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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \rightarrow P_{2k+1}\left(x\right) = -P_{2k+1}\left(-x\right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\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
<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; background-color: #FFDDDD; text-align: left;"> 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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.2.b)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

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

A_n = \frac{2n + 1}{2} <f, P_n> $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 4a.10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle A_{2k+1} = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 4a.12)
 * }
 * }

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

f\left(\mu\right) = \left(1-\mu^2\right)^3 , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 4a.15)
 * }
 * }

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

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

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

$$\displaystyle \begin{align} & {{A}_{0}}=T_0 1.173108 \\ & {{A}_{2}}=T_0 0.639393 \\ & {{A}_{4}}=T_0 0.714901 \\ & {{A}_{6}}=T_0 0.726179 \\ \end{align} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * <p style="text-align:right;">$$\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"


 * $${{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.1)$$
 * }
 * }
 * }


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


 * $${{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.2)$$
 * }
 * }
 * }

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"


 * $$\tanh (\alpha )=\frac{{{e}^{\alpha }}-{{e}^{-\alpha }}}{{{e}^{\alpha }}+{{e}^{-\alpha }}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3)$$
 * }
 * }
 * }


 * 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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.4)$$
 * }
 * }


 * 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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.5)$$
 * }
 * }

.

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; background-color: #FFDDDD; text-align: left;"> 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} $$ $$
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle Eqt (7.2)
 * }
 * }


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

\begin{align} (2) P_{2k+1}(x) = [odd] \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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} $$ $$
 * <p style="text-align:right;">$$\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} $$ $$
 * <p style="text-align:right;">$$\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} $$ $$
 * <p style="text-align:right;">$$\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} $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Reference
<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; background-color: #FFDDDD; text-align: left;"> 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"

$$r_{PQ}^{2}=\sum\limits_{i=1}^{3}{\left( x_{Q}^{i}-x_{P}^{i} \right)^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1) $$
 * }
 * }

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

$$\left( r_{PQ} \right)^{2}=r_{Q}^{2}\left( \left( \frac{r_{P}}{r_{Q}} \right)^{2}+1-2\left( \frac{r_{P}}{r_{Q}} \right)\cos \gamma, \right)$$
 * <p style="text-align: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
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3) $$
 * }
 * }
 * {| 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:
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4) $$
 * }
 * }
 * $$x_{P}^{1}=r_{P}\cos \left( \theta _{P} \right)\cos \left( \phi _{P} \right)$$
 * $$x_{P}^{2}=r_{P}\cos \left( \theta _{P} \right)\sin \left( \phi _{P} \right)$$
 * $$x_{P}^{3}=r_{P}\sin \left( \theta _{P} \right)$$
 * $$x_{Q}^{1}=r_{Q}\cos \left( \theta _{Q} \right)\cos \left( \varphi _{Q} \right)$$
 * $$x_{Q}^{2}=r_{Q}\cos \left( \theta _{Q} \right)\sin \left( \varphi _{Q} \right)$$
 * $$x_{Q}^{3}=r_{Q}\sin \left( \theta _{Q} \right)$$

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
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5) $$
 * }
 * }


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




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


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


 * <p style="text-align: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
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7) $$
 * }
 * }


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

$$\left( r_{PQ} \right)^{2}=r_{Q}^{2}-2r_{Q}r_{P}\cos \gamma +r_{P}^{2}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8) $$
 * }
 * }

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" |

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


 * <p style="text-align: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"

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

Where the Binomial coefficient


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

$$ \binom{r}{k}= \frac{r \left( r-1 \right) \left( r-2 \right).... \left( r-k+1 \right)}{k!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 8.6)
 * style= |
 * }
 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

\binom{ -\frac{1}{2}}{k}=  { \alpha }_{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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"


 * $$\left( n+1 \right){{P}_{n+1}}-\left( 2n+1 \right)x{{P}_{n}}+n{{P}_{n-1}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.1)$$
 * }
 * }
 * }

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"


 * $${{P}_{n+1}}=\frac{\left( 2n+1 \right)x{{P}_{n}}-n{{P}_{n-1}}}{\left( n+1 \right)}$$
 * <p style="text-align: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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.3)$$
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.4)$$
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.5)$$
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.6)$$
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.7)$$
 * }
 * }

$$\displaystyle Eq. 9.3 - Eq. 9.5$$ 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
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.1) $$
 * }
 * }
 * {| style="width:100%" border="0"



$$A^{-\frac{1}{2}}=\sum_{i=0}^\infty \alpha_i x^i $$ Where
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.2) $$
 * }
 * }


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

$$x=2\mu \rho -\rho ^{2}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.3) $$
 * }
 * }

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
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.4) $$
 * }
 * }
 * $$x^{0}=1$$
 * $$ x^{1}=2\mu\rho-\rho^2 $$
 * $$ x^{2}=4\mu^2\rho^2-4\mu\rho^3+\rho^4$$
 * $$ x^{3}=8\mu^3 \rho^3-12\mu^2\rho^4+6\mu\rho^5-\rho^6$$
 * $$ x^{4}=16\mu^4\rho^4-32\mu^3\rho^5+24\mu^2\rho^6-8\mu\rho^7+\rho^8 $$
 * $$ x^{5}32\mu^5 \rho^5-80\mu^4\rho^6+80\mu^3\rho^7-40\mu^2\rho^8+10\mu\rho^9-\rho^{10}= $$
 * $$ x^{6}= 64\mu ^{6}\rho ^{6}-192\mu ^{5}\rho ^{7}+240\mu ^{4}\rho ^{8}-160\mu ^{3}\rho ^{9}+60\mu ^{2}\rho ^{10}-12\mu \rho ^{11}+\rho ^{12}$$

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
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.5) $$
 * }
 * }


 * {| 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
 * <p style="text-align:right;">$$\,(Eq. 10.6) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 10.7) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 10.8) $$
 * style= |
 * }


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


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

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


 * <p style="text-align: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"

$$P_{n}\left( x \right)=\sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor }{\ }\left( -1 \right)^{i}\ \frac{\left( 2n-2i \right)!}{2^{n}\ i!\ \left( n-i \right)!\ \left( n-2i \right)!}x^{n-2i}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.10) $$
 * }
 * }

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 $$ P_{6}\left( x \right)$$
 * <p style="text-align: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
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.12) $$
 * }
 * }


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

$$P_{n}\left( x \right)=\sum\limits_{i=0}^{\left\lfloor n/2 \right\rfloor }{\ }\ \frac{1\cdot 3\cdot \ldots \cdot \left( 2n-2i-1 \right)}{2^{i}\ i!\ \left( n-2i \right)!}\left( -1 \right)^{i}x^{n-2i}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.13) $$
 * }
 * }

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 $$ P_{6}\left( x \right)$$
 * <p style="text-align: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.
 * <p style="text-align:right;">$$\displaystyle (Eq. 10.14) $$
 * }
 * }

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