User:Egm6321.f09.team5/HW7

Problem Statement
Proof that $$ Q_{1}(x) = \frac{1}{2}x\log \frac{(1+x)}{(1-x)} - 1 $$

Solution
$$ Q_{1}(x) = \frac{1}{2}x \log \frac{(1+x)}{(1-x)} -1 $$

Given the Legendre Equation as follows:

$$ F = (1-x^{2})y^{''} - 2xy^{'} + n(n+1)y = 0 $$

And one homogeneous solution:

$$ P_{1}(x) = x $$

The Legendre Equation can be rearranged in the following manner:

$$ F = y^{''} - (\frac{2x}{1-x^{2}})y^{'} + \frac{n(n+1)}{1-x^{2}}y = 0 $$

Using the known homogeneous solution the following can be calculated:

$$ y = x = U_{1} $$

$$ y^{'} = 1 $$

$$ y^{''} = 0 $$

Using the undeterminate factor method the following can be written as the second homogeneous solution:

$$ U_{2} = U_{1} \int_{}^{x} \frac{1}{t^{2}} \exp ( \int_{}^{t} (x - (\frac{2s}{1-s^{2}}ds)dt $$

$$ U_{2} = U_{1} \int_{}^{x} \frac{1}{t^{2}} (1-t^{2})^{-1}dt $$

$$ U_{2} = U_{1} \int_{}^{x} \frac{1}{t^{2}} \frac{1}{(1-t^{2})}dt $$

$$ U_{2} = U_{1} \int_{}^{x} \frac{1}{t^{2}-t^{4}} dt $$

$$ U_{2} = x \left [ \frac{1}{2} \ln (\frac{t + t^{2}}{t-t^{2}}) \right ] = \frac{1}{2}x \ln(\frac{x+x^{2}}{x-x^{2}}) $$

$$ U_{2} = \frac{1}{2}x \log \frac{(1+x)}{(1-x)} -1 $$

Problem Statement
Use Equation 2 from page 37-1 to show that $$ Q_{n} \qquad $$ is even or odd,depending on "n"

Equation 2:

$$Q_n(x)=P_n(x)tanh^{-1}(x)-2 \sum_{j=1,3,5}^{J} \frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)$$

$$J=1+2[\frac{n-1}{2}]$$

ODD: $$Q_0(x)=tanh^{-1}(x)$$

Solution
From Problem 10 in Homework#6, we know that $$P_n$$ is odd when n is odd ( $$n=2k+1$$ ), and even when n is even ($$n=2k$$).

We also know that $$tanh^{-1}(x)$$ is odd when n is even and even when n is odd.

So when n is odd $$P_n(x)tanh^{-1}(x)$$ is odd times even which equals odd, and when n is even $$P_n(x)tanh^{-1}(x)$$ is even times odd which is odd.

Regardless of n, both of the first terms yeild an odd value. The second term, containing $$P_{n-j}(x)$$ is even when n is odd, and odd when n is even. This subtracted by the odd first term will yield:

odd-odd=even when n is even

odd-even=odd when n is odd

$$Q_n(x)=odd$$ when n is odd $$Q_n(x)=even$$ when n is even

Problem Statement
Plot the Following:

$$ P_{0}, P_{1} , P_{2} , P_{3} , P_{4} $$

$$ Q_{0}, Q_{1} , Q_{2} , Q_{3} , Q_{4} $$

Solution
Plots of the Legendre Polynomial:

Plots of the Legendre Functions:



Problem Statement
Pg. 37-2.

What would be the result for

 =  ?

Solution
We know that when Legendre functions P_n and Q_n share the same indices they are opposite in terms of being odd or even functions. The product of and odd and even function is always odd.

Therefore:

$$  = \int_{-1}^{+1} [odd]dx$$

Based upon symmetry, the definite integral of an odd function centered about the origin must be equal to zero.

As a result:

$$  =  = 0$$

Problem Statement
Pg. 37-2 and 37-3

$$ \int_{-1}^{+1} [ (1-x^{2})L_{n}^{'} ]^{'}dx + n(n+1) \int_{-1}^{+1}L_{n}L_{m}dx = 0 $$

Integrate the first term to obtain a simplified expression.

Solution
$$\alpha= \int_{-1}^{+1} [ (1-x^{2})L_{n}^{'} ]^{'}dx $$

Use integration by parts:

$$ \int udv = uv - \int vdu $$

Where:

$$u=L_m$$

$$du=L_m^{'}dx$$

$$dv=[(1-x^{2})L_{n}^{'}]^{'}dx$$

$$v=(1-x^{2})L_{n}^{'}$$

$$\alpha=L_m(1-x^{2})L_{n}^{'}-\int (1-x^{2})L_{n}^{'}L_m^{'}dx$$

Evaluating from -1 to +1 makes the "uv" term equal to zero, so we are left with:

$$\alpha= -\int_{-1}^{+1} (1-x^{2})L_{n}^{'}L_{m}^{'}dx $$

Problem Statement
Pg 38-2.

Use Equation (2) and (3) in (1) to find $$ (rpq)^{2} $$

$$ (rpq)^{2} = \sum_{i=1}^{3} (x_{Q}^{i} - x_{P}^{i})^{2}   \qquad  $$ EQUATION 1

EQUATION 2

$$ x_{P}^{1} = r_{P} \cos \theta_{P} \cos \phi_{P} $$

$$ x_{P}^{2} = r_{P} \cos \theta_{P} \sin \phi_{P} $$

$$ x_{P}^{3} = r_{P} \sin \theta_{P} $$

EQUATION 3

$$ x_{Q}^{1} = r_{Q} \cos \theta_{Q} \cos \phi_{Q} $$

$$ x_{Q}^{2} = r_{Q} \cos \theta_{Q} \sin \phi_{Q} $$

$$ x_{Q}^{3} = r_{Q} \sin \theta_{Q} $$

Solution
$$ (rpq)^{2} = (X_{Q}^{1} - X_{P}^{1})^{2} + (X_{Q}^{2} - X_{P}^{2})^{2} + (X_{Q}^{3} - X_{P}^{3})^{2} $$

$$ = (r_{Q} \cos \theta_{Q} \cos \phi_{Q} - r_{Q} \cos \theta_{Q} \cos \phi_{Q})^{2} + (r_{Q} \cos \theta_{Q} \sin \phi_{Q} - r_{P} \cos \theta_{P} \sin \phi_{P})^{2} + (r_{Q} \sin \theta_{Q} - r_{P} \sin \theta_{P})^{2} $$

$$ = r_{Q}^{2} (\cos^{2} \theta_{Q} \cos^{2} \phi_{Q} + \cos^{2} \theta_{Q} \sin^{2} \phi_{Q} + \sin^{2}\theta_{Q}) + r_{P}^{2} (\cos^{2} \theta_{P} \cos^{2} \phi_{P} + \cos^{2} \theta_{P} \sin^{2} \phi_{P} + \sin^{2}\theta_{P}) - 2r_{Q}r_{P} (\cos \theta_{Q} \cos \phi_{Q} \cos \theta_{P} \cos \phi_{Q} + \cos \theta_{Q} \sin \phi_{Q} \cos \theta{p} \sin \phi_{P} + \sin \theta_{Q} \sin \theta_{P}) $$

Where the terms inside the parentheses of the third term will be called $$ \cos \gamma$$

The following result is then obtained:

$$= r_{Q}^{2} ( \cos^{2}\theta_{Q} + \sin^{2} \theta_{Q}) + r_{P}^{2} (\cos^{2} \theta_{P} + \sin^{2} \theta_{P}) - 2r_{Q}r_{P} \cos \gamma $$

Problem Statement
From Pg. 38-4 use Equations 4 and 5 to obtain Equations 6 and 7 from Pg. 38-3

Equation 4 from 38-4:

$$(x+y)^4=\sum_{k=0}^\infty [\begin{matrix} r \\ k \end{matrix}]x^{r-k}y^k $$

Equation 5 from 38-4:

$$[\begin{matrix} r \\ k \end{matrix}]=\frac{r(r-1)...(r-k+1)}{k!}$$

Equation 6 from 38-3:

$$(1-x)^{-\frac{1}{2}}=\sum_{i=0}^\infty \alpha_ix^i$$

Equation 7 from 38-3:

$$\alpha_i=\frac{1*3*...*(2i-1)}{2*4*...*(2i)} $$

Solution
Combining Equation 4 and 5:

$$(x+y)^4=\sum_{k=0}^\infty \frac{r(r-1)...(r-k+1)}{k!}x^{r-k}y^k $$

If $$r=-\frac{1}{2}$$

Then $$(x+y)^4=\sum_{k=0}^\infty \frac{(-\frac{1}{2})(-\frac{3}{2})...(\frac{1}{2}-k)}{k!}x^{-\frac{1}{2}-k}y^k $$

Because of the factors in the numerator, which correspond to k, we can move a $$2^k$$ to the denominator, and we can rearrange terms to eliminate the y terms and use this format: $$(1-x)^{r}=\sum_{k=0}^\infty [\begin{matrix} r \\ k \end{matrix}]x^k$$ to get:

$$(1-x)^{-\frac{1}{2}}=\sum_{k=0}^\infty \frac{(1)(3)...(2k-1)}{2^kk!}x^{k} $$

Setting k=i and expanding $$2^ii!$$ we get

$$(1-x)^{-\frac{1}{2}}=\sum_{i=0}^\infty \frac{(1)(3)...(2i-1)}{(2)(4)...(2i)}x^{i} $$

Setting

$$\alpha_i=\frac{1*3*...*(2i-1)}{2*4*...*(2i)} $$

We get

$$(1-x)^{-\frac{1}{2}}=\sum_{i=0}^\infty \alpha_ix^i$$

Problem Statement
Pg. 39-2 Continue the power series development to find P3,P4,P5 and compare the results to those obtained by Equations 6 and 7 from Pg. 31-3

$$P_0(\mu)=\alpha_0=1$$

$$P_1(\mu)=2\mu \alpha_1=\mu$$

$$P_2(\mu)=\frac{1}{2}(3\mu^2 -1)$$

Solution
We can use these equations to find the polynomials:

$$A(\mu, p) = 1-2\mu p + p^2$$

$$(1-x)^{-\frac{1}{2}}=\sum_{i=0}^\infty \alpha_ix^i$$

$$\frac{1}{\sqrt{A(\mu, p)}}=\sum_{i=0}^\infty \alpha_i(2\mu p - p^2)^i$$

$$\frac{1}{\sqrt{A(\mu, p)}}=\alpha_0+\alpha_1(2\mu p -p^2)+\alpha_2(2\mu p -p^2)^2+\alpha_3(2\mu p -p^2)^3+\alpha_4(2\mu p -p^2)^4+\alpha_5(2\mu p -p^2)^5...$$

$$\alpha_0=1$$, $$\alpha_1=\frac{1}{2}$$, $$\alpha_2=\frac{3}{8}$$, $$\alpha_3=\frac{5}{16}$$, $$\alpha_4=\frac{35}{128}$$, $$\alpha_5=\frac{63}{256}$$

$$\alpha_0(2\mu p -p^2)^0 = \alpha_1(2\mu p -p^2)^0 $$

$$\alpha_1(2\mu p -p^2)^1 = \alpha_1(2\mu p -p^2)^1 $$

$$\alpha_2(2\mu p -p^2)^2 = \alpha_2(p^4-4 p^3 u+4 p^2 u^2) $$

$$\alpha_3(2\mu p -p^2)^3 = \alpha_3(-p^6+6 p^5 \mu-12 p^4 \mu^2+8 p^3 \mu^3) $$

$$\alpha_4(2\mu p -p^2)^4 = \alpha_4(p^8-8 p^7 u+24 p^6 u^2-32 p^5 u^3+16 p^4 u^4) $$

$$\alpha_5(2\mu p -p^2)^5 = \alpha_5(-p^{10}+10 p^9 u-40 p^8 u^2+80 p^7 u^3-80 p^6 u^4+32 p^5 u^5) $$

Pulling the polynomials out:

$$P_3(\mu)=( -\alpha_24\mu+\alpha_38\mu^3 )= -\frac{3}{2}\mu+\frac{5}{2}\mu^3$$

$$P_4(\mu)=( \alpha_2-\alpha_312\mu^2+\alpha_416\mu^4 )=\frac{3}{8}-\frac{15}{4}\mu^2+\frac{35}{8}\mu^4$$

$$P_5(\mu)=( \alpha_36\mu-\alpha_432\mu^3+\alpha_532\mu^5 )=\frac{15}{8}\mu-\frac{35}{4}\mu^3+\frac{63}{8}\mu^5 )$$

This matches with answers derived in Problem 6 in Homework#6

Contributing Authors
--Egm6321.f09.team5.risher 19:00, 9 December 2009 (UTC) Egm6321.f09.team5.bear 19:48, 9 December 2009 (UTC) --Egm6321.f09.team5.GV 22:51, 9 December 2009 (UTC)