User:Egm6321.f09.team4/HW7

= Homework #7: Team 4 =

Problem #1:
Lecture (37-1):


 * Given:


 * Non-polynomial homogenous solution to Legendre equation for $$n=1$$:


 * $$ \begin{align}

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


 * Find :


 * Show that this definition is equivalent to:


 * $$ \begin{align}

Q_{1}\left(x\right) = x \tanh^{-1}\left(x\right)-1 \end{align} $$


 * Solution :


 * First, remember the solution to Problem #13 in HW 6:


 * 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} $$


 * Then, plug this result into the second definition for $$Q_{1}$$:


 * $$ \begin{align}

Q_{1}\left(x\right) &= x \tanh^{-1}x-1 \\ &= x \left[ \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \right] - 1 \\ &= \frac{1}{2}x\ln \left( \frac{1+x}{1-x} \right) - 1 \qquad \text{Proof.} \end{align} $$

Contributors:


 * Egm6321.f09.team4.sls 18:12, 1 December 2009 (UTC)

Problem #2:
Lecture (37-1):


 * Given:


 * The general formula for non-polynomial, homogenous solutions to the Legendre equation:


 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ \frac{2n-2j+1}{\left(2n-j+1\right)j}P_{n-j}\left(x\right) \right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$


 * Find :


 * Whether $$Q_{n}$$ is even or odd, for any $$n$$. In other words, for which $$n$$ values are $$Q_{n}$$ odd, and for which $$n$$ are $$Q_{n}$$ even?


 * Solution :


 * From HW 6, Problem 9, we know that:


 * $$ \begin{align}

&P_{2k}\left(x\right) \quad \text{ is even.} \\ &P_{2k+1}\left(x\right) \quad \text{ is odd.} \end{align} $$


 * So, this solution will involve two cases: when $$n$$ is an even number (i.e., $$n=2k$$ for some pos. integer $$k$$), and when $$n$$ is an odd number (i.e., $$n=2k+1$$ for some pos. integer k).


 * Note: values that do not depend on $$x$$ will not affect the evenness or oddness of the result. So, to simplify the derivation, the equation for $$Q_{n}$$ will be rewritten as:


 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{n-j}\left(x\right) \right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$


 * Case 1 : $$\ n=2k$$ (even #)


 * $$ \begin{align}

Q_{2k}\left(x\right) &= P_{2k}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{2k-j}\left(x\right) \right] \end{align} $$


 * $$\begin{align} P_{2k}\left(x\right) \end{align}$$ is even, and $$\tanh^{-1}\left(x\right)$$ is known to be odd. Since the product of an odd and an even function is odd:


 * $$ \begin{align}

P_{2k}\left(x\right)\tanh^{-1}\left(x\right) \Rightarrow \left[\text{even}\right]\cdot\left[\text{odd}\right] \Rightarrow \left[\text{odd}\right] \end{align} $$


 * Inside the summation, the index ($$2k-j$$) must be an odd number, since the difference between an even number ($$2k$$) and an odd number ($$j$$) is always odd. Since ($$P_{n} \Rightarrow \left[odd\right]$$) for odd numbered $$n$$, and since the sum of odd numbers is itself odd, we get the following result:


 * $$ \begin{align}

-2\sum_{j=1,3,5\dots}^{J}\left\{ C\ P_{2k-j}\left(x\right)\right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left\{-2C\ \left[\text{odd}\right] \right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left[\text{odd}\right] \Rightarrow \left[\text{odd}\right] \end{align} $$


 * Since both terms are odd, the sum of the two terms must also be odd. Therefore, we can make the following conclusion:


 * $$ \begin{align}

Q_{2k}\left(x\right) \Rightarrow \left[\text{odd}\right], \text{ for } k\geq 0, k \in \mathbb{Z} \quad \text{(even numbered } n \text{)} \end{align} $$


 * Case 2 : $$\ n=2k+1$$ (odd #)


 * $$ \begin{align}

Q_{2k+1}\left(x\right) &= P_{2k+1}\left(x\right)\tanh^{-1}\left(x\right) - 2 \sum_{j=1,3,5,\dots}^{J}\left[ C\ P_{2k+1-j}\left(x\right) \right] \end{align} $$


 * Remember, $$\begin{align} P_{2k+1}\left(x\right) \Rightarrow \left[\text{odd}\right] \end{align}$$, and $$\tanh^{-1}\left(x\right) \Rightarrow \left[\text{odd}\right]$$. Since the product of two odd functions is even, the first term is even:


 * $$ \begin{align}

P_{2k+1}\left(x\right)\tanh^{-1}\left(x\right) \Rightarrow \left[\text{odd}\right]\cdot\left[\text{odd}\right] \Rightarrow \left[\text{even}\right] \end{align} $$


 * Inside the summation, the index ($$2k+1-j$$) must be an even number, since the difference between an odd number ($$2k+1$$) and another odd number ($$j$$) is always even. Since ($$P_{n} \Rightarrow \left[even\right]$$) for even numbered $$n$$, and since the sum of even numbers is itself even, we get the following result:


 * $$ \begin{align}

-2\sum_{j=1,3,5\dots}^{J}\left\{ C\ P_{2k+1-j}\left(x\right)\right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left\{-2C\ \left[\text{even}\right] \right\} \Rightarrow \sum_{j=1,3,5,\dots}^{J}\left[\text{even}\right] \Rightarrow \left[\text{even}\right] \end{align} $$


 * Since both terms are even, the sum of the two terms must also be even. Therefore, we can make the following conclusion:


 * $$ \begin{align}

Q_{2k+1}\left(x\right) \Rightarrow \left[\text{even}\right], \text{ for } k\geq 0, k \in \mathbb{Z} \quad \text{(odd numbered } n \text{)} \end{align} $$


 * In summary, we have shown that the polynomial solutions to the Legendre equation always have the opposite "oddness" to the non-polynomial solutions. In other words:


 * For even numbered $$n$$: $$P_{n}$$ is even, and $$Q_{n}$$ is odd.


 * For odd numbered $$n$$: $$P_{n}$$ is odd, and $$Q_{n}$$ is even.

Contributors:
 * Egm6321.f09.team4.sls 18:32, 2 December 2009 (UTC)

Egm6321.f09.Team4.meyerat 01:59, 8 December 2009 (UTC)

Problem #3:
Lecture (37-1):


 * Given:


 * Polynomial Legendre solutions (Legendre functions of the first kind) given as:


 * $$ \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} $$


 * Non-Polynomial Legendre solutions (Legendre functions of the second kind) given as:


 * $$ \begin{align}

Q_{n}\left(x\right) &= P_{n}\left(x\right)\tanh^{-1}\left(x\right) - 2\sum_{j=1,3,5,\dots}^{J}\left[ \frac{2n-2j+1}{\left(2n-j+1\right)j} P_{n-j}\left(x\right)\right] \\ &\text{ where } J = 1 + 2\left\lfloor \frac{n-1}{2} \right\rfloor \end{align} $$


 * Find :

 Plot $$\left\{P_{0},P_{1},P_{2},P_{3},P_{4}\right\}$$. Plot $$\left\{Q_{0},Q_{1},Q_{2},Q_{3},Q_{4}\right\}$$. 


 * Solution :


 * Graphs of the Legendre polynomials of the first kind $$P_{n}$$:


 * [[Image:egm6321.f09.sls.hw7.leg_P.png|450px|$$P_{n}$$]]


 * Graphs of the Legendre polynomials of the second kind $$Q_{n}$$:


 * [[Image:egm6321.f09.sls.hw7.leg_Q.png|450px|$$Q_{n}$$]]

Contributors:
 * Egm6321.f09.team4.sls 20:46, 1 December 2009 (UTC)

Problem #4:
Lecture (37-2):


 * Given:


 * Definition of the inner product of two functions on the interval $$\left[a,b\right]$$:


 * $$ \begin{align}

\left\langle g,h \right\rangle = \int_{a}^{b}g\left(x\right)h\left(x\right)\text{d}x \end{align} $$


 * Find :


 * The inner product between the $$n^{th}$$ polynomial and non-polynomial homogenous solution to the Legendre equation, on the interval $$\left[-1,1\right]$$:


 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = \int_{-1}^{1} P_{n}\left(x\right) Q_{n}\left(x\right) \text{d}x \end{align} $$


 * Solution :


 * From Problem #3, we know that $$P_{n}$$ and $$Q_{n}$$ always have opposite "oddness". Because the product of an odd function and an even function is always odd:


 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = \int_{-1}^{1} \left[\text{odd}\right] \text{d}x \end{align} $$


 * By symmetry, we know that any integral of an odd function over a finite integral centered on the origin must equal zero. Therefore:


 * $$ \begin{align}

\left\langle P_{n},Q_{n} \right\rangle = 0 \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:36, 2 December 2009 (UTC)

Problem #5:
Lecture (37-3):


 * Given:


 * The following portion of the proof of $$\left\langle P_{n},Q_{n}\right\rangle = 0$$, for $$n\neq m$$:


 * $$ \begin{align}

\alpha = \int_{-1}^{+1} L_{m}\left[ \left(1-x^{2}\right)L_{n}'\right]'\text{d}x \end{align} $$


 * and the definition of integration by parts:


 * $$ \begin{align}

\int_{a}^{b}f\left(x\right)g'\left(x\right)\text{d}x = \left[f\left(x\right)g\left(x\right)\right]_{a}^{b} - \int_{a}^{b}f'\left(x\right)g\left(x\right)\text{d}x \end{align} $$


 * Find :


 * Show that the following is true, by performing integration by parts:


 * $$ \begin{align}

\alpha = -\int_{-1}^{+1} \left(1-x^{2}\right)L_{n}'L_{m}'\text{d}x \end{align} $$


 * Solution :


 * First, make the following definitions:


 * $$ \begin{align}

f\left(x\right) &= L_{m} \\ g\left(x\right) &= \left(1-x^{2}\right)L_{n}' \\ a = -1, b = 1 \end{align} $$


 * Then, substitute these values into the integration by parts formula:


 * $$ \begin{align}

\int_{-1}^{1} L_{m}\left[\left(1-x^{2}\right)L_{n}'\right]'\text{d}x = \left[L_{m}\left(1-x^{2}\right)L_{n}'\right]_{-1}^{1} - \int_{-1}^{1} L_{m}'\left(1-x^{2}\right)L_{n}'\text{d}x \end{align} $$


 * Examining the first term, note that $$(1-x^{2})=0$$ for both -1 and 1. So, this term evaluates to zero: $$\left[L_{m}\left(1-x^{2}\right)L_{n}'\right]_{-1}^{1} = 0$$. Simplifying:


 * $$ \begin{align}

\int_{-1}^{1}L_{m}\left[\left(1-x^{2}\right)L_{n}'\right]'\text{d}x = \alpha = -\int_{-1}^{1} L_{m}'L_{n}'\left(1-x^{2}\right)\text{d}x \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:53, 2 December 2009 (UTC)

Problem #6:
Lecture (38-2):


 * Given:


 * Points $$P$$ and $$Q$$ in spherical coordinates, such that:


 * $$ \begin{align}

P &= \left(r_{p},\phi_{p},\theta_{p}\right) \\ Q &= \left(r_{q},\phi_{q},\theta_{q}\right) \end{align} $$


 * Let the cartesian coordinates of these points be given as:


 * $$ \begin{align}

P &= \left(x_{p1}, x_{p2}, x_{p3}\right) \\ Q &= \left(x_{q1}, x_{q2}, x_{q3}\right) \end{align} $$


 * The relationship between these two coordinate systems is given as:


 * $$ \begin{align}

x_{p1} &= r_{p}\cos\left(\theta_{p}\right)\cos\left(\phi_{p}\right) \\ x_{p2} &= r_{p}\cos\left(\theta_{p}\right)\sin\left(\phi_{p}\right) \\ x_{p3} &= r_{p}\sin\left(\theta_{p}\right) \\ \\ x_{q1} &= r_{q}\cos\left(\theta_{q}\right)\cos\left(\phi_{q}\right) \\ x_{q2} &= r_{q}\cos\left(\theta_{q}\right)\sin\left(\phi_{q}\right) \\ x_{q3} &= r_{q}\sin\left(\theta_{q}\right) \end{align} $$


 * Find :


 * Convert the following equation for $$\left(r_{pq}\right)^{2}$$ to spherical coordinates, by substituting in the relationships detailed above:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} = ||\vec{PQ}||^{2} = \sum_{i=1}^{3}\left(x_{qi}-x_{pi}\right)^{2} \end{align} $$


 * Solution :


 * Start off by inserting the definitions of $$x_{pi}\ \forall i$$ and $$x_{qi}\ \forall i$$ into the formula for $$\left(r_{pq}\right)^{2}$$:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} = &\left[r_{q}\cos\left(\theta_{q}\right)\cos\left(\phi_{q}\right) - r_{p}\cos\left(\theta_{p}\right)\cos\left(\phi_{p}\right)\right]^{2} \\ &+ \left[r_{q}\cos\left(\theta_{q}\right)\sin\left(\phi_{q}\right) - r_{p}\cos\left(\theta_{p}\right)\sin\left(\phi_{p}\right)\right]^{2} \\ &+ \left[ r_{q}\sin\left(\theta_{q}\right) - r_{p}\sin\left(\theta_{p}\right)\right]^{2} \end{align} $$


 * Expanding each squared term:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)\cos^{2}\left(\phi_{q}\right) - 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{p}\right)\cos\left(\phi_{q}\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_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\sin\left(\phi_{p}\right)\sin\left(\phi_{q}\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_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \end{align} $$


 * Then, rearranging the terms for simplification:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)\left[ \sin^{2}\left(\phi_{q}\right) + \cos^{2}\left(\phi_{q}\right)\right] + r_{p}^{2}\cos^{2}\left(\theta_{p}\right)\left[\sin^{2}\left(\phi_{p}\right)+\cos^{2}\left(\phi_{p}\right)\right]\\ &+ r_{q}^{2}\sin^{2}\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \\ &- 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right) \left[ \cos\left(\phi_{p}\right)\cos\left(\phi_{q}\right) + \sin\left(\phi_{p}\right)\sin\left(\phi_{q}\right) \right] \\ &- 2r_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) \end{align} $$


 * Using the trig. identities ($$\sin^{2}\left(x\right) + \cos^{2}\left(x\right)=1$$) and ($$\cos\left(x\right)\cos\left(y\right)+\sin\left(x\right)\sin\left(y\right)=\cos\left(x-y\right)$$), we can further simplify the equation:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{q}^{2}\cos^{2}\left(\theta_{q}\right)+r_{p}^{2}\cos^{2}\left(\theta_{p}\right)+r_{q}^{2}\sin^{2}\left(\theta_{q}\right) + r_{p}^{2}\sin^{2}\left(\theta_{p}\right) \\ &- 2r_{p}r_{q}\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{q}-\phi_{p}\right) \\ &- 2r_{p}r_{q}\sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right) \end{align} $$


 * Performing another regrouping allows further simplification using the first trig identity:


 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ &r_{p}^{2}\left[\sin^{2}\left(\theta_{p}\right) + \cos^{2}\left(\theta_{p}\right)\right] + r_{q}^{2} \left[\sin^{2}\left(\theta_{q}\right)+\cos^{2}\left(\theta_{q}\right)\right] \\ &- 2r_{p}r_{q}\left[\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right) \cos\left(\phi_{q}-\phi_{p}\right) + \sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right)\right] \end{align} $$


 * After performing the last simplification, the final form of the solution is reached.


 * $$ \begin{align}

\left(r_{pq}\right)^{2} =\ r_{p}^{2}+r_{q}^{2}-2r_{p}r_{q}\left[\cos\left(\theta_{p}\right)\cos\left(\theta_{q}\right)\cos\left(\phi_{q}-\phi_{p}\right) + \sin\left(\theta_{p}\right)\sin\left(\theta_{q}\right)\right] \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 19:49, 2 December 2009 (UTC)

Problem #7:
Lecture (38-4):


 * Given:


 * The binomial theorem, generalized to real values of $$r$$ (as opposed to integers):


 * $$ \begin{align}

\left(a+b\right)^{r} &= \sum_{k=0}^{\infty}\left( \begin{array}{c} r \\ k \end{array} \right) a^{r-k}b^{k} \\ &\text{ where } \left( \begin{array}{c} r\\ k \end{array} \right) = \frac{r\left(r-1\right)\dots\left(r-k+1\right)}{k!} \end{align} $$


 * Find :


 * Show that the following equation is true:


 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\alpha_{i}x^{i} \\ &\text{ where } \alpha_{i} = \frac{1 \cdot 3 \cdot \ldots \cdot \left(2i-1\right)}{2 \cdot 4 \cdot \ldots \cdot \left(2i\right)} \end{align} $$


 * Solution :


 * Define the following values:


 * $$ \begin{align}

&a = 1 \\ &b = -x \\ &r = -\frac{1}{2} \\ &k = i \end{align} $$


 * Substitute these values into the binomial theorem:


 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\frac{\left(-0.5\right)\left(-0.5-1\right)\cdots \left(-0.5-i+1\right)}{i!} \left(1\right)\left(-1\right)^{i}\left(x\right)^{i} \\ &= \sum_{i=0}^{\infty}\frac{\left(0.5\right)\left(1+0.5\right)\left(2+0.5\right)\cdots\left(i-1+0.5\right)}{\left(1\right)\left(2\right)\left(3\right)\cdots\left(i\right)}x^{i} \end{align} $$


 * Now, multiply both the top and the bottom of the fraction by ($$2^{i}$$):


 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\frac{\left(1\right)\left(2+1\right)\left(4+1\right) \cdots \left(2i-1\right)}{\left(2\right)\left(4\right)\cdots\left(2i\right)}x^{i} \\ &= \sum_{i=0}^{\infty}\frac{\left(1\right)\left(3\right)\left(5\right)\cdots\left(2i-1\right)}{\left(2\right)\left(4\right)\cdots\left(2i\right)}x^{i} \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 20:24, 2 December 2009 (UTC)

Egm6321.f09.Team4.meyerat 02:07, 8 December 2009 (UTC)

Problem #8:
Lecture (39-2):


 * Given:


 * The following result from the generalized binomial theorem (see last problem):


 * $$ \begin{align}

\left(1-x\right)^{-\frac{1}{2}} &= \sum_{i=0}^{\infty}\alpha_{i}x^{i} \\ &\text{ where } \alpha_{i} = \frac{1 \cdot 3 \cdot \ldots \cdot \left(2i-1\right)}{2 \cdot 4 \cdot \ldots \cdot \left(2i\right)} \end{align} $$


 * The primary definitions for the Legendre polynomials given in Problem #5 of HW 6:


 * $$ \begin{align}

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


 * The following alternative definition for the Legendre polynomials $$P_{n}\left(\mu\right)$$:


 * $$ \begin{align}

\left(1 - 2\mu\rho + \rho^{2}\right)^{-\frac{1}{2}} = \sum_{n=0}^{\infty}P_{n}\left(\mu\right)\rho^{n} \end{align} $$


 * Find :


 * Use the alternate definition of the Legendre polynomials to find $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$. Compare these results to those obtained from one of the primary definitions.


 * Solution :


 * Using the primary definition, we have the following values for $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$:


 * $$ \begin{align}

P_{3}\left(\mu\right) &= \frac{1}{2}\left(5\mu^{3}-3\mu\right) \\ P_{4}\left(\mu\right) &= \frac{1}{8}\left(35\mu^{4}-30\mu^{2}+3\right) \\ P_{5}\left(\mu\right) &= \frac{1}{8}\left(63\mu^{5}-70\mu^{3}+15\mu\right) \end{align} $$


 * Continuing the power series developed in class, we obtain:


 * $$\begin{align}

\frac{1}{\sqrt{1-2\mu p+p^2}}=\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+... \end{align} $$



\begin{align} =\alpha_0+\alpha_1(2\mu p-p^2)+\alpha_2(4\mu^2p^2-4\mu p^3+p^4)+\alpha_3(8\mu^3p^3-12\mu^2p^4+6\mu p^5-p^6) +\alpha_4(16\mu^4p^4-32\mu^3p^5+24\mu^2p^6-8\mu p^7+p^8)\\ +\alpha_5(32\mu^5p^5-80\mu^4p^6+80\mu^3p^7-40\mu^2p^8+10\mu p^9-p^{10})+... \end{align} $$


 * The terms can be grouped according to $$p^n$$ as:



\begin{align} \frac{1}{\sqrt{1-2\mu p+p^2}}=\alpha_0p^0+(2\mu \alpha_1)p^1+(4\mu^2 \alpha_2 -\alpha_1)p^2+(8\mu^3 \alpha_3-4\mu\alpha_2)p^3+(16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2)p^4\\ +(32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3)p^5+... \end{align} $$


 * Where,


 * $$P_0\left(\mu\right)=\alpha_0$$


 * $$P_1\left(\mu\right)=2\mu \alpha_1$$


 * $$P_2\left(\mu\right)=4\mu^2 \alpha_2 -\alpha_1$$


 * $$P_3\left(\mu\right)=8\mu^3 \alpha_3-4\mu\alpha_2$$


 * $$P_4\left(\mu\right)=16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2$$


 * $$P_5\left(\mu\right)=32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3$$


 * The values for $$\alpha_i$$ can be determined from:


 * $$\alpha_i=\frac{1\cdot 3 \cdot(...)\cdot(2i-1)}{2\cdot4\cdot (...)\cdot(2i)}$$


 * As such, we obtain:


 * $$\alpha_0=1$$


 * $$\alpha_1=\frac{1}{2}$$


 * $$\alpha_2=\frac{3}{8}$$


 * $$\alpha_3=\frac{5}{16}$$


 * $$\alpha_4=\frac{35}{128}$$


 * $$\alpha_5=\frac{63}{256}$$


 * Substituting in yields:


 * $$P_0\left(\mu\right)=\alpha_0=1$$


 * $$P_1\left(\mu\right)=2\mu \alpha_1=\mu$$


 * $$P_2\left(\mu\right)=4\mu^2 \alpha_2 -\alpha_1=\frac{1}{2}(3\mu^2-1)$$


 * $$P_3\left(\mu\right)=8\mu^3 \alpha_3-4\mu\alpha_2=8\mu^3\left(\frac{5}{16}\right)-4\mu\left(\frac{3}{8}\right)=\frac{1}{2}(5\mu^3-3\mu)$$


 * $$P_4\left(\mu\right)=16\mu^4\alpha_4-12\mu^2\alpha_3+\alpha_2=16\mu^4 \left ( \frac{35}{128} \right )-12\mu^2\left ( \frac{5}{16} \right )+\left ( \frac{3}{8} \right )=\frac{1}{8}(35\mu^4-30\mu^2+3)$$


 * $$\begin{align}

P_5\left(\mu\right)=32\mu^5\alpha_5-32\mu^3\alpha_4+6\mu\alpha_3=32\mu^5\left ( \frac{63}{256} \right )-32\mu^3\left ( \frac{35}{128} \right )+6\mu\left ( \frac{5}{16} \right ) &= \\ \frac{63}{8}\mu^5-\frac{35}{4}\mu^3+\frac{15}{8}\mu &= \\ \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \end{align}$$


 * Therefore the alternative definition yields the same values for $$P_{3}\left(\mu\right),\ P_{4}\left(\mu\right),\ P_{5}\left(\mu\right)$$

Contributors:

Egm6321.f09.team4.palubin.d 17:48, 6 December 2009 (UTC)