User:Egm6321.f09.team4/HW6

= Homework #6: Team 4 =

Problem #1:
Lecture 31-1:


 * Given :


 * A circular cylindrical coordinate system (see lecture slide for graphic):


 * $$ \begin{align}

\xi_{1} &= r \\ \xi_{2} &= \theta \\ \xi_{3} &= z \end{align} $$


 * This system is related to cartesian coordinates as follows:


 * $$ \begin{align}

x &= r \text{ cos}\left(\theta\right) &= \xi_{1} \text{ cos}\left(\xi_{2}\right) \\ y &= r \text{ sin}\left(\theta\right) &= \xi_{1} \text{ sin}\left(\xi_{2}\right) \\ z &= z                               &= \xi_{3} \end{align} $$


 * Find :

  Obtain $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of     $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$.   Find $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2} = \sum_{k}\left(h_{k}\right)^{2}\left(\text{d}\xi_{k}\right)^{2}$$. Identify $$\left\{h_{i}\right\}$$ in terms of $$\left\{\xi_{j}\right\}$$.   Find Laplacian in this coordinate system.  


 * Solution :


 * First we solve for the components of $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ individually:


 * $$\text{d}x_1$$ is a projection of changes in $$\xi_1$$ and $$\xi_2$$ onto the x axis. Changes in $$\xi_3$$ have no effect on $$\text{d}x_1$$.


 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 + \lim_{\text{d}\xi_2 \to 0}\left [ \xi_1 \text{cos}\left ( \xi_2 + \text{d}\xi_2 \right )-\xi_1 \text{cos }\xi_2 \right ]$$


 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 + \frac{\text{d}}{\text{d}\xi_2}\left (\xi_1\text{cos }\xi_2 \right )$$


 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 - \xi_1\text{sin }\xi_2 \text{ d}\xi_2$$


 * Similarly, for $$\text{d}x_2$$ we project onto the y axis instead of the x axis:


 * $$\text{d}x_2=\text{sin }\xi_2 \text{ d}\xi_1 + \lim_{\text{d}\xi_2\to 0}\left [ \xi_1 \text{sin}\left ( \xi_2+\text{d}\xi_2 \right )-\xi_1\text{sin }\xi_2 \right ]$$


 * $$\text{d}x_2=\text{sin }\xi_2 \text{ d}\xi_1 + \xi_1\text{cos }\xi_2\text{ d}\xi_2$$


 * For $$\text{d}x_3$$, only changes in $$\xi_3$$ have an influence, since changes in $$\xi_1$$ and $$\xi_2$$ are constrained to a single plane perpendicular to z:


 * $$\text{d}x_3=\text{d}\xi_3$$


 * Therefore $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}=\left\{(\text{cos }\xi_2 \text{ d}\xi_1 - \xi_1\text{sin }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_2 \text{ d}\xi_1 + \xi_1\text{cos }\xi_2\text{ d}\xi_2),\text{d}\xi_3 \right\}$$


 * Next we solve for $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}$$ using the solution for $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$above:


 * $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}=\left ( \text{cos }\xi_2\text{ d}\xi_1-\xi_1\text{sin }\xi_2 \text{d}\xi_2 \right )^2+\left ( \text{sin }\xi_2\text{ d}\xi_1+\xi_1\text{cos }\xi_2 \text{ d}\xi_2 \right )^2+\left ( \text{ d}\xi_3 \right )^2$$


 * $$=\text{cos}^2\xi_2\left (\text{ d}\xi_1 \right )^2-\cancel{2\xi_1 \text{cos }\xi_2 \text{ sin }\xi_2 \text{ d}\xi_1 \text{d}\xi_2}+\xi_1^2\text{sin}^2\xi_2 (\text{d}\xi_2)^2+\text{sin}^2\xi_2(\text{d}\xi_1)^2+\cancel{2\xi_1 \text{cos }\xi_2 \text{ sin }\xi_2 \text{ d}\xi_1 \text{d}\xi_2}+\xi_1^2\text{cos}^2\xi_2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$


 * $$=\text{cos}^2\xi_2\left (\text{ d}\xi_1 \right )^2+\xi_1^2\text{sin}^2\xi_2 (\text{d}\xi_2)^2+\text{sin}^2\xi_2(\text{d}\xi_1)^2+\xi_1^2\text{cos}^2\xi_2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$


 * $$=(\text{cos}^2\xi_2+\text{sin}^2\xi_2)\left (\text{ d}\xi_1 \right )^2+(\xi_1^2\text{sin}^2\xi_2 +\xi_1^2\text{cos}^2\xi_2)(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$


 * Using the trigonometric relation $$\text{sin}^2+\text{cos}^2=1$$, we obtain:


 * $$\text{ d}s^{2} =\left (\text{ d}\xi_1 \right )^2+\xi_1^2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$


 * Where:


 * $$h_1=1$$


 * $$h_2=\xi_1$$


 * $$h_3=1$$


 * Next we find the Laplacian in these coordinates for a function $$\Psi$$:


 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$


 * For i=1 we get:


 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}\left [ \xi_1\frac{\partial \Psi}{\partial \xi_1} \right ]$$


 * For i=2 we get:


 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_2}\left [ \frac{1}{\xi_1}\frac{\partial \Psi}{\partial \xi_2} \right ]$$


 * For i=3 we get:


 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_3}\left [ \xi_1 \frac{\partial \Psi}{\partial \xi_3} \right ]$$


 * Therefore the overall Laplacian is as follows:


 * $$\Delta \Psi =\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}\left [ \xi_1\frac{\partial \Psi}{\partial \xi_1} \right ]+\frac{1}{\xi_1}\frac{\partial}{\partial \xi_2}\left [ \frac{1}{\xi_1}\frac{\partial \Psi}{\partial \xi_2} \right ]+\frac{1}{\xi_1}\frac{\partial}{\partial \xi_3}\left [ \xi_1 \frac{\partial \Psi}{\partial \xi_3} \right ]$$

Contributors:

Egm6321.f09.team4.palubin.d 17:52, 16 November 2009 (UTC)

Problem #2:
Lecture 31-1:


 * Given :


 * A spherical coordinate system (see Lecture 29-3):


 * $$ \begin{align}

\xi_{1} &= r \\ \xi_{2} &= \phi \\ \xi_{3} &= \theta \end{align} $$


 * This system is related to cartesian coordinates as follows:


 * $$ \begin{align}

x &= r \text{ cos}\left(\theta\right) \text{ cos}\left(\phi\right) &= \xi_{1} \text{ cos}\left(\xi_{3}\right) \text{ cos}\left(\xi_{2}\right) \\ y &= r \text{ cos}\left(\theta\right) \text{ sin}\left(\phi\right) &= \xi_{1} \text{ cos}\left(\xi_{3}\right) \text{ sin}\left(\xi_{2}\right)\\ z &= r \text{ sin}\left(\theta\right) &= \xi_{1} \text{ sin}\left(\xi_{3}\right) \end{align} $$


 * Find :

  Obtain $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of      $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$. 

 Find $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2} = \sum_{k}\left(h_{k}\right)^{2}\left(\text{d}\xi_{k}\right)^{2}$$. Identify $$\left\{h_{i}\right\}$$ in terms of $$\left\{\xi_{j}\right\}$$. 

 Find Laplacian in this coordinate system.  </ol></ul>


 * Solution :


 * First we solve for the the individual components of $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$


 * To start, we assume that $$\text{d}x_1$$ is equal to the projection of an infinitesimal distance s onto the x-axis in spherical coordinates:


 * $$\text{d}x_1=\text{d}\xi_1\text{cos }\xi_2 \text{cos }\xi_3+ \text{cos }\xi_2\lim_{\text{d}\xi_3\to 0}\left [ r\text{cos}(\xi_3+\text{d}\xi_3)-r\text{cos }\xi_3 \right ]+\text{cos }\xi_3\lim_{\text{d}\xi_2\to 0}\left [ r\text{cos}(\xi_2+\text{d}\xi_2)-r\text{cos }\xi_2 \right ]$$


 * $$=\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2$$


 * Similarly, for $$\text{d}x_2$$ as projected on the y-axis:


 * $$\text{d}x_2=\text{d}\xi_1\text{sin }\xi_2 \text{cos }\xi_3+ \text{sin }\xi_2\lim_{\text{d}\xi_3\to 0}\left [ r\text{cos}(\xi_3+\text{d}\xi_3)-r\text{cos }\xi_3 \right ]+\text{cos }\xi_3\lim_{\text{d}\xi_2\to 0}\left [ r\text{sin}(\xi_2+\text{d}\xi_2)-r\text{sin }\xi_2 \right ]$$


 * $$=\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2$$


 * For $$\text{d}x_3$$, the differentials are projected onto the z-axis, yielding:


 * $$\text{d}x_3=\text{d}\xi_1\text{sin }\xi_3+\lim_{\text{d}\xi_3\to 0}\left [ r\text{sin }(\xi_3 + \text{d}\xi_3)-r\text{sin }\xi_3\right ]$$


 * $$=\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3$$


 * Therefore:


 * $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}=\left\{(\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3)\right\}$$


 * Next we solve for $$\text{ d}s^{2}$$:


 * $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}=(\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2)^2+(\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2)^2+(\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3)^2$$


 * $$=\text{cos}^2\xi_3\text{cos}^2\xi_2(dr)^2+\xi_1^2\text{cos}^2\xi_2\text{sin}^2\xi_3(d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3\text{sin}^2\xi_2(d\xi_2)^2-2\xi_1\text{cos}^2\xi_2\text{cos}\xi_3\text{sin}\xi_3 (d\xi_1d\xi_3)-\cancel{2\xi_1\text{cos}^2\xi_3\text{sin}\xi_2\text{cos}\xi_2(d\xi_1d\xi_2)}+\cancel{2\xi_1^2\text{cos}\xi_2\text{sin}\xi_3\text{cos}\xi_3\text{sin}\xi_2(d\xi_3d\xi_2)}$$


 * $$+\text{sin}^2\xi_2\text{cos}^2\xi_3(d\xi_1)^2+\xi_1^2\text{sin}^2\xi_2\text{sin}^2\xi_3(d\xi_3)^2+\xi_1^2\text{cos}^2\xi_2\text{cos}^2\xi_3(d\xi_2)^2-2\xi_1\text{sin}^2\xi_2\text{sin}\xi_3\text{cos}\xi_3(d\xi_3d\xi_1)+\cancel{2\xi_1\text{sin}\xi_2\text{cos}\xi_2\text{cos}^2\xi_3(d\xi_2d\xi_1)} -\cancel{2\xi_1^2\text{sin}\xi_2\text{sin}\xi_3\text{cos}\xi_2\text{cos}\xi_3(d\xi_3d\xi_2)}$$


 * $$+\text{sin}^2\xi_3(d\xi_1)^2+\xi_1^2\text{cos}^2\xi_3(d\xi_3)^2+2\xi_1\text{cos}\xi_3\text{sin}\xi_3(d\xi_1d\xi_3)$$


 * $$=[(\text{cos}^2\xi_3\text{cos}^2\xi_2+\text{sin}^2\xi_2\text{cos}^2\xi_3)+\text{sin}^2\xi_3](d\xi_1)^2+[\xi_1^2\text{cos}^2\xi_2\text{sin}^2\xi_3+\xi_1^2\text{sin}^2\xi_2\text{sin}^2\xi_3+\xi_1^2\text{cos}^2\xi_3](d\xi_3)^2$$


 * $$+[\xi_1^2\text{cos}^2\xi_3\text{sin}^2\xi_2+\xi_1^2\text{cos}^2\xi_2\text{cos}^2\xi_3](d\xi_2)^2+[-2\xi_1\text{cos}^2\xi_2\text{cos}\xi_3\text{sin}\xi_3-2\xi_1\text{sin}^2\xi_2\text{sin}\xi_3\cos\xi_3+2\xi_1\cos\xi_3\sin\xi_3](d\xi_1d\xi_3)$$


 * $$=[\text{cos}^2\xi_3(\text{cos}^2\xi_2+\text{sin}^2\xi_2)+\text{sin}^2\xi_3](d\xi_1)^2+\xi_1^2[\text{sin}^2\xi_3(\text{cos}^2\xi_2+\text{sin}^2\xi_2)+\text{cos}^2\xi_3](d\xi_3)^2$$


 * $$+\xi_1^2\text{cos}^2\xi_3[\text{sin}^2\xi_2+\text{cos}^2\xi_2](d\xi_2)^2+2\xi_1\cos\xi_3\sin\xi_3[-\text{cos}^2\xi_2-\text{sin}^2\xi_2+1](d\xi_1d\xi_3)$$


 * Applying the trigonometric relation $$\sin^2 + \cos^2=1$$:


 * $$=[\text{cos}^2\xi_3(1)+\text{sin}^2\xi_3](d\xi_1)^2+\xi_1^2[\text{sin}^2\xi_3(1)+\text{cos}^2\xi_3](d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3[1](d\xi_2)^2+\cancel{2\xi_1\cos\xi_3\sin\xi_3[-1+1](d\xi_1d\xi_3)}$$


 * $$=[1](d\xi_1)^2+\xi_1^2[1](d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3[1](d\xi_2)^2$$


 * Therefore, $$\text{ d}s^{2}=(d\xi_1)^2+\xi_1^2\text{cos}^2\xi_3(d\xi_2)^2+\xi_1^2(d\xi_3)^2$$


 * Where:


 * $$h_1=1$$


 * $$h_2=\xi_1\cos\xi_3$$


 * $$h_3=\xi_1$$


 * The Laplacian for any function $$\Psi$$ in these coordinates can be determined from:


 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$


 * For i=1,


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{1}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{1} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{1}} \right] &= \frac{1}{\xi_1^{2}\text{ cos } \xi_3}\ \frac{\partial}{\partial \xi_1}\left[ \frac{\xi_1^{2}\text{ cos } \xi_3}{1^{2}} \frac{\partial \Psi}{\partial \xi_1}\right] \\

&= \frac{1}{\xi_1^{2}\text{ cos } \xi_3}\ \frac{\partial}{\partial \xi_1}\left[ \xi_1^{2}\text{ cos } \xi_3\ \frac{\partial \Psi}{\partial \xi_1}\right] \\

&= \frac{1}{\xi_1^{2}} \frac{\partial}{\partial \xi_1}\left[ \xi_1^2\frac{\partial \Psi}{\partial \xi_1}\right] \end{align} $$


 * For i=2:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{2}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{2} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{2}} \right] &=\frac{1}{\xi_1^2\text{ cos }\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{\xi_1^2 \text{ cos }\xi_3 }{\xi_1^2 \text{ cos}^2 \xi_3 }\cdot \frac{d\Psi }{d\xi_2} \right ] \\

&=\frac{1}{\xi_1^2\text{ cos }\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{1}{\text{ cos }\xi_3} \cdot \frac{d\Psi }{d\xi_2} \right ] \\

&=\frac{1}{\xi_1^2\text{ cos}^2\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{d\Psi }{d\xi_2} \right ] \end{align} $$


 * for i=3:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{3}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{3} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{3}} \right] &= \frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \frac{\xi_1^2 \text{ cos }\xi_3}{ \xi_1^{2}}\ \frac{ \partial \Psi }{\partial \xi_3} \right] \\

&=\frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \text{ cos }\xi_3 \cdot \frac{ \partial \Psi }{\partial \xi_3} \right] \end{align} $$


 * Therefore the overall Laplacian is as follows:


 * $$\Delta \Psi =\frac{1}{\xi_1^{2}} \frac{\partial}{\partial \xi_1}\left[ \xi_1^2\frac{\partial \Psi}{\partial \xi_1}\right]+\frac{1}{\xi_1^2\text{ cos}^2\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{d\Psi }{d\xi_2} \right ]+\frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \text{ cos }\xi_3 \cdot \frac{ \partial \Psi }{\partial \xi_3} \right]$$

Contributors: Egm6321.f09.team4.palubin.d 00:22, 18 November 2009 (UTC)

Problem #3:
Lecture 31-2:


 * Given :


 * Spherical coordinates in standard convention; uses $$\bar{\theta}$$ instead of $$\theta$$ (see Lecture 29-3). The only difference is that $$\bar{\theta}$$ is measured from the positive z-axis instead of the x-y plane.  In other words, $$\bar{\theta} = \frac{\pi}{2}$$ and $$\theta = 0$$ at the equator.


 * This introduces the following changes:


 * $$ \begin{align}

\bar{\theta} &= \frac{\pi}{2} - \theta \\ \\ x &= r \text{ sin}\left(\bar{\theta}\right) \text{ cos}\left(\phi\right) \\ y &= r \text{ sin}\left(\bar{\theta}\right) \text{ sin}\left(\phi\right) \\ z &= r \text{ cos}\left(\bar{\theta}\right) \end{align} $$


 * Find :


 * Obtain an expression for the Laplacian in this coordinate system.


 * Solution :


 * It has been previously determined that for spherical coordinates using $$\theta$$ rather than $$\bar{\theta}$$,


 * $$\text{d}s^2=\text{d}r^2+r^2\cos\theta(\text{d}\phi)^2+r^2(\text{d}\theta)^2$$


 * And since $$\bar{\theta} = \frac{\pi}{2} - \theta $$:


 * $$\text{d}\bar{\theta}=\text{d}(\frac{\pi}{2})-\text{d}\theta=0-\text{d}\theta=-\text{d}\theta$$


 * $$\cos\theta=\sin\bar{\theta}$$


 * Converting to the new coordinates,


 * $$\text{d}s^2=\text{d}r^2+r^2\sin\bar{\theta}(\text{d}\phi)^2+r^2(\text{d}\bar{\theta})^2$$


 * And:


 * $$h_1=1$$


 * $$h_2=r\sin\bar{\theta}$$


 * $$h_3=r$$


 * The Laplacian for any function $$\Psi$$ in these coordinates can be determined from:


 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$


 * For i=1:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{1}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{1} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{1}} \right] &= \frac{1}{r^{2}\text{ sin } \bar{\theta}}\ \frac{\partial}{\partial r}\left[ \frac{r^{2}\text{ sin } \bar{\theta}}{1^{2}} \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}\text{ sin } \bar{\theta}}\ \frac{\partial}{\partial r}\left[ r^{2}\text{ sin } \bar{\theta}\ \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}} \frac{\partial}{\partial r}\left[ r^2\frac{\partial \Psi}{\partial r}\right] \end{align} $$


 * For i=2:


 * Using a similar procedure as above, we find that:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{2}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{2} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{2}} \right] &=\frac{1}{r^2\text{ sin }\bar{\theta} }\frac{d}{d\phi }\left [ \frac{r^2 \text{ sin }\bar{\theta} }{r^2 \text{ sin}^2 \bar{\theta} }\cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ sin }\bar{\theta} }\frac{d}{d\phi }\left [ \frac{1}{\text{ sin }\bar{\theta}} \cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ sin}^2\bar{\theta} }\frac{d}{d\phi }\left [ \frac{d\Psi }{d\phi} \right ] \end{align} $$


 * For i=3:


 * As above, we find that:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{3}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{3} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{3}} \right] &= \frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \theta}\left[ \frac{r^2 \text{ sin }\bar{\theta}}{ r^{2}}\ \frac{ \partial \Psi }{\partial \bar{\theta}} \right] \\

&=\frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \bar{\theta}}\left[ \text{ sin }\bar{\theta} \cdot \frac{ \partial \Psi }{\partial \bar{\theta}} \right] \end{align} $$


 * The overall Laplacian is then:


 * $$\Delta \Psi =\frac{1}{r^{2}} \frac{\partial}{\partial r}\left[ r^2\frac{\partial \Psi}{\partial r}\right]+\frac{1}{r^2\text{ sin}^2\bar{\theta} }\frac{d}{d\phi }\left [ \frac{d\Psi }{d\phi} \right ]+\frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \bar{\theta}}\left[ \text{ sin }\bar{\theta} \cdot \frac{ \partial \Psi }{\partial \bar{\theta}} \right]$$

Contributors: Egm6321.f09.team4.palubin.d 01:55, 18 November 2009 (UTC)

Problem #4:
Lecture 31-2:


 * Given :


 * $$ \begin{align}

\lambda\left( \lambda + 1 \right) = n\left( n+1\right) \end{align} $$


 * Find :


 * Two solutions for $$\lambda$$ in terms of $$n$$ (use the quadratic equation).


 * Solution :


 * $$\lambda\left( \lambda + 1 \right) = n\left( n+1\right)$$
 * $$\lambda^2+\lambda- n\left( n+1\right)=0$$
 * $$\lambda_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{1-4(1)(-n(n+1))}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{4n^2+4n+1}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{(2n+1)^2}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm (2n+1)}{2}$$
 * $$\lambda_{1,2}=-\frac{1}{2}\pm \left (n+\frac{1}{2} \right )$$


 * Therefore,
 * $$\lambda_{1}=n$$
 * $$\lambda_{2}=-n-1=-(n+1)$$

Contributors: Egm6321.f09.Team4.meyerat 00:27, 16 November 2009 (UTC)

Problem #5:
Lecture 31-3:


 * Given :


 * $$ \begin{align}

P_{n}\left(x\right) &= \sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} \\ &\text{where } \left\lfloor n/2\right\rfloor = \ \text{floor of } n/2 \end{align} $$


 * Find :


 * Show that the above expression is equivalent to the following:


 * $$ \begin{align}

P_{n}\left(x\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$


 * Solution :


 * Setting these two equations equal to each other, we can simplify what we're trying to prove:


 * $$ \begin{align}

\sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} &= \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \\

\cancel{x^{n-2i}} \cancel{\left(-1\right)^{i}}\ \frac{\left(2n-2i\right)!}{2^{n}\ \cancel{i!}\ \left(n-i\right)!\ \cancel{\left(n-2i\right)!}} &= \cancel{x^{n-2i}} \cancel{\left(-1\right)^{i}}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ \cancel{i!}\ \cancel{\left(n-2i\right)!}} \qquad \forall\ i \\

\frac{\left(2n-2i\right)!}{2^{n}\left(n-i\right)!} &= \frac{1\times 3\times \dots \times \left(2n-2i-1\right)}{2^{i}} \qquad \forall\ i \end{align} $$


 * Expanding the left-hand side of this equation:


 * $$ \begin{align}

\frac{\left(2n-2i\right)!}{2^{n}\left(n-i\right)!} &= \frac{\left(2n-2i\right)!}{2^{n-i}2^{i}\left(n-i\right)!} \\ &= \frac{\left(\prod_{k=n-i+1}^{2n-2i} k\right)\left(\prod_{k=1}^{n-i} k\right) \frac{1}{2^{n-i}}}{2^{i}\left(\prod_{k=1}^{n-i} k \right)} \\ &= \frac{\frac{1}{2^{n-i}} \left(\prod_{k=n-1+1}^{2n-2i} k\right) }{2^{i}} \\ &= \frac{\prod_{k=1}^{n-i} \left(2k-1\right)}{2^{i}} \\ &= \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}} \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 19:26, 16 November 2009 (UTC)

Problem #6:
Lecture 31-3:


 * Given :


 * Either of the equations from Problem #5:


 * $$ \begin{align}

P_{n}\left(x\right) &= \sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} \\ &\text{where } \left\lfloor n/2\right\rfloor = \ \text{floor of } n/2 \end{align} $$


 * OR


 * $$ \begin{align}

P_{n}\left(x\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1,3,\dots,\left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$


 * Find :


 * $$P_{i}\left(x\right)$$, for $$i=0,1,2,3,4$$.


 * Solution :


 * $$ \begin{align}

P_{0}\left(x\right) &= \sum_{i=0}^{0}\left(\dots\right) = x^{0}\left(-1\right)^{0}\frac{0!}{2^{0}\left(0!\right)\left(0-0\right)!\left(0-0\right)!} \\ &= 1 \\ \\  P_{1}\left(x\right) &= \sum_{i=0}^{0}\left(\dots\right) = x^{1-2\left(0\right)}\left(-1\right)^{0}\frac{\left(2-0\right)!}{2\left(0\right)!\left(1-0\right)!\left(1-0\right)!} \\ &= x \\ \\ P_{2}\left(x\right) &= \sum_{i=0}^{1}\left(\dots\right) = x^{2-2\left(0\right)}\left(-1\right)^{0}\frac{4!}{2^{2}0!\left(2\right)!\left(2\right)!} + x^{2-2}\left(-1\right)^{1}\frac{\left(4-2\right)!}{2^{2}\left(1\right)!\left(1\right)!\left(0\right)!} \\ &=\frac{3}{2}x^{2}-\frac{1}{2} \\ &=\frac{1}{2}\left(3x^{2}-1\right) \\ \\ P_{3}\left(x\right) &= \sum_{i=0}^{1}\left(\dots\right) = x^{3-2\left(0\right)}\left(-1\right)^{0}\frac{6!}{2^{3}\left(0\right)!\left(3\right)!\left(3\right)!} + x^{3-2}\left(-1\right)^{1}\frac{\left(6-2\right)!}{2^{3}\left(1\right)!\left(2\right)!\left(1\right)!} \\ &= x^{3} \frac{\cancel{6}\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}}{2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{3}\cdot \cancel{2}} - x \frac{\cancel{4}\cdot 3\cdot \cancel{2}}{2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{2}} \\ &= x^{3}\frac{5}{2} - x\frac{3}{2} \\ &= \frac{1}{2}\left(5x^{3}-3x\right) \\ \\ P_{4}\left(x\right) &= \sum_{i=0}^{2}\left(\dots\right) = x^{4-2\left(0\right)}\left(-1\right)^{0}\frac{8!}{2^{4}\left(0\right)!\left(4\right)!\left(4\right)!} + x^{4-2}\left(-1\right)^{1}\frac{6!}{2^{4}\left(1\right)!\left(3\right)!\left(2\right)!} + x^{4-4}\left(-1\right)^{2}\frac{4!}{2^{4}\left(2\right)!\left(2\right)!\left(0\right)!} \\ &= x^{4}\frac{\cancel{8}\cdot 7\cdot \cancel{6}\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot}{\cancel{2}\cdot \cancel{2}\cdot \cancel{2}\cdot 2\cdot 4\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot} - x^{2}\frac{6\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot}{\cancel{2}\cdot 2\cdot 2\cdot 2\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{2}\cdot} + \frac{\cancel{4}\cdot 3\cdot \cancel{2}}{2\cdot 2\cdot 2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{2}} \\ &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:19, 16 November 2009 (UTC)

Problem #7:
Lecture 32-1:


 * Given :


 * The Legendre polynomials $$P_{0},\dots,P_{4}$$ verified in Problem #6 (see Lecture 31-3):


 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$


 * Find :


 * Show that these polynomials are solutions to the Legendre equation (given below; also see Lectures 14-2 and 30-4).


 * $$ \begin{align}

\left(1-x^{2}\right)y'' - 2xy' + n\left(n+1\right)y = 0 \end{align} $$


 * Note: Legendre polynomial $$P_{k}$$ is a solution to the specific Legendre equation given by $$ n = k $$.


 * Solution :


 * k = 0 :


 * Let $$y = P_{0} = 1$$. This results in the following derivative values:


 * $$ \begin{align}

y' &= 0 \\ y'' &= 0 \end{align} $$


 * Plugging these values, along with $$n=0$$, into the Legendre equation:


 * $$ \begin{align}

\left(1-x^{2}\right)\left(0\right) - 2x\left(0\right) + 0\left(0+1\right)\left(1\right) = 0 \\ 0 = 0 \quad \text{Proof.} \end{align} $$


 * k = 1 :


 * Let $$y = P_{1} = x$$. This results in the following derivative values:


 * $$ \begin{align}

y' &= 1 \\ y'' &= 0 \end{align} $$


 * Plugging these values, along with $$n=1$$, into the Legendre equation:


 * $$ \begin{align}

\left(1-x^{2}\right)\left(0\right) - 2x\left(1\right) + 1\left(1+1\right)x &= 0 \\ -2x + 2x &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$


 * k = 2 :


 * Let $$\begin{align} y = P_{2} = \frac{1}{2}\left(3x^{2}-1\right) \end{align}$$. This results in the following derivative values:


 * $$ \begin{align}

y' &= 3x \\ y'' &= 3 \end{align} $$


 * Plugging these values, along with $$n=2$$, into the Legendre equation:


 * $$ \begin{align}

\left(1-x^{2}\right)\left(3\right) - 2x\left(3x\right) + \left(6\right)\left(\frac{1}{2}\right)\left(3x^{2}-1\right) &= 0 \\ 3 - 3x^{2} - 6x^{2} + 9x^{2} - 3 &= 0 \\ 3 - 3 - 9x^{2} + 9x^{2} &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$


 * k = 3 :


 * Let $$\begin{align} y = P_{3} = \frac{1}{2}\left(5x^{3}-3x\right) \end{align}$$. This results in the following derivative values:


 * $$ \begin{align}

y' &= \frac{1}{2}\left(15x^{2} - 3\right) \\ y'' &= \frac{1}{2}\left(30x\right) = 15x \end{align} $$


 * Plugging these values, along with $$n=3$$, into the Legendre equation:


 * $$ \begin{align}

\left(1-x^{2}\right)\left(15x\right) - 2x\left(\frac{1}{2}\right)\left(15x^{2}-3\right) + 12\frac{1}{2}\left(5x^{3}-3x\right) &= 0 \\ 15x - 15x^{3} - 15x^{3} + 3x + 30x^{3} - 18x &= 0 \\ \left(15x + 3x - 18x\right) + \left(30x^{3}-15x^{3}-15x^{3}\right) &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$


 * k = 4 :


 * Let $$\begin{align} \end{align}$$. This results in the following derivative values:


 * $$ \begin{align}

y' &= \frac{35}{2}x^{3} - \frac{15}{2}x \\ y'' &= \frac{105}{2}x^{2} - \frac{15}{2} \end{align} $$


 * Plugging these values, along with $$n=4$$, into the Legendre equation:


 * $$ \begin{align}

\left(1-x^{2}\right)\left(\frac{105}{2}x^{2}-\frac{15}{2}\right) - 2x\left(\frac{35}{2}x^{3} -\frac{15}{2}x\right) + 20 \left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) &= 0 \\ \frac{105}{2}x^{2} - \frac{15}{2} - \frac{105}{2}x^{4} + \frac{15}{2}x^{2} - \frac{70}{2}x^{4} + \frac{30}{2}x^{2} + \frac{175}{2}x^{4} - \frac{150}{2}x^{2} + \frac{15}{2} &= 0 \\ \left(-\frac{105}{2}-\frac{70}{2}+\frac{175}{2}\right)x^{4} + \left(\frac{105}{2}+\frac{15}{2}+\frac{30}{2} - \frac{150}{2}\right)x^{2} + \cancel{\left(\frac{15}{2}-\frac{15}{2}\right)} &= 0 \\ \left(-\frac{175}{2}+\frac{175}{2}\right)x^{4} + \left(\frac{150}{2} - \frac{150}{2}\right)x^{2} &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 20:26, 16 November 2009 (UTC)

Problem #8:
Lecture 33-3:


 * Given :


 * Functions $$f\left(\right)$$ and $$\left\{g_{i}\left(\right)\right\}$$ such that:


 * $$ \begin{align}

f\left(\right) = \sum_{i}\ g_{i}\left(\right) \end{align} $$


 * Find :

<ul> <li> Show that, if $$\left\{g_{i}\right\}$$ is odd, then $$f$$ is odd. </li> <li> Show that, if $$\left\{g_{i}\right\}$$ is even, then $$f$$ is even. </li> </ol></ul>


 * Solution :


 * Start with the definitions of even and odd functions. If a function is even, than


 * $$f(x)=f(-x)$$


 * If the function is odd, than,


 * $$-f(x)=f(-x)$$


 * So, for our problem,


 * $$f(x)=\sum_{i}^{\infty }g_{i}(x)$$
 * $$f(x)=g_{0}(x)+g_{1}(x)+g_{2}(x)+...$$


 * If $$g(x)$$ is odd, than


 * $$g_{0}(-x)+g_{1}(-x)+g_{2}(-x)+...$$
 * $$=-g_{0}(x)-g_{1}(x)-g_{2}(x)-...$$
 * $$=-\left (g_{0}(x)+g_{1}(x)+g_{2}(x)+... \right )$$
 * $$=-f(x)$$


 * Therefore, if $$g_{i}(x)$$ is odd, than $$f(x)$$ is also odd. The same concept applies if $$g(x)$$ is an even function.

Contributors: Egm6321.f09.Team4.meyerat 01:46, 17 November 2009 (UTC)

Problem #9:
Lecture 33-3:


 * Given :


 * Functions $$P_{2k}(x)$$ and $$P_{2k+1}(x)$$


 * Find :


 * Show that $$P_{2k}(x)$$ is even and $$P_{2k+1}(x)$$ is odd.


 * Solution :


 * The equation for $$P_{n}(x)$$ is given as:


 * $$P_{n}(x)=\sum_{i=0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-1)!(n-2i)!}$$


 * Substitute in n=2k,


 * $$P_{2k}(x)=\sum_{i=0}^{[2k/2]}(-1)^{i}\frac{(4k-2i)!x^{2k-2i}}{2^{2k}i!(2k-1)!(2k-2i)!}$$


 * This can be thought of as


 * $$P_{2k}(x)=\sum_{i=0}^{[k]}C_kx^{2k-2i}$$


 * where $$C_k$$ is just a coefficient. Therefore, we can rearrange the above as


 * $$P_{2k}(x)=\sum_{i=0}^{[k]}C_k\left (x^{k-i} \right )^2$$


 * Since $$k$$ and $$i$$ are whole, positive integers, and $$k\geq i$$, than $$k-i$$ will always be a positive integer. Therefore, regardless of whether $$x^{k-i}$$ is an odd or even function, it will always be squared, resulting in an even function:


 * Therefore, $$P_{2k}(x)$$ is even.


 * Similarly,


 * $$P_{2k+1}(x)=\sum_{i=0}^{[(2k+1)/2]}A_k\left (x^{2k+1-2i} \right )$$


 * $$P_{2k+1}(x)=\sum_{i=0}^{[(2k+1)/2]}A_k\left (x^{2k-2i} \right )x$$


 * Which is an even function times an odd function $$x$$. And as always, an even function times and odd function is always an odd function.


 * Therefore, $$P_{2k+1}(x)$$ is odd.

Contributors: Egm6321.f09.Team4.meyerat 01:29, 16 November 2009 (UTC)

Problem #10:
Lecture (33-4):


 * Given :


 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ c_{i} x^{i} \quad \in \mathcal{P}_{4} \\ &\text{where:} \\ &\qquad c_{0} = 3\\ &\qquad c_{1} = 10\\ &\qquad c_{2} = 15\\ &\qquad c_{3} = -1\\ &\qquad c_{4} = 5 \end{align} $$


 * Find :


 * Obtain $$\left\{a_{i}\right\}$$ such that:


 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} P_{i} = q\left(x\right) \end{align} $$


 * Note: $$ P_{i} $$ are the Legendre polynomials verified in Problem #6 (see Lecture 31-3):


 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$


 * Solution :


 * Plug the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:


 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$


 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$, obtaining a system of five equations:


 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$


 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:


 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$


 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:


 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$


 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:


 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$


 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):


 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$


 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:


 * $$ \begin{align}

a_{0} &= 9, \quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7}, \quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 15:04, 16 November 2009 (UTC)

Problem #11:
Lecture (34-2):


 * Given :


 * $$\Psi (1,\theta )=f(\theta )=T_ocos(\theta)$$
 * $$f(\mu )=T_o\sqrt{1-\mu ^2}$$
 * $$A_n=\frac{2n+1}{2}<f,P_n>$$


 * Find :


 * a) Without calculation, find the property of $$A_n$$.
 * b) Complete 3 nonzero coefficients of $$A_n$$ analytically or numerically using Gaussian-Legendre accurately to within 5%.


 * Solution :


 * a)We see that $$f(\mu)=T_o\sqrt{1-\mu ^2}$$ is an even function and we know that $$P_n(\mu)$$ is an even function when $$n$$ is an even integer. Therefore,


 * $$A_n=\frac{2n+1}{2}<f,P_n>=0$$ when n is odd, therefore,
 * $$A_1=A_3=A_5...=0$$.


 * b)The first three coefficients are as follows:


 * $$A_n=\frac{2n+1}{2}\int_{-1}^{1}T_o\sqrt{1-\mu^2}P_n(\mu)d\mu$$


 * $$A_0=\frac{T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$


 * From the integration tables, we see that this is,


 * $$A_0=\frac{T_o}{2}\left (\frac{\mu}{2}\sqrt{1-\mu^2}+\frac{1}{2}sin^{-1}(\mu) \right )_{-1}^{1}$$


 * $$A_0=\frac{T_o}{2}\left (\frac{\pi}{4}-\frac{3\pi}{4}\right )$$


 * $$A_0=-\frac{T_o\pi}{4}$$


 * $$A_2=\frac{5T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}\left ( \frac{1}{2}(3\mu^2-1) \right )d\mu$$
 * $$A_2=\frac{5T_o}{2}\left [\int_{-1}^{1}\frac{3}{2}\mu^2\sqrt{1-\mu^2}d\mu-\int_{-1}^{1}\frac{1}{2}\sqrt{1-\mu^2}d\mu \right ]$$
 * $$A_2=\frac{5T_o}{2}\left [\frac{3}{2}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu-\frac{1}{2}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu \right ]$$
 * $$A_2=\frac{15T_0}{4}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu-\frac{5T_0}{4}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$


 * From integral tables this becomes,


 * $$A_2=\frac{15T_0}{4}\left [ \frac{\mu^2}{8}(2\mu^2-1)\sqrt{1-\mu^2}+\frac{1}{8}sin^{-1}\mu \right ]_{-1}^{1}-\frac{5T_0}{4}\left [\frac{\mu}{2}\sqrt{1-\mu^2}+\frac{1}{2}sin^{-1}(\mu) \right ]_{-1}^{1}$$


 * $$A_2=\frac{15T_0}{4}\left [\frac{\pi}{16}-\frac{3\pi}{16}\right ]-\frac{5T_0}{4}\left [\frac{\pi}{4}-\frac{3\pi}{4}\right ]$$


 * $$A_2=-\frac{15T_0\pi}{32}+\frac{5T_0\pi}{8}$$


 * $$A_2=\frac{5T_0\pi}{32}$$


 * $$A_4=\frac{9T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}\left ( \frac{25}{8}\mu^4-\frac{15}{4}\mu^2+\frac{3}{8} \right )d\mu$$


 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu-\frac{135T_0}{8}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu+\frac{27T_0}{16}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$


 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu-\frac{135T_0}{8}\left [\frac{\pi}{16}-\frac{3\pi}{16}\right ]+\frac{27T_0}{16}\left [\frac{\pi}{4}-\frac{3\pi}{4}\right ]$$


 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$


 * Substitute in $$\mu=cos(x)$$ and we obtain


 * $$A_4=-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)sin^2(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$


 * $$A_4=-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)dx+\frac{225T_0}{16}\int_{0}^{\pi}cos^6(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$


 * From the integration tables


 * $$\int cos^n(x)dx=\frac{1}{n}cos^{n-1}(x)sin(x)-\frac{n-1}{n}\int cos^{n-2}(x)dx$$


 * Therefore, we have,


 * $$A_4=\frac{225T_0}{16}\left [\frac{1}{6}cos^5(x)sin(x) \right ]_{0}^{\pi}+\frac{225T_0}{16}\frac{5}{6}\int_{0}^{\pi}cos^4(x)dx-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$


 * Which reduces to,


 * $$A_4=0-\frac{225T_0}{96}\int_{0}^{\pi}cos^4(x)dx+\frac{81T_0\pi}{64}$$


 * $$A_4=0-\frac{225T_0}{96}\left [\left (\frac{1}{4}cos^3(x)sin(x) \right )_{0}^{\pi}+\frac{3}{4}\int_{0}^{\pi} cos^2(x)dx  \right ]+\frac{81T_0\pi}{64}$$


 * $$A_4=0-\frac{225T_0}{96}\left [\left (\frac{1}{4}cos^3(x)sin(x) \right )_{0}^{\pi}+\frac{3}{4}\left (\frac{x}{2}+\frac{ \sin 2x}{4}  \right )_{0}^{\pi} \right ]+\frac{81T_0\pi}{64}$$


 * $$A_4=0-\frac{225T_0}{96}\left [\frac{3\pi}{8}\right ]+\frac{81T_0\pi}{64}$$


 * $$A_4=-\frac{675T_0\pi}{768}+\frac{81T_0\pi}{64}$$


 * $$A_4=\frac{297T_0\pi}{768}=\frac{99T_0\pi}{256}$$


 * In summary,


 * $$A_0=-\frac{T_o\pi}{4}$$
 * $$A_2=\frac{5T_0\pi}{32}$$
 * $$A_4=\frac{99T_0\pi}{256}$$

Contributors: Egm6321.f09.Team4.meyerat 04:20, 18 November 2009 (UTC)

Problem #12:
Lecture (35-3):


 * Given :


 * Wikipedia table for Gaussian Legendre Quadrature


 * Find :


 * Verify the Gaussian Legendre Quadrature table in wikipedia for analytical expressions of $$\left \{ x_{j} \right \}$$ and $$\left \{ w_{j} \right \}$$, $$j=1,2...,n$$ and $$n=1,...5$$ after completing homework on page 31-3. Evaluate numerically $$\left \{ x_{j} \right \}$$ and $$\left \{ w_{j} \right \}$$ and compare results with Abramowitz and Stegun.


 * Solution :


 * For reference only, the wikipedia table is shown below:

Contributors:

Problem #13:
Lecture (36-4):


 * Given :


 * $$ \begin{align}

\tanh \left(x\right) = \frac{\exp\left(x\right) - \exp\left(-x\right)}{\exp\left(x\right) + \exp\left(-x\right)} \end{align} $$


 * Find :


 * Show that:


 * $$ \begin{align}

\tanh^{-1} \left(x\right) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$.


 * Solution :


 * Start with the following statement:


 * $$ \begin{align}

\tanh\left( \ \tanh^{-1}\left(x\right)\ \right) &= x \\ \tanh\left(\theta\right) &= x, \quad \text{(where } \theta = \tanh^{-1}\left(x\right) \text{)} \end{align} $$


 * Insert $$\text{ } \theta$$ into the definition of $$tanh(x)$$:


 * $$ \begin{align}

\left(\frac{\exp\left(\theta\right)-\exp\left(-\theta\right)}{\exp\left(\theta\right) + \exp\left(-\theta\right)}\right) &= x \\ \exp\left(\theta\right)-\exp\left(-\theta\right) &= x\exp\left(\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right) - x\exp\left(\theta\right) &= \exp\left(-\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right)\left(1-x\right) &= \exp\left(-\theta\right)\left(1+x\right)\\ \exp\left(2\theta\right) &= \frac{1+x}{1-x} \\ 2 \theta &= \ln\left(\frac{1+x}{1-x}\right) \\ \theta &= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$


 * Since $$\theta = \tanh^{-1}\left(x\right)$$:


 * $$ \begin{align}

\tanh^{-1}\left(x\right) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 17:20, 16 November 2009 (UTC)