User:Egm6321.f09.Team2/HW6

Homework Assignment #6 - due Wednesday, 11/18, 22:00 UTC

Problem 1
p.31-1 For Circular Cylinder Coordinates find:

1) $$(dx_i)\ in\ terms\ of\ (\xi x_j)$$

2) Find $$ds^2=\sum_i^{}(dx_i)^2=\sum_k^{}(h_k)^2(d\xi_k)^2$$ Identify $$(h_i)\ in\ terms\ of\ (\xi_j)$$

3) Find the Laplacian in curvilinear coordinates.

4) See Problem 8

Problem 2
p.31-1 For Spherical Coordinates find:

1) $$(dx_i)\ in\ terms\ of\ (\xi x_j)$$

2) Find $$ds^2=\sum_i^{}(dx_i)^2=\sum_k^{}(h_k)^2(d\xi_k)^2$$ Identify $$(h_i)\ in\ terms\ of\ (\xi_j)$$

3) Find the Laplacian in spherical coordinates.

Use Astronomy Convention

Problem 3
p. 31-2 Find the Laplacian (from problem 2) in spherical coordinates but use the Math/Physics Convention.



Problem 4
p.31-2 Show that the solutions for $$\lambda (\lambda +1)=n(n+1)$$ are $$\lambda = n\ $$ and $$ \lambda=-(n+1)$$

$$\lambda (\lambda +1)=n(n+1)$$

$$\frac{(\lambda +1)}{n} = \frac{(n+1)}{\lambda}$$

$$\frac{\lambda}{n}+\frac{1}{n} = \frac{n}{\lambda}+\frac{1}{\lambda}$$

Assuming n and $$\lambda$$ are sufficiently large, $$\frac{\lambda}{n} = \frac{n}{\lambda}$$

Cross multiply and $$\lambda^2 = n^2$$   or

 $$\lambda = n\ $$

Likewise, $$\lambda (\lambda +1)=n(n+1)$$

$$(\lambda)^2 + \lambda=(n^2)+n$$

$$(\lambda^2 - (n^2) = n - \lambda$$

$$(\lambda + n)(\lambda - n) = (-1)(\lambda - n)$$

$$\lambda + n = -1 $$

$$\lambda = -1 - n = (-1)(n + 1)$$

 $$ \lambda=-(n+1)$$

Problem 5
p. 31-3 Show that $$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i{(2n-2i)!x^{n-2i}\over 2^ni!(n-i)!(n-2i)!}$$ can be reduced to $$\sum_{i=0}^{[n/2]}{1\times 3\times...(2n-2i-1)\over 2^ii!(n-2i)!}(-1)^ix^{n-2i}$$

Problem 6
p.31-3 Verify that equations (1)-(5) on p.31-3 can be written as Equation (6)or(7) from p.31-3

As seen from Problem 5, Equation 6 and Equation 7 are equivalent; therefore, anything that can be written as Equation 6 can be written as Equation 7.

Eq. (1): $$P_0(x) = 1$$ Eq. (2): $$P_1(x) = x$$ Eq. (3): $$P_2(x) = \frac{1}{2}(3x^2 - 1)$$ Eq. (4): $$P_3(x) = \frac{1}{2}(5x^3 - 3x)$$ Eq. (5): $$P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + {3}{8}$$

can be written as: Eq. (6): $$P_n(x) = \sum_{i=0}^{\frac{n}{2}} (-1)^i(\frac{(2n-2i)! x^{n-2i}}{(2^ni!)(n-i)!(n-2i)!})$$

For Eq. (1), $$n=0$$ and $$i=o$$, so the summation is only 1 number:
 * $$ = (-1)^0(\frac{(0)! x^{0}}{(2^00!)(0)!(0)!}) = 1(\frac{1*1}{(1*1)(1)(1)}) = 1$$

For Eq. (2), $$n=1$$ and i goes from 0 to $$\frac{1}{2}$$.
 * $$P_2(x) = \sum_{i=0}^{\frac{1}{2}} (-1)^i(\frac{(2-2i)! x^{1-2i}}{(2^1i!)(1-i)!(1-2i)!} = x + \frac{i}{2*pi} = x$$

For Eq. (3),$$P_2(x) = \sum_{i=0}^{1} (-1)^i(\frac{(4-2i)! x^{2-2i}}{(2^2i!)(2-i)!(2-2i)!}) = (\frac{1}{2}3x^2) - (\frac{1}{2}) = \frac{1}{2}(3x^2 - 1)$$

For Eq. (4),$$P_3(x) = \sum_{i=0}^{\frac{3}{2}} (-1)^i(\frac{(6-2i)! x^{3-2i}}{(2^3i!)(3-i)!(3-2i)!}) = (\frac{1}{2}(5x^3) - (\frac{1}{2}(3x) = \frac{1}{2}(5x^3 - 3x)$$

And Eq. (5), $$P_4(x) = \sum_{i=0}^{2} (-1)^i(\frac{(8-2i)! x^{4-2i}}{(2^4i!)(4-i)!(4-2i)!}) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + {3}{8}$$

Problem 7
p. 32-1 Verify that equations (1)-(5) on p.31-3 are solutions of the Legendre equation (1) on p.14-2 and p.30-4

Legendre equation is $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$

Eq. (1): $$P_0(x) = 1$$ Eq. (2): $$P_1(x) = x$$ Eq. (3): $$P_2(x) = \frac{1}{2}(3x^2 - 1)$$ Eq. (4): $$P_3(x) = \frac{1}{2}(5x^3 - 3x)$$ Eq. (5): $$P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$

So $$P_n(x) = y$$ and n is given in each equation,

Eq. (1): $$y = 1$$, $$ y' = 0$$, and $$ y'' = 0$$ with $$n = 0$$
 * $$ (1-x^2)(0) - 2x(0) + 0(0+1)1 = 0$$

Eq. (2): $$y = x$$, $$y' = 1$$, $$y'' = 0$$, $$n = 1$$
 * $$ (1-x^2)(0) - 2x(1) + 1(1+1)x = 0$$

Eq. (3): $$y = \frac{1}{2}(3x^2 - 1)$$, $$ y' = 3x$$, and $$ y'' = 3$$ with $$n = 2$$
 * $$ (1-x^2)(3) - 2x(3x) + 2(2+1)\frac{1}{2}(3x^2 - 1)$$
 * $$ 3 - 3x^2 - 6x^2 + 9x^2 - 3 = 0$$

Eq. (4): $$y = \frac{1}{2}(5x^3 - 3x)$$, $$ y' = \frac{15}{2}x^2-\frac{3}{2}$$, and $$ y'' = 15x$$ with $$n = 3$$
 * $$ (1-x^2)(15x) - 2x(\frac{15}{2}x^2-\frac{3}{2}) + 3(3+1)\frac{1}{2}(5x^3 - 3x)$$
 * $$ 15x - 15x^3 - 15x^3 + 3x + 30x^3 - 18x = 18x - 30x^3 + 30x^3 - 18x = 0$$

And Eq. (5): $$y = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$, $$ y' = \frac{35}{2}x^3-\frac{15}{2}x$$, and $$ y'' = \frac{105}{2}x^2 - \frac{15}{2}$$ with $$n = 4$$
 * $$ (1-x^2)(\frac{105}{2}x^2-\frac{15}{2}) - 2x(\frac{35}{2}x^3-\frac{15}{2}x) + 4(4+1)\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8} = 0$$

Problem 8
p.32-1 Continuation of Problem 1. Obtain the separate equations for the Laplace equation on circular cylinder coordinates an identify the Bessel differential equation. $$x^2y''+xy'-(x^2-\nu^2)y=0$$ Compare to King p.84

Given:
 * $$\nabla^2 \psi = \frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial \psi}{\partial r})+\frac{1}{r^2}\frac{\partial^2 \psi}{\partial \theta^2}+\frac{\partial^2 \psi}{\partial \xi^2}$$

Assumptions:
 * 1) $$\psi \!$$ is not dependent on $$\xi \!$$
 * 2) The equation is separable, $$\psi=R(r)\Theta(\theta) \!$$

This makes the Laplacian,
 * $$\nabla^2\psi=\frac{\Theta}{r}\frac{\partial}{\partial r}(r\frac{\partial R}{\partial r})+\frac{R}{r^2}\frac{\partial^2 \Theta}{\partial \theta^2}$$

Multiplying by $$r^2 \!$$ and dividing by $$R\Theta \!$$,
 * $$\nabla^2\psi=\frac{r}{R}\frac{d}{dr}(r\frac{dR}{dr})+\frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2}$$

Or,


 * $$\alpha(r)+\beta(\theta)=0 \!$$
 * $$\Rightarrow \alpha(r)=-\beta(\theta)=k^2$$

The two separated equations are then,


 * 1) $$r^2\frac{d^2R}{dr^2}+r\frac{dR}{dr}-k^2R=0$$
 * 2) $$\frac{d^2\Theta}{d\theta^2}+k^2\Theta=0$$

For (1), we can use the trial solution, $$R=r^{\lambda} \!$$, $$R'=\lambda r^{\lambda-1} \!$$, $$R''=\lambda(\lambda-1)r^{\lambda-2} \!$$

Plugging in yields,
 * $$\lambda(\lambda-1)+\lambda-k^2=0 \!$$
 * $$\Rightarrow \lambda^2=k^2 \!$$

For (2), we can use the trial solution $$\Theta = \exp\sigma \!$$, $$\Theta'=\sigma\exp\sigma \!$$, $$\Theta''=\sigma^2\exp\sigma \!$$

Plugging in yields,
 * $$\sigma^2+\lambda^2=0 \! $$
 * $$\Rightarrow \sigma^2=-\lambda^2 \!$$

Problem 9
p. 33-3 For $$f=\sum_i g_i$$ 1) show if {gi} is odd then f is odd 2) show if {gi} is even then f is even

A continuous function f can be expressed as $$f(x)=\sum_{n=0}^\infty A_nP_n(x)$$

We have observed that when n is EVEN $$P_n$$ is EVEN and when n is ODD $$P_n$$ is ODD

If we say $$g_i=A_nP_n$$

Then $$g_i$$ will be EVEN for all EVEN values of n.

And $$g_i$$ will be ODD for all ODD values of n.

Problem 10
p. 33-3 Show $$P_{2k}(x)$$ is even for k=0,1,2,... Show $$P_{2k+1}(x)$$ is odd for k=0,1,2,... Use equation (6) from p.31-3

For $$P_{2k}(x)$$ 'n' will always be an even number for all values of 'k' i.e. k=0,1,2,3,4,5......; n=0,2,4,6,8,10....

For $$P_{2k+1}(x)$$ 'n' will always be an odd number for all values of 'k' i.e. k=0,1,2,3,4,5......; n=1,3,5,7,9,11....

From the equation $$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i{(2n-2i)!x^{n-2i}\over 2^ni!(n-i)!(n-2i)!}$$ where [n/2] is only the integer part of the quotient i.e. [3/2]=1.5=1

$$\ x^{n-2i}$$ will always be an even number for all even values of 'n'.

$$\ x^{n-2i}$$ will always be an odd number for all odd values of 'n'.

Problem 11
p. 33-4 For $$q(x)=\sum_{i=0}^4c_ix^i$$ where $$c_0=3,\ c_1=10,\ c_2=15,\ c_3=-1,\ c_4=5$$

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

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

$$q(x)=\sum_{i=0}^4c_ix^i$$ can be shown as a column matrix:

$$\begin{bmatrix} c_0x^0\\c_1x^1\\c_2x^2\\c_3x^3\\c_4x^4 \end{bmatrix}=\begin{bmatrix} 3x^0\\10x^1\\15x^2\\-1x^3\\5x^4 \end{bmatrix}$$

$$q(x)=\sum_{i=0}^4a_iP_i=\begin{bmatrix} a_0&a_1&a_2&a_3&a_4 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4 \end{bmatrix}$$

Setting $$q(x)=\sum_{i=0}^4c_ix^i=\sum_{i=0}^4a_iP_i(x)$$ yields

$$\begin{bmatrix} a_0&a_1&a_2&a_3&a_4 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4 \end{bmatrix}=\begin{bmatrix} 3x^0\\10x^1\\15x^2\\-1x^3\\5x^4 \end{bmatrix}$$

$$a_0P_0=3x^0 \Rightarrow a_0={3/P_0}$$

$$a_1P_1=10x^1 \Rightarrow a_1={10x/P_1}$$

$$a_2P_1=15x^2 \Rightarrow a_2={15x^2/P_2}$$

$$a_3P_3=-1x^3 \Rightarrow a_3={-x^3/P_3}$$

$$a_4P_4=5x^4 \Rightarrow a_4={5x^4/P_4}$$

From lecture p.31-3

$$P_0(x)=1$$

$$P_1(x)=x$$

$$P_2(x)=1/2(3x^2-1)$$

$$P_3(x)=1/2(5x^3-3x)$$

$$P_4(x)=25/8x^4-15/4x^2+3/8$$

 $$a_0=3$$

$$a_1={10x\over x}=10$$

$$a_2={15x^2\over 1/2(3x^2-1)}$$

$$a_3={-x^3\over 1/2(5x^3-3x)}$$

$$a_4={5x^4\over 25/8x^4-15/4x^2+3/8}$$

Fig. 1 Fig. 2

Problem 12
Complete homework from p.22-2 if not done on previous HW#4

See HW#4 Problem 5

Problem 13
p. 34-2 Given the b.c.$$\psi(1,\theta)=f(\theta)=T_0cos\theta$$

1) Without calculation find the properties of An 2)  Compute three (3) non-zero coefficients for An using the following methods: 2a)Analytically (integration table or direct integration) use either $$\theta$$ or $$\mu$$.  2b)Numerically using Gaus-Legendre accurately within 5%

1) Since $$f \!$$ is even, $$A_n \!$$ will be zero when $$P_n \!$$ is odd or $$n=2k+1 \!$$

2a) The first 3 non-zero coefficients are $$ n=0,2,4 \!$$ calculated as follows:


 * $$A_n=\frac{2n+1}{2}\int_{-1}^{1}f(x)P_n(x)dx$$ with, $$f(x)=T_0\sqrt{1-x^2} \!$$.

To facilitate integration the equation can be converted to the variable $$\theta \!$$ by,


 * $$x=\sin{\theta} \!$$, $$dx=\cos{\theta}d\theta \!$$
 * $$f(\theta)=T_0\cos{\theta} \!$$
 * $$x=-1, \theta = -\frac{\pi}{2} \!$$, $$x=1, \theta = \frac{\pi}{2} \!$$

The new integral is then,


 * $$A_n=\frac{2n+1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}T_0\cos{\theta}P_n(\sin{\theta})\cos\theta d\theta \!$$

For $$n=0,2,4 \!$$ the equations are,


 * $$A_0=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}T_0\cos^2{\theta}P_0(\sin{\theta}) d\theta \!$$
 * $$A_2=\frac{5}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}T_0\cos^2{\theta}P_2(\sin{\theta}) d\theta \!$$
 * $$A_4=\frac{9}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}T_0\cos^2{\theta}P_4(\sin{\theta}) d\theta \!$$

And


 * $$P_0(\sin{\theta})=1 \!$$
 * $$P_2(\sin{\theta})=\frac{1}{2}(3\sin^2{\theta}-1) \!$$
 * $$P_4(\sin{\theta})=\frac{35}{8}\sin^4{\theta}-\frac{15}{4}\sin^2{\theta}+\frac{3}{8} \!$$

The identity to aid in integration is,


 * $$\int\cos^n{\theta}d\theta = \frac{\cos^{n-1}{\theta}\sin{\theta}}{n}+\frac{n-1}{n}\int\cos^{n-2}{\theta}d\theta \!$$

Making sure to convert any sine terms into cosine, the results are as follows,


 * $$A_0=\frac{\pi T_0}{4} \!$$
 * $$A_2=\frac{-5\pi T_0}{32} \!$$
 * $$A_4=\frac{-9\pi T_0}{256} \!$$

2b) Numerically solving for An can be done using the Gauss-Legendre quadrature. An integral from [-1,1] can be approximated by,


 * $$\sum_{j=1}^{m}w_jg(x_j) \!$$ where $$x_j \!$$ are the roots of the Legendre polynomial $$P_m \!$$
 * $$w_j = \frac{2}{(m+1)P'_m(x_j)P_{m+1}(x_j)}\!$$

For our problem,


 * $$g(x_j)=T_0\sqrt{1-x^2_j} P_n(x_j) \!$$

Therefore, $$A_n \!$$ can be written,


 * $$A_n = \frac{2n+1}{2}T_0\sum_{j=1}^{m}w_j\sqrt{1-x_j^2}P_n(x_j) \!$$

Writing out each function for n=0,2,4,


 * $$A_0 = \frac{1}{2}T_0\sum_{j=1}^{m}\frac{2\sqrt{1-x_j^2}P_0(x_j)}{(m+1)P'_m(x_j)P_{m+1}(x_j)} \!$$
 * $$A_2 = \frac{5}{2}T_0\sum_{j=1}^{m}\frac{2\sqrt{1-x_j^2}P_2(x_j)}{(m+1)P'_m(x_j)P_{m+1}(x_j)} \!$$
 * $$A_4 = \frac{9}{2}T_0\sum_{j=1}^{m}\frac{2\sqrt{1-x_j^2}P_4(x_j)}{(m+1)P'_m(x_j)P_{m+1}(x_j)} \!$$

Choosing $$m=10 \!$$ gives us the accuracy we need (5%) in determining each $$A_n\!$$,


 * $$A_0 = 0.7858 T_0 \!$$
 * $$A_2 =-0.4890 T_0 \!$$
 * $$A_4 =-0.1067 T_0 \!$$

Problem 14
p. 35-3 Verify the table for the Gaus-Legendre quadrature in wikepedia, analytic expression of {xj} and {wj} for j=1,...,n and n=1,...,5. You must first verify the expression for Pn for n=5,6. Evaluate numerically {xj} and {wj} and compare results with Abramovitz & Stegun, Table 25.4 p.916. 

First we will verify $$P_n\ when\ n=5,6$$

$$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i{(2n-2i)!x^{n-2i}\over 2^ni!(n-i)!(n-2i)!}$$

When n=5

$$P_{5(i=0)}(x)=(-1)^0{10!x^5\over 2^5(0!)(5!)(5!)}={(1)3628800x^5\over 460800}={63\over 8}x^5$$

$$P_{5(i=1)}(x)=(-1)^1{8!x^3\over 2^5(1!)(4!)(3!)}={(-1)40320x^3\over 4608}={-70\over 8}x^3$$

$$P_{5(i=2)}(x)=(-1)^2{6!x\over 2^5(2!)(3!)(1!)}={(1)720x\over 384}={15\over 8}x$$

$$P_5(x)=P_{5(i=0)}+P_{5(i=1)}+P_{5(i=2)}={63\over 8}x^5-{70\over 8}x^3+{15\over 8}x$$  $$P_5(x)={1\over 8}(63x^5-70x^3+15x)$$ When n=6

$$P_{6(i=0)}(x)=(-1)^0{12!x^6\over 2^6(0!)(6!)(6!)}={(1)479001600x^6\over 33177600}={231\over 16}x^6$$

$$P_{6(i=1)}(x)=(-1)^1{10!x^3\over 2^6(1!)(5!)(4!)}={(-1)3628800x^4\over 184320}={-315\over 16}x^4$$

$$P_{6(i=2)}(x)=(-1)^2{8!x^2\over 2^6(2!)(4!)(2!)}={(1)40320x^2\over 6144}={105\over 16}x^2$$

$$P_{6(i=3)}(x)=(-1)^3{6!x^0\over 2^6(3!)(3!)(0!)}={(-1)720\over 2304}=-5$$

$$P_6(x)=P_{6(i=0)}+P_{6(i=1)}+P_{6(i=2)}+P_{6(i=3)}={231\over 16}x^6-{315\over 16}x^4+{105\over 16}x^2-5$$  $$P_6(x)={1\over 16}(231x^6-315x^4+105x^2-5)$$

$$P_1,\ P_2,\ P_3,\ P_4$$ have been shown earlier in problem #6 of HW#6

Solve for the roots of $$(x_i)\ for\ n=1,..,5$$

$$P_1(x)=x\Rightarrow roots\ x_1=0$$

$$P_2(x)=1/2(3x^2-1)\Rightarrow x^2=1/3\Rightarrow roots\ x_2=+/-\sqrt {1/3}$$

$$P_3(x)=1/2(5x^3-3x)\Rightarrow (5/2)x^3=(3/2)x\Rightarrow x^2=3/5\Rightarrow roots\ x_3=0,\ x_3=+/-\sqrt{3/5}$$

$$P_4(x)=1/8(35x^4-30x^2+3)\Rightarrow roots\ x_4=+/-\sqrt {(3-2\sqrt 6/5)\over 7},\ x_4=+/-\sqrt {(3+2\sqrt 6/5)\over 7}$$

$$P_5(x)=1/8(63x^5-70x^3-15x)\Rightarrow roots\ x_5=0, x_5=+/-1/3\sqrt {5-2\sqrt 10/7},\ x_5=+/-1/3\sqrt {5+2\sqrt 10/7}$$

Solve for {wj} for j=1,...n and n=1,...5

where $$w_j={-2\over (n+1)P'_n(x_j)P_{n+1}(x_j)}$$

n=1

$$P_1=x;\ P'_1=1;\ P_2=1/2(3x_j^2-1)$$  $$w_1={-2\over (2)(1)(-1/2)}=2$$

n=2

$$P_2=1/2(3x_j^2-1);\ P'_2=3x_j;\ P_3=1/2(5x_j^3-3x)$$  $$w_{2+}={-2\over (3)(3)\sqrt{1/3}(1/2)[5\sqrt{1/3}^3-3\sqrt{1/3}]}={-2\over 9(1/3)^{1/2}[5/2(1/3)^{3/2}-3/2(1/3)^{1/2}]}={-2\over -2}=1$$

$$w_{2-}={-2\over (3)(3)(-\sqrt{1/3})(1/2)[5(-\sqrt{1/3}^3)-3(-\sqrt{1/3})]}={-2\over -9(1/3)^{1/2}[5/2(-1/3)^{3/2}-3/2(-1/3)^{1/2}]}={-2\over -2}=1$$

n=3

$$P_3=1/2(5x_j^3-3x);\ P'_3=15/2x_j^2-3/2;\ P_4=1/8(35x^4-30x^2+3)$$  $$w_{3-0}={-2\over (4)[(15/2)(0)-3/2](1/8)[35(0)-30(0)+3]}={-2\over (4)(-3/2)(3/8)}={-2\over -9/4}=8/9$$

$$w_{3+}={-2\over (4)[(15/2)\sqrt{3/5}^2-3/2](1/8)[35\sqrt{3/5}^4-30\sqrt{3/5}^2+3]}={-2\over (4)(3)(-18/5)}={-2\over -3.6}=5/9$$

$$w_{3-}={-2\over (4)[(15/2)(-\sqrt{3/5}^2)-3/2](1/8)[35(-\sqrt{3/5}^4)-30(-\sqrt{3/5}^2)+3]}={-2\over (4)(3)(-18/5)}={-2\over -3.6}=5/9$$

n=4

$$P_4=1/8(35x^4-30x^2+3);\ P'_4=1/8(140x^3-60x);\ P_5=1/8(63x^5-70x^3+15x)$$  $$w_{4+-}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3-60\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3+15\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]}={18+\sqrt 30\over 36}$$

$$w_{4--}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3-60-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3+15\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg)\Bigg]}={18+\sqrt 30\over 36}$$

$$w_{4++}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3-60\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3+15\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]}={18-\sqrt 30\over 36}$$

$$w_{4-+}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3-60-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3+15\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg)\Bigg]}={18-\sqrt 30\over 36}$$

n=5

$$P_5=1/8(63x^5-70x^3+15x);\ P'_5=1/8(315x^4-210x^2+15);\ P_6=1/16(231x^6-315x^4+105x^2-5)$$

 $$w_{5-0}={-2\over (6)\Big[1/8\Big(315(0)-210(0)+15\Big)\Big]\Big[1/16\Big(231(0)-315(0)-5\Big)\Big]}={-2\over (6)(15/8)(-5/16)}=128/225$$

$$w_{5+-}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4-210\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^6-315\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4+105\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332+13\sqrt70\over 900}$$

$$w_{5--}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4-210\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^6-315\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4+105\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332+13\sqrt70\over 900}$$

$$w_{5++}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4-210\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^6-315\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4+105\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332-13\sqrt70\over 900}$$

$$w_{5-+}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4-210\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^6-315\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4+105\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332-13\sqrt70\over 900}$$

Numerical values of {xj} and {wj}

$$\begin{bmatrix} {} & x_i & w_i \\ n=2 &

0.577350269189626 & 1.000000000000000 \\ n=3 & 0.000000000000000 & 0.888888888888889 \\ n=3 & 0.774596669241483 & 0.555555555555556 \\ n=4 & 0.339981043584856 & 0.652145154862546 \\ n=4 & 0.861136311594053 & 0.347854845137454\\ n=5 & 0.000000000000000 & 0.568888888888889 \\ n=5 & 0.538469310105683 & 0.478628670499366 \\ n=5 & 0.906179845938664 & 0.236926885056189 \end{bmatrix}$$

Problem 15
p. 36-4 show:

$$Q_0(x)=\frac{1}{2}log\Bigg({1+x\over 1-x}\Bigg)=tanh^{-1}(x)$$

If


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

then


 * $$\tanh{\Bigg(\tanh^{-1}x\Bigg)} = x $$

or


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

where


 * $$\tanh{x}=\frac{e^{2x}-1}{e^{2x}+1} $$

Plugging in gives


 * $$\frac{e^{\log{\frac{1+x}{1-x}}}-1}{e^{\log{\frac{1+x}{1-x}}}+1} \Rightarrow \frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1} $$


 * $$\frac{1+x-1+x}{1+x+1-x} = \frac{2x}{2}$$


 * $$\Rightarrow x \!$$

Contributing Team Members
--Joe Gaddone 20:59, 16 November 2009 (UTC) Kumanchik 21:13, 16 November 2009 (UTC) --Matt Walker 1400, 18 November 2009 (UTC) Egm6321.f09.Team2.sungsik 20:08, 18 November 2009 (UTC)