^{4}}-21{{x}^{2}}-9{{x}^{2}}+3}{8}$$
 * }
 * }
 * }

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

$${{P}_{4}}=\frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.5)$$
 * }
 * }

Generate $$\displaystyle {{P}_{5}}$$

 * Put $$\displaystyle n=4$$ in $$\displaystyle Eq. 9.2$$ implies:

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

 * $${{P}_{5}}=\frac{\left( 2\times 4+1 \right)x{{P}_{4}}-4{{P}_{3}}}{\left( 4+1 \right)}$$
 * }
 * }
 * }

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

 * $${{P}_{5}}=\frac{9x\cdot \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)-4\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)}{5}$$
 * }
 * }
 * }

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

 * $${{P}_{5}}=\frac{63{{x}^{5}}}{8}-\frac{54{{x}^{3}}}{8}+\frac{27x}{8\times 5}-2{{x}^{3}}+\frac{6x}{5}$$
 * }
 * }
 * }

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

$${{P}_{5}}=\frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.6)$$
 * }
 * }

Generate $$\displaystyle {{P}_{6}}$$

 * substitute $$\displaystyle n=5$$ in $$\displaystyle Eq. 9.2$$ yields:

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

 * $${{P}_{6}}=\frac{\left( 2\times 5+1 \right)x{{P}_{5}}-5{{P}_{4}}}{\left( 5+1 \right)}$$
 * }
 * }
 * }

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

 * $${{P}_{6}}=\frac{11x\cdot \frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)-5\times \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)}{6}$$
 * }
 * }
 * }

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

 * $${{P}_{6}}=\frac{231{{x}^{6}}}{16}-\frac{770{{x}^{4}}}{48}+\frac{55{{x}^{2}}}{16}-\frac{175{{x}^{4}}}{48}+\frac{50{{x}^{2}}}{16}-\frac{5}{16}$$
 * }
 * }
 * }

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

$${{P}_{6}}=\frac{1}{16}\left( 231{{x}^{6}}-315{{x}^{4}}+105{{x}^{2}}-5 \right)$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.7)$$
 * }
 * }

$$\displaystyle Eq. 9.3 - Eq. 9.5$$ are same as the[[media:2010_11_09_15_00_14.djvu|Eqs.4-6 in page-2 Mtg-36]] respectively.

Contributing members

 * Solved and posted by Egm6321.f10.team3.Sudheesh 21:27, 5 December 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 01:05, 7 December 2010 (UTC)

= Problem 10 Continued Power Series Expansion=

Given
Given the power series expansion in equations 2, 3, and 4 on page 41-3 of the notes

Find
The power series expansion to find $$\left\{ P_{3}...P_{6} \right\}$$ and compare to equations 7 and 8 on page 36-2 of the lecture notes, and to HW 7.9

Solution
Given in lecture

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

$$ \alpha_i =\frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ and
 *
$$\displaystyle (Eq. 10.1) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$A^{-\frac{1}{2}}=\sum_{i=0}^\infty \alpha_i x^i $$ Where
 *
$$\displaystyle (Eq. 10.2) $$
 * }
 * }

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

$$x=2\mu \rho -\rho ^{2}$$
 *
$$\displaystyle (Eq. 10.3) $$
 * }
 * }

Expanding the terms of 7.2
 * {| style="width:100%" border="0"

$$A^{-\frac{1}{2}}=\alpha _{0}x^{0}+\alpha _{1}x^{1}+\alpha _{2}x^{2}+\alpha _{3}x^{3}+\alpha _{4}x^{4}+\alpha _{5}x^{5}+\alpha _{6}x^{6}$$ The $$x^i$$ portions become
 *
$$\displaystyle (Eq. 10.4) $$
 * }
 * }
 * $$x^{0}=1$$
 * $$ x^{1}=2\mu\rho-\rho^2 $$
 * $$ x^{2}=4\mu^2\rho^2-4\mu\rho^3+\rho^4$$
 * $$ x^{3}=8\mu^3 \rho^3-12\mu^2\rho^4+6\mu\rho^5-\rho^6$$
 * $$ x^{4}=16\mu^4\rho^4-32\mu^3\rho^5+24\mu^2\rho^6-8\mu\rho^7+\rho^8 $$
 * $$ x^{5}32\mu^5 \rho^5-80\mu^4\rho^6+80\mu^3\rho^7-40\mu^2\rho^8+10\mu\rho^9-\rho^{10}= $$
 * $$ x^{6}= 64\mu ^{6}\rho ^{6}-192\mu ^{5}\rho ^{7}+240\mu ^{4}\rho ^{8}-160\mu ^{3}\rho ^{9}+60\mu ^{2}\rho ^{10}-12\mu \rho ^{11}+\rho ^{12}$$

The $$\alpha_i$$ portions become reduced to
 * $$\alpha_0 = 1 $$
 * $$\alpha_1 = \frac{1}{2} $$
 * $$\alpha_2 = \frac{3}{8} $$
 * $$\alpha_3 = \frac{5}{16} $$
 * $$\alpha_4 = \frac{35}{128}$$
 * $$\alpha_5 = \frac{63}{256}$$
 * $$\alpha_6 = \frac{231}{1024}$$

Substitute into equation 7.4 and group by $$\rho^i$$
 * {| style="width:100%" border="0"

$$1+\frac{1}{2}\rho +\left( -\frac{1}{2}+\frac{3}{2}\mu ^{2} \right)\rho ^{2}+\left( -\frac{3}{2}\mu +\frac{5}{2}\mu ^{3} \right)\rho ^{3}+\left( \frac{3}{8}-\frac{15}{4}\mu ^{2}+\frac{35}{8}\mu ^{4} \right)\rho ^{4}+\left( \frac{15}{8}\mu -\frac{35}{4}\mu ^{3}+\frac{63}{8}\mu ^{5} \right)\rho ^{5}+\left( -\frac{5}{16}+\frac{165}{16}\mu ^{2}-\frac{315}{16}\mu ^{4}+\frac{231}{16}\mu ^{6} \right)\rho ^{6}$$ Thus
 *
$$\displaystyle (Eq. 10.5) $$
 * }
 * }

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$P_{3}(\mu )=-\frac{3}{2}\mu +\frac{5}{2}\mu ^{3}$$

and
 *
$$\,(Eq. 10.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$P_{4}(\mu )=\frac{3}{8}-\frac{15}{4}\mu ^{2}+\frac{35}{8}\mu ^{4}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

and
 *
$$\,(Eq. 10.7) $$
 * style= |
 * }

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$P_{5}(\mu )=\frac{15}{8}\mu -\frac{35}{4}\mu ^{3}+\frac{63}{8}\mu ^{5}$$

and
 *
$$\,(Eq. 10.8) $$
 * style= |
 * }

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$P_{6}(\mu )=\left( -\frac{5}{16}+\frac{165}{16}\mu ^{2}-\frac{315}{16}\mu ^{4}+\frac{231}{16}\mu ^{6} \right)$$

 *
$$\,(Eq. 10.9) $$
 * style= |
 * }

These compare similarly to problem 7.9. Also, the first four Legendre equations match up with lecture notes page 36-2, equations 2 - 6. Comparing to equation 7, page 36-2 of the lecture notes, which is given by:

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

$$P_{n}\left( x \right)=\sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor }{\ }\left( -1 \right)^{i}\ \frac{\left( 2n-2i \right)!}{2^{n}\ i!\ \left( n-i \right)!\ \left( n-2i \right)!}x^{n-2i}$$
 *
$$\displaystyle (Eq. 10.10) $$
 * }
 * }

Plugging in $$ P_{5}\left( x \right)$$
 * {| style="width:100%" border="0"

$$P_{5}\left( x \right)=\frac{10!}{2^{5}\cdot 1\cdot 5!\cdot 5!}x^{5}-\frac{8!}{2^{5}\cdot 1\cdot 4!\cdot 4!}x^{3}+\frac{6!}{2^{5}\cdot 2!\cdot 3!\cdot 1!}x$$ Which yields equation 10.8. Plugging in $$ P_{6}\left( x \right)$$
 *
$$\displaystyle (Eq. 10.11) $$
 * }
 * }

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

$$P_{6}\left( x \right)=\frac{12!}{2^{6}\cdot 1\cdot 6!\cdot 6!}x^{6}-\frac{10!}{2^{6}\cdot 1\cdot 5!\cdot 5!}x^{4}+\frac{8!}{2^{6}\cdot 2!\cdot 4!\cdot 2!}x^{2}+\frac{6!}{2^{6}\cdot 3!\cdot 3!\cdot 1!}$$ Which yields equation 10.9. Checking equation 8, which is given by
 *
$$\displaystyle (Eq. 10.12) $$
 * }
 * }

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

$$P_{n}\left( x \right)=\sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor }{\ }\ \frac{1\cdot 3\cdot \ldots \cdot \left( 2n-2i-1 \right)}{2^{i}\ i!\ \left( n-2i \right)!}\left( -1 \right)^{i}x^{n-2i}$$
 *
$$\displaystyle (Eq. 10.13) $$
 * }
 * }

Plugging in $$ P_{5}\left( x \right)$$

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

$$P_{5}\left( x \right)=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2^{0}\cdot 0!5!}x^{5}-\frac{1\cdot 3\cdot 5\cdot 7}{2^{1}\cdot 1!3!}x^{3}+\frac{1\cdot 3\cdot 5}{2^{2}\cdot 2!1!}x$$ Which yields equation 10.8. Plugging in $$ P_{6}\left( x \right)$$
 *
$$\displaystyle (Eq. 10.14) $$
 * }
 * }

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

$$P_{6}\left( x \right)=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11}{2^{0}\cdot 0!6!}x^{6}-\frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2^{1}\cdot 1!4!}x^{4}+\frac{1\cdot 3\cdot 5\cdot 7}{2^{2}\cdot 2!2!}x^{2}-\frac{1\cdot 3\cdot 5}{2^{3}\cdot 3!1!}$$ Which yields equation 10.9.
 *
$$\displaystyle (Eq. 10.14) $$
 * }
 * }

Contributing Members

 * Solved and posted by James Roark 02:30, 7 December 2010 (UTC), all original content.

=Contributing Members=
 * Adam franklin Solved and posted problems 3 and 4.
 * Schulze solved problem 5.
 * James Roark 04:15, 7 December 2010 (UTC) Solved and posted 7.7 and 7.10

 * --Egm6321.f10.team3.cook 13:30, 7 December 2010 (UTC) Solved and Posted problem 8

title: User:Egm6321.f10.team03/Hwk7


text: Template:Motivation and emotion/Textbook/Chapter/Recording

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


text:
Student online safety guidelines

It can be a challenge to talk with your kids about the content they post online — especially if you don't feel like a computer-savvy parent or teacher.

students should consider five things before posting online:

1. Audience: Who will read or see the information? 2. Anonymity: Nothing they post online is truly anonymous. 3. Permanence: Like it or not, information posted online can stay around for a very long time. 4. Copyright: It's illegal to copy others' work and pass it off as their own. 5. Free speech: This doesn't mean they can publish threats or try to ruin another person's reputation without legal consequences.

Things to consider

Audience: Anything you publish could eventually be read and seen by anyone. Even though you may think that only certain people are able to access your information, pictures, or videos published, you may be wrong because we do not have complete control over it once we write, print, send, or post something. You never know who is looking over your friends’ shoulders, or if those friends might take the content, spread it around online, or use it against you.

Anonymity: Using a pseudonym (fake name) can attempt to disguise your identity while developing your writing or artistic skills. Technology has made it much easier to appear to publish something anonymously because you can use a screen name, alias, or temporary email address. That said, just about everything that is published can be traced back to a specific person since everything done online has what is called a “digital footprint.” This footprint can enable authorities to discover where every piece of information on the Internet came from and who wrote/posted it.

Permanence: What you publish may be seen by others for many years to come. This can be a huge benefit, but it can also be problematic if you publish something inappropriate, embarrassing, or hurtful to others. Think about how you would feel if your parents, future college admissions office, or employer saw what you published?

Copyright: It is wrong to copy, use, or publish anything under your name that you did not personally create without getting permission from the person who did. For example, if you are using someone else’s words or pictures, you should give them credit by acknowledging what they wrote or said, or get permission from the person who originally took the picture. This is more than just common courtesy; it is also the law.

Free Speech: Publishing our thoughts or ideas is a form of speech that is often protected by the First Amendment. Having the right to free speech, however, doesn’t mean that we can say whatever we want, whenever we want. You can’t, for example, publish threats against someone else or ruin their reputation.

When it comes to using the internet, it doesn't matter how old you are, you should always be cautious when posting, emailing, talking to people and even browsing websites. We have posted some internet Do's and Don'ts points below

DO
 * Do ask your parent before joining any chat sites or social networks
 * Do use an anonymous e-mail address if possible
 * Do use graphics instead of pictures of yourself
 * Do use monitored chat and social sites

Don't
 * Don't use your real name
 * Don't put photos or videos of people without asking their permission
 * Don't put your phone number, email address or real address on your website or social network
 * Don't give your phone number, email address or real address to anyone you don't know online
 * Don't leave a website without logging out first
 * Don't purchase anything with a credit card
 * Don't agree to meet with anybody you don't know personally

Above all the biggest thing to remember is be SMART! If you wouldn't give someone certain information in person, then don't do it online. And if you aren't sure about something ASK your parents or a teacher!

Get Net Wise has lots great screen capture tutorials. An online service of companies and non-profit groups concerned about child safety on the Internet. The Web site provides a comprehensive "Web-wide" resource with safety tips, ways to report online trouble, tech tools for families, great Web sites for kids and a glossary of Internet terms.

Security/Tools

If you like watching videos rather than reading safety guidelines the link below is the place to be

[]

Take the Quiz !!!

How smart are you with online safety? Link provided []

For Parents

If you aren't the most tech savvy person don't worry, there are still ways you can make sure your children are safe on the internet. Here are some helpful tips for you:

here is a website that lists family safe search engines: http://www.wisekids.org.uk/Kids_safe_search_engines.htm here is one such example: http://www.labnol.org/internet/how-parents-can-protect-kids-from-adult-content/9267/
 * Make sure the computer is in a family room, so you can see the computer
 * Make sure they take a break every 30 minutes
 * Make sure they ask you before they download anything
 * Make sure they ask you before they sign up for anything
 * Try and have them use family safe search engines
 * Use software that allows you to set parental controls

title: Student online safety guidelines


text: Motivation and emotion/Book/2010/Procrastination/Feel like procrastinating


text:
User talk:Jbeauli4

 Hello Jbeauli4, 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 03:44, 2 December 2010 (UTC)

title: User talk:Jbeauli4


text: Motivation and emotion/Book/2010/Procrastination/Are you a procrastinator


text:
Canada in the Cold War

The Cold War had such far reaching effects that it made Canadians stop and consider our Arctic north as more than a frontier, but as part of our homeland and identity. The shortest land route to Russia (formerly the USSR) was across the North Pole. It is my intent to tie together several themes of Canada’s north in the context of the Cold War: issues of Canadian sovereignty, security and the effects this had on Inuit peoples. Looking at such issues historically will put perspective on the increasingly controversial topic of northern sovereignty. This is not only because of the viability of an emerging Northwest Passage, but because of the potential natural resources of the area. I would like to link a book review of Melanie McGrath’s The Long Exile (2006) because it highlights a particular relocation program for the purpose of establishing Canadian sovereignty in the High Arctic. Lastly is the importance of Canada’s north in terms of security, especially during the Cold War. This includes the improvements of weapons technologies and the responses in both the north and in Ottawa.

Who has Claim, What is at Stake
A highly relevant discourse to the issue of sovereignty in the Arctic has been taking place for the last decade and shows no signs of diminishing. Global warming and climate change have become prominent political and popular debates as the November 2007 issue of The Walrus illustrates. But before we dive into the ramifications of climate change we should identify the main actors involved in the question of Arctic sovereignty. The three most relevant are Canada, the United States and Russia, but many other nations intrude into the Arctic: Finland, Iceland, Denmark, Sweden, and Norway. Thus the potential complexities of jurisdictions, land claims, claims of sovereignty and the resources therein are far beyond the scope of this study. However, consult the further reading segment located below for relevant articles and books.

Natural Resources
The other controversial topic touched upon is the hunt for energy. Never before has energy featured so prominently in geopolitical discourse, “its promotion of the development of drilling within the Arctic National Wildlife Refuge (ANWR) in the north of Alaska is legendary.” However we are speaking of natural resources on an industrial scale. There are other important resources of the Arctic which have been a point of contention over the last century: animal pelts. As Melanie McGrath points out, “the occupation of the island by Canadian Eskimos will remove any excuse Greenlanders may presently have for crossing over and hunting there.” In respect to Inuit economy, trapping and hunting are the main sources of energy and income. As climate change encroaches on traditional hunting grounds and trapping lines, the Inuit are forced to put out more energy with less chance of profitable return. The increasingly more attractive alternative is to abandon traditional ways of life and join contemporary social settings.

Climate Change
Returning to climate change, I wish to avoid all topics save those that relate to the opening of the Northwest passage. Although this passage has been crossed numerous times it is still fraught with danger and unpredictable weather conditions. The interesting point to the Northwest passage is a long tradition of claiming exploratory firsts. Which country has its flag on the moon? Which country first placed its flag on the North Pole? The South Pole? Mt. Everest? It seems illogical but the legitimacy of a nations claims to something is often no more complicated than discovering it, or establishing regular use of it. Thus, in order to substantiate claims to the Northwest passage, it is important to place resources there; ships, patrols, people, equipment, laboratories, bases, ports, etc. It is this mentality that initiated the relocation of Inuit families to the High Arctic.

McGrath’s The Long Exile has dealt extensively with this topic (albeit in a popular historical account), and my review of it is linked. However, consult the further readings for more information, specifically Unikkaaqatigiit Putting the Human Face on Climate Change: Perspectives from Inuit in Canada (2005).

Effects of the Cold War
The most dramatic event to affect the Arctic is the Cold War. One of the most inhospitable geographies on the planet was, for a brief time, a hive of activity and significance. The area became the focus of militarization to such an extent the Rob Huebert has stated: “the Arctic became one of the most militarized regions during the cold war, in the early days of which both sides developed long-range bombers as their main delivery system for nuclear weapons.” The increase of weapons technologies greatly affected the perceived importance of the Arctic coast and ocean. Improved submarines and their payload capabilities made the Arctic “front” a significant area of concern.

Security in the North
However, Whitney P. Lackenbauer points to a less grandiose path Canada took in its Cold War security. In terms of policing the Arctic, we have already seen how cost prohibitive it would be. In addition the main pressures to intensify military presence in the north comes from the United States. As Lackenbauer states, “if Canada was neither able nor willing to defend the northern approaches to the continent, the Americans would be compelled to take unilateral measures to defend themselves and could thus become a security threat themselves.” Whatever steps the Canadian government took to bolster its northern boarders it would invariable fail to meet the United States’ expectations of defense. Although, going back to an earlier point of legitimacy through use and population of an area, the following cost effective solution also serves to substantiate Canada’s claim to the north. That solution was the Canadian Rangers, our civilian militia which has its roots in the Second World War. The Rangers not only served to actively populate our sparsely inhabited Arctic, but did so with an official air.

Connection to the Diefenbunker
The last theme I wish to connect to the Canadian Arctic and the Cold War is the historical evidence left behind. Specifically cities of memory, which could be located far north in unused or derelict military outposts, but more specifically in the Emergency Government Headquarters located in Carp Ontario. Also known as the largest “Diefenbunker” it has been converted into a public museum. Inside is a direct connection (no longer operational) to the Distant Early Warning (DEW) line located in the Arctic. One of the Diefenbunker main purposes was to detect and relay warnings of a nuclear attack through the DEW radar line in the High Arctic. I will not delve deeper into the Diefenbunker, but provide the link to my somewhat theoretical review of the site.

Conclusion
Climate change and natural resources have been and continue to increase the political significance of Canada’s Arctic. However, this politicization has a long history, most significantly during the tense years of the Cold War. Evidence of these effects can be seen in the various Diefenbunkers spread across Canada, but especially in the central bunker located just outside of Ottawa. Culture and society in the north has traditionally taken a back seat against the politics surrounding the Arctic and further study of the role the Inuit played in these events is needed. This brief summary of several related threads is by no means exhaustive. It should rather be viewed as a starting point to particular studies regarding Canada’s northern boarders.

title: Canada in the Cold War


text:
Book Reviews/The Long Exile

McGrath, Melanie. The Long Exile: a True Story of Deception and Survival among the Inuit of the Canadian Arctic. London: Fourth Estate, 2006. Pp. 302.

Melanie McGrath’s book The Long Exile is a descriptive account of the hardships and challenges facing Canada’s Inuit as a whole, and of the specific examples of these hardships. The narrative weaves a poignant and emotive path starting with the background to the famous documentary, Nanook of the North, and ending with a relocation program of several families to substantiate Canada’s claim to its northern boundaries.

As a historical work and its ability to contribute to a relevant discourse on Canada’s Cold War experience, The Long Exile is insignificant. However, if the goal of the book is to broaden Canadian awareness of the turbulent history of our Inuit peoples and the depth of Canadian history then this book is certainly worth noting.

I will attempt to take contrasting views of The Long Exile; on the one hand analyzing its effectiveness to historical scholarship, looking at sources and methodology. On the other hand I will consider the major themes of the book, a larger historical context of Canada during the Cold War, and the books contribution to the history of Canada Inuit.

To take a very broad approach to The Long Exile’s contribution to historical scholarship one must define what is meant by historical scholarship, or even history. There are numerous opinions on what history is, how it is conducted, what it should accomplish, who is qualified, and why bother at all. In very basic terms history is to re-convey the past through some medium of communication.

The next hurdle is to do so with as much veracity or legitimacy as possible (unless one is seeking to prove the impossibility of such an action).

The conventional way of doing this is by using evidence, some of which is factual and some is deductive-analytical. Even facts become disputed however, as E.H. Carr discusses in What is History? The general questions Carr asks such as whether history is a bunch of compiled, objective facts, or an interpretation of past events by a historian, are relevant to The Long Exile because of the large absence of facts.

The evidence used by McGrath is problematic from a scholarly standpoint. The way she presents her “facts” do not resemble an objective, scientific reflection of historical research. This is not to say that this is how history should be done, because a compilation of inert facts and statistics leaves many historical works in an inaccessible prison of their own numbers. McGrath has clearly gone another route, which is more storytelling and lively, designed to engage the audience intellectually and emotionally.

It is a fitting course of action given the impression one receives of her main subjects: the Inuit. Their preference for storytelling as a means of history mirrors McGrath’s process. However, the essence of scholarly work is its engagement with the discipline, its members, and their work and opinion.

A “scholarly” work is something that has been read, analyzed, chewed over, criticized and commented on, before it is even published. Then it goes through a much more vigorous and public debate all over again. Furthermore, such a work relies on other historians and their works which have undergone the same process.

It is unclear to what extent The Long Exile has engaged in such a process. Searches of her name and book title reveal no additional published works or articles in any scholarly journals. It does not appear that this book or any other published works have received any scholarly attention that I have found. This criticism is not an attack on her work, but an observation of McGrath’s contribution to a more traditional academic milieu.

Whether or not one accepts this interpretation of what a historical work should accomplish, or has set out to accomplish, changes the argument completely. For example, if it was McGrath’s intention to reveal the abuse and misguided treatment of a group of Inuit on an emotive and personal level, then it is hard to find fault with her methods. In addition, the degree to which one agrees with “traditional” historical methodology also renders the above criticisms moot.

A significant issue in this book is the use of oral history, which may help account for the lack of supporting evidence. Before we look at the pros and cons of oral history, it should be noted that McGrath does not set the reader straight with regards to her use of evidence anywhere other than the “Acknowledgments” in the last two pages (p. 301-302). It is possible to deduce her use of oral interviews from the “High Arctic Exiles” as they are known, but she never explicitly says so.

One of the ways of dealing with oral testimony in any form is by using corroborating evidence. This is done because of the subjective nature of memory, the form of questions being asked, and the perspectives of both interviewer and interviewee. The inherent problems multiply when dealing with controversial or intensely emotional subjects. However, McGrath raises an excellent point in the meticulous way the Inuit conduct oral history, ways which seem unfathomable to those of us who have just to tap our fingers a few times to obtain “objective” information.

McGrath recounts the story of a British rescue voyage in the Arctic which recorded their conversations with the Inuit. When they asked them about the explorer Martin Frobisher’s expeditions in 1576, the Inuit responses where nearly verbatim what was recorded back in England (p. 271-272). This struck me as ironic because she does not present the reader with evidence of where this story came from. However, it does illustrate both the significance and sophistication of the Inuit oral tradition.

Another major source of evidence McGrath uses is the documentary. The most relevant is a famous film of life in Canada’s Arctic, Nanook of the North, directed by Robert Flaherty. As McGrath presents it the subject matter of the documentary is of secondary importance to its director and his illegitimate son Josephie Flaherty: the books main characters if you will.

Nanook’s depiction of life in the Arctic has informed such a vast audience and has created long standing stereotypes which perhaps lead McGrath to disregard the particulars of the film. She may be using documentaries in this way in order to undercut a “popular” historical representation from her perspective. The other side to this interpretation is that both the book and documentary are very similar ways of representing the past. If one considers both sources as popular mediums which are largely designed to reach a wider audience, then they complement each other.

Moreover, one can do away with many historical issues and theories. We may also get a glimpse into the motivations of the author by looking at the motivations of the documentarian. McGrath writes of Robert Flaherty: “in Nanook he had wanted to capture the struggle for survival, because, for him, it was at the heart of Inuit life. How he achieved this was of less concern. ‘Sometimes you have to lie,’ he said.

‘One often has to distort a thing to catch its true spirit’” (p. 198). McGrath could be looking at the Inuit from Flaherty’s perspective, a perspective which considers the Inuit way of life, its people and its history as something that has no need of embellishment, exaggeration, or glamorization. Similarly, she may have taken this same idea as justification for writing a historical account without conventional historical methods. One proposal of what McGrath is doing is an attempt to capture the essence of the Inuit and their struggles; an essence which might supersede a perceived need for methodology.

A very glaring problem with The Long Exile is its contribution to Canada’s Cold War experience. McGrath makes very few references to the issue and those references concern the need for military presence in the far north (p. 91). The reasons she gives for relocating Inuit further north is two-fold: firstly to establish Canadian authority to claim sovereignty on this territory (p. 91, 95, 297), and secondly for socio-economic reasons.

It was felt (by bureaucrats in Ottawa) that the Inuit’s reliance on fur trading was at the heart of the social and economic downturn in their ways of life, and it was concluded that the Inuit needed to resume a traditional, self-sufficient lifestyle of hunting and living (p. 92-95).

This is the rationale behind the relocation as it was conceived, and again in 1995 as reported by the Royal Commission on Aboriginal Peoples, “the concern was with the ability of the fur trade to sustain the income levels to which Inukjuak Inuit hand become accustomed…Greater reliance on hunting would substitute for the income that fur trading would, in the long term, be unable to provide” (p. 295). It was not the Cold War which initiated the Inuit relocation, but as a way of dealing with the “Eskimo Problem” (p. 92).

Finally, The Long Exile is not a history of the Inuit in and of itself, although the emotive narrative lends itself to such an interpretation. When one asks oneself, what it the purpose of this book, one may always fall back on the larger context: Canada’s national consciousness. Perhaps this is not an explicit goal of the book, but it remains a significant factor for those Canadians who are unaware of the depth and multifaceted history of this nation. By focusing on the people, their desires, beliefs, attitudes and actions.

McGrath makes them very real; perhaps more real than the existing evidence and a “traditional” historical work would lead us to believe. This book informs a Canadian identity in new or unthought-of ways and lends credence to some unorthodox ways of conveying history: oral tradition, documentary, landscape and monument.

In conclusion, The Long Exile presents and unconventional, but no less legitimate form of historical inquiry. As a form of popular history McGrath does an excellent job of describing the nuances of social life in an environment that is not suited for a more conventional historical process.

This emotive description, while not the objective, unbiased perspective of, what British historian Lawrence Stone called, “number crunching idiot bastards,” allows the reader to connect in a meaningful way to a unique people.

Furthermore the Inuit are an intrinsic part of what Canada is and what it means to be Canadian, an idea that is best served through a popular, engaging narrative such as the one Melanie McGrath has offered.

title: Book Reviews/The Long Exile


text:
Diefenbunker

The Emergency Government Headquarters or “Diefenbunker” located in Carp, Ontario is the largest such shelter in Canada. I will take a balanced approach to the possible uses of this historic site, whether, on the one hand, it should be used in order to claim certain facts and experiences of Canada during the Cold War, or whether such a site presents grounds for skepticism in terms of its use as public history. It should be noted that these arguments will take place strictly in the context of the Diefenbunker and issues surrounding this particular site or museum. There are obvious merits to accepting the Diefenbunker as an objective representation of the Cold War, foremost among them as a source of national and even personal identity. Because so many people still living today experienced the Cold War the Diefenbunker constitutes a source of personal memory and experience. Conversely there are arguments against the use of the Diefenbunker as an objective source of public history and memory. Such arguments take issue with the concepts of public identity, public consensus, and the overarching narratives we tell ourselves as a nation.

The Diefenbunker presents, literally, a concrete primary source of Canada’s Cold War experience. This particular shelter may resonate even more because of its relative importance being the primary fallback position of the top tier of government given a nuclear attack. There is a tangible aura of authority and gravity to the site, an atmosphere that has been self-consciously maintained and enhanced through its conversion into a public museum. For those of us who did not live through the Cold War, the Diefenbunker brings the narrative of this international conflict into reality. It is evidence, or proof, of what has heretofore only been words on a page, or second hand verbal accounts. For others that have lived through these events, there is a similar reaction to the reality of it, but augmented by personal memory. Personal memories of the Cold War could either reinforce or conflict within the Diefenbunker itself, which raises a host of problems we shall get to in a minute.

As a representation of historical events, even if one lived through them, the Diefenbunker serves different purposes to different audiences. Here we must differentiate between the historically untrained public and historians. For the non-historian audience, the Diefenbunker serves to locate national identity. The shelter answers the questions of what Canada as a whole went through, felt, reacted to, the steps taken to safeguard the country in perilous times. The concepts national narratives are constructed around are all present in the Diefenbunker. The visitor leaves the site with a shared sense of empowerment that certain individuals (representing Canada as a whole) did what was necessary to protect this country. This is the civic or national faith that David Glassberg says undercuts ethnic and class divisions. He is speaking of the politics of public history and the responsibility this history has to cultural knowledge. Thus the merits of the Diefenbunker lie in unifying an overarching national identity.

The consequences of not allowing for an objective view towards sites such as the Diefenbunker not only take away from a collective Canadian identity, but from an authoritative source of public history as well. Again, we must delineate between historians and non-historians. For the non-historian an accepted, objective view of the past makes things much simpler. There is no reason to doubt, nor could there be, the existence and purpose of a site such as the Diefenbunker. Some of the purposes that this site delivers are the fear and reality of nuclear attack, the foresight and deliberation through which the bunker was created, and Canada’s ability to combat this fear and uncertainty. As long as we accept these messages we can accept a more unified, cohesive concept of nationhood.

However, things become much more problematic for the historian, as we are now in a position to critically analyze the messages of the Diefenbunker, and its use a source of public history. The notion of an objective view of the past as represented by public history has been criticized by many, “the meaning of a historical book, film, or display is not intrinsic, determined solely by the intention of the author, but changes as audiences actively reinterpret what they see and hear by placing it in alternative contexts derived from their diverse social backgrounds.” This relativism applies to all history, but what is particular to the Diefenbunker is the way it is institutionalized; that is, how the concept of subjectivity is situated in the context of the Diefenbunker. As a building or monument, the Diefenbunker can be read as part of the landscape, but like landscape there are countless ways to convey or communicate the presence and purpose of the object according to each individual.

One of the main reasons for skepticism about the Diefenbunker is because of memory. Here, the use of memory is not relative to those that lived through the Cold War, and those that did not; it is simply used in reference to all past events. Natalie Zemon Davis and Randolph Starn wrote in the introduction to Representations’ special issue on memory, "But this is Pierre Nora’s point: we distinguish lieux de mémoire, he insists, because we no longer live in a world suffused with memory or fully committed to overarching ideological narratives - so, for example, The Triumph of Western Civilization, of the Nation-State, of the Proletariat, etc. - defining what is supposed to be memorable." Thus the Diefenbunker can be seen as an attempt to “define what is supposed to be memorable,” because we have lost a universal and superimposed grand narrative of nation. However, David Neufeld raises the point that, “the Cold War has yet to become an integral part of Canadian national identity.” Although Neufeld in talking about national identity in an official, institutionalized sense, it begs the question, if we have lost an accepted version of Canada’s narrative, will the Diefenbunker prove a valid substitute for public memory.

Steven Knapp’s main question is about the viability of designated sites of memory such as the Diefenbunker. He poses it thus, “why aren't narratives of events in our actual past simply replaceable by other narratives that meet our criteria of symbolic relevance?” This goes hand in hand with Glassberg’s idea (mentioned above) that there is no intrinsic meaning in historical phenomenon, but only the meaning we take from it. If one accepts Glassberg’s and others’, relativism, than the Diefenbunker loses is authority as a representative narrative of Canada’s Cold War experience. The messages conveyed by the Diefenbunker are no longer (or have never been) under the power of the site itself. This problem is compounded by collective memory of which the Diefenbunker represents, but does not create. When discussing power, it is important to note the separation between the origination and the representation. The narrative of the Cold War experience is not contingent on the Diefenbunker; rather, it is the other way around.

Furthermore, what authority we can derive from the Diefenbunker as a site of collective memory is inherently political because it is public and shared. Pierre Nora says of the relationship between history and memory, “this national definition of the present imperiously demanded justification through the illumination of the past.” The ways in which we define the present, especially the collective or shared present, have unavoidable political connotations. And one of the Diefenbunker’s purposes is to illuminate the past in order to define present day Canada.

If we accept the Diefenbunker as a representation of public history and memory then we must do so by giving consideration to what these concepts mean. For some, an objective explanation of the Cold War makes the messages of the Diefenbunker easy to understand. However, if we raise doubts about the nature or viability of public history and especially memory then these doubts should be considered. We should not read the purpose of the Diefenbunker in a narrow sense, but rather open up the discussion to broader concepts of how to represent history and why. The Diefenbunker is a good example of how the narrative of nation has been institutionalized, but it is only one example of a type of history that commits itself to a narrow view. The question remains whether we should accept this narrow view of nation in order to make a solid claim of a “Canadian” experience, or allow for critical analysis.

title: Diefenbunker


text:
User talk:Kurtis5732~enwikiversity

 Hello Kurtis5732, 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:17, 2 December 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:17, 2 December 2010 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Kurtis5732. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Kurtis5732~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:36, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:14, 19 April 2015 (UTC)

title: User talk:Kurtis5732~enwikiversity


text:
Talk:Motivation and emotion/Textbook/Motivation/Alcohol consumption and sexual motivation

-- Jtneill - Talk - c 12:19, 2 December 2010 (UTC)

-- Jtneill - Talk - c 04:26, 14 December 2010 (UTC)