# Problem 1: Odd and Even Solutions of Legendre DE

From Lecture Slide 37-1.

## Given

From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 $\displaystyle P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i},$ $\displaystyle (Eq. 1)$

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

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

where $\displaystyle J$ is given by,

 $\displaystyle J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor .$ $\displaystyle (Eq. 3)$

## Find

Using Equation (2), to show when $\displaystyle Q_n$ is even or odd, depending on whether "$\displaystyle n$" is even or odd.

## Solution

We know, from Odd and Even property of $\displaystyle P_n$, that $\displaystyle P_n$ is odd, when $\displaystyle n$ is odd and even, when $\displaystyle n$ is even and the fact that

 $\displaystyle \tanh^{-1}(x),$ $\displaystyle (Eq. 4)$

is odd.

(i). When $\displaystyle n$ is odd.

When $\displaystyle n$ is odd, $\displaystyle P_n$ is odd and so,

 $\displaystyle P_n(x)\tanh^{-1}(x),$

is even.

 So, when $\displaystyle n$ is odd, first term in Equation (2) is even.

When $\displaystyle n$ is odd, value of $\displaystyle n - j$ is even since $\displaystyle j$ values are odd, and so, $\displaystyle P_{n-j}$ is even.

 So, when $\displaystyle n$ is odd, all the terms in the summation of Equation (2) are even.

And so,

 $\displaystyle \Rightarrow$ So, when $\displaystyle n$ is odd, $\displaystyle Q_n$ is even.

(ii). When $\displaystyle n$ is even.

When $\displaystyle n$ is even, $\displaystyle P_n$ is even and so,

 $\displaystyle P_n(x)\tanh^{-1}(x),$

is odd.

 So, when $\displaystyle n$ is even, first term in Equation (2) is odd.

When $\displaystyle n$ is even, value of $\displaystyle n - j$ is odd since $\displaystyle j$ values are odd, and so, $\displaystyle P_{n-j}$ is odd.

 So, when $\displaystyle n$ is even, all the terms in the summation of Equation (2) are odd.

And so,

 $\displaystyle \Rightarrow$ So, when $\displaystyle n$ is even, $\displaystyle Q_n$ is odd.

# Problem 2: Legendre Solutions Plot

From Lecture Slide 37-1.

## Given

From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 $\displaystyle P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} ,$ $\displaystyle (Eq. 1)$

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

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

where $\displaystyle J$ is given by,

 $\displaystyle J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor .$ $\displaystyle (Eq. 3)$

## Find

Plot {$\displaystyle P_0, P_1, P_2, P_3, P_4$} and {$\displaystyle Q_0, Q_1, Q_2, Q_3, Q_4$} using Matlab.

## Solution

Please refer Problem 6 Solution of HW6, for $\displaystyle P_0(x),\ P_1(x),\ P_2(x),\ P_3(x),\ P_4(x)$.

Please refer [1], p. 33, for expressions of $\displaystyle Q_0(x), Q_1(x), Q_2(x), Q_3(x)$.

The expression for $\displaystyle Q_4(x)$is given by,

 $\displaystyle Q_4= P_4(x) \tanh^{-1}(x) - 2 \sum_{j=1,3}^{} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x)$
 $\displaystyle \Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] \tanh^{-1}(x) - 2 \left[ \frac{2(4)-2+1}{(2(4)-1+1)1} P_{3}(x) \right] - 2 \left[ \frac{2(4)-2(3)+1}{(2(4)-3+1)3} P_{1}(x) \right]$
 $\displaystyle \Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] \tanh^{-1}(x) - \frac{7}{8} \left[ 5 x ^ 3 - 3 x\right] - \frac{1}{3} x$
.

Matlab Code:

clear all;
x = -1:0.01:1;
P_0 = 1*ones(1,length(x));
P_1 = x;
P_2 = (1/2)*(3*x.^2 - 1);
P_3 = (1/2)*(5*x.^3 - 3*x);
P_4 = (35/8)*x.^4 - (15/4)*x.^2 + (3/8);
Q_0 = atanh(x);
Q_1 = x.*atanh(x) - 1;
Q_2 = (1/2)*(3*x.^2 - 1).*atanh(x) - (3/2)*x;
Q_3 = (1/2)*(5*x.^3 - x).*atanh(x) - (5/2)*x.^2 + (2/3);
Q_4 = ((35/8)*x.^4 - (15/4)*x.^2 + (3/8)).*atanh(x) -(7/8)*(5*x.^3 - 3*x) - (1/3)*x;
figure(1);
plot(x,P_0,'--r',x,P_1,'--b',x,P_2,'--g',x,P_3,'--c',x,P_4,'--m');
legend('P_0','P_1','P_2','P_3','P_4');
figure(2);
plot(x,Q_0,'--r',x,Q_1,'--b',x,Q_2,'--g',x,Q_3,'--c',x,Q_4,'--m');
legend('Q_0','Q_1','Q_2','Q_3','Q_4');


Plots:

The plots for $\displaystyle P_n(x)$ with $\displaystyle n = 0,1,2,3,4$ are shown below:

The plots for $\displaystyle Q_n(x)$ with $\displaystyle n = 0,1,2,3,4$ are shown below:

# Problem 3: Orthogonality of Legendre Solutions

From Lecture Slide 37-2.

## Given

From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 $\displaystyle P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} ,$ $\displaystyle (Eq. 1)$

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

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

where $\displaystyle J$ is given by,

 $\displaystyle J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor$ $\displaystyle (Eq. 3)$

## Find

Show that, $\displaystyle < P_n, Q_n > = 0$.

## Solution

From lecture slide 33-1, the inner (scalar) product is given by,

 $\displaystyle = \int_{-1}^{+1} G(x)F(x)dx,$ $\displaystyle (Eq. 4)$

and Equation (4) becomes zero, when $\displaystyle G(x)$ is even and $\displaystyle F(x)$ is odd or when $\displaystyle G(x)$ is odd and $\displaystyle F(x)$ is even.

For $\displaystyle P_n(x)$ and $\displaystyle Q_n(x)$ using Equation (4),

 $\displaystyle = \int_{-1}^{+1} P_n(x)Q_n(x)dx.$ $\displaystyle (Eq. 5)$

From Solution to Problem 1 above, $\displaystyle Q_n(x)$ is odd and $\displaystyle P_n(x)$ is even, when $\displaystyle n$ is even. When $\displaystyle n$ is odd, $\displaystyle Q_n(x)$ is even and $\displaystyle P_n(x)$ is odd.

So, for all values of $\displaystyle n$, $\displaystyle Q_n(x)$ is opposite to $\displaystyle P_n(x)$ in oddness and evenness, which means,

 $\displaystyle \int_{-1}^{+1} P_n(x)Q_n(x)dx = 0.$

And so,

 $\displaystyle = 0$

# Problem 4: Integration by Parts: Legendre Solutions

From lecture slide 37-3.

## Given

From lecture slide 37-2,

 $\displaystyle \alpha = \int_{-1}^{+1} L_m \left[ (1-x^2) L'_n \right]' dx$ $\displaystyle (Eq. 1)$

where $\displaystyle L_n$ and $\displaystyle L_m$ are solutions to the Legendre differential equation.

## Find

Show that

 $\displaystyle \alpha = -\int_{-1}^{+1} (1-x^2) L'_n L'_m dx,$ $\displaystyle (Eq. 2)$

after integration by parts of $\displaystyle \alpha$ in Equation (1).

## Solution

Let,

 $\displaystyle u = L_m \Rightarrow du = L'_m dx,$ $\displaystyle (Eq. 3)$

and,

 $\displaystyle dv = \left[ (1-x^2) L'_n \right]' dx.$ $\displaystyle (Eq. 4)$
 $\displaystyle \Rightarrow \int dv = \int\left[ (1-x^2) L'_n \right]' dx \Rightarrow v = \left[ (1-x^2) L'_n \right].$

We know,

 $\displaystyle \int_{a}^{b} udv = \left[uv \right]_{a}^{b} - \int_{a}^{b} vdu ,$ $\displaystyle (Eq. 5)$

and so, with $\displaystyle a = -1$ and $\displaystyle b = +1$

 $\displaystyle \alpha = \int_{-1}^1 L_m \left[ (1-x^2) L'_n \right]' dx = \left[ L_m(1-x^2) L'_n\right]_{-1}^{+1} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx ,$ $\displaystyle (Eq. 6)$

For values of $\displaystyle x = -1$ and $\displaystyle x = +1$,

 $\displaystyle (1-x^2) = 0.$ $\displaystyle (Eq. 7)$

Substituting Equation (7) in first term of Equation (6), first term becomes zero.

 $\displaystyle \alpha = \cancelto{0}{\left[ L_m(1-x^2) L'_n\right]_{-1}^{+1}} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx.$

So,

 $\displaystyle \alpha = - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx$

# Problem 5: Attraction of Spheres

From lecture slide 38-2.

## Given

From lecture slide 38-2,

 $\displaystyle (r_{PQ})^2 = \sum_{i=1}^{3} \left(x_Q^i - x_P^i\right)^2,$ $\displaystyle (Eq. 1)$

where,

$\displaystyle x_P^1 = x_P,\ x_P^2 = y_P,\ x_P^3 = z_P,\$

and,

$\displaystyle x_Q^1 = x_Q,\ x_Q^2 = y_Q,\ x_Q^3 = z_Q,\$

given by,

 $\displaystyle x_P = r_P \cos(\theta_P) \cos(\psi_P);\ y_P = r_P \cos(\theta_P) \sin(\psi_P);\ z_P = r_P \sin(\theta_P),$ $\displaystyle (Eq. 2)$

and,

 $\displaystyle x_Q = r_Q \cos(\theta_Q) \cos(\psi_Q);\ y_Q = r_Q \cos(\theta_Q) \sin(\psi_Q);\ z_Q = r_Q \sin(\theta_Q).$ $\displaystyle (Eq. 3)$

## Find

Show that,

 $\displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) \cos\gamma,$ $\displaystyle (Eq. 4)$

where,

 $\displaystyle \cos\gamma = \cos(\theta_Q) \cos(\theta_P) \cos(\psi_Q-\psi_P) + \sin(\theta_Q)\sin(\theta_P)$ $\displaystyle (Eq. 5)$

## Solution

Equation (1) can be written as,

 $\displaystyle (r_{PQ})^2 = \left(x_Q - x_P\right)^2 + \left(y_Q - y_P\right)^2 + \left(z_Q - z_P\right)^2.$ $\displaystyle (Eq. 6)$

Substituting Equations (2) and (3) in (6),

 \displaystyle \begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q \cos(\theta_Q) \cos(\psi_Q) - r_P \cos(\theta_P) \cos(\psi_P)\right]^2 + \\ & \left[r_Q \cos(\theta_Q) \sin(\psi_Q) - r_P \cos(\theta_P) \sin(\psi_P)\right]^2 + \\ & \left[r_Q \sin(\theta_Q) - r_P \sin(\theta_P)\right]^2 \end{align}
 \displaystyle \begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q^2 \cos^2(\theta_Q) \cos^2(\psi_Q) + r_P^2 \cos^2(\theta_P) \cos^2(\psi_P) - 2 r_P r_Q \cos(\theta_Q) \cos(\psi_Q) \cos(\theta_P) \cos(\psi_P) \right] + \\ & \left[r_Q^2 \cos^2(\theta_Q) \sin^2(\psi_Q) + r_P^2 \cos^2(\theta_P) \sin^2(\psi_P) - 2 r_P r_Q \cos(\theta_Q) \sin(\psi_Q) \cos(\theta_P) \sin(\psi_P)\right] + \\ & \left[r_Q^2 \sin^2(\theta_Q) + r_P^2 \sin^2(\theta_P) - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P)\right] \end{align}
 \displaystyle \begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 \cos^2(\theta_Q) \cancelto{1}{\left[ \cos^2(\psi_Q) + \sin^2(\psi_Q) \right]} + r_Q^2 \sin^2(\theta_Q) - 2 r_P r_Q \cos(\theta_Q) \cos(\theta_P) \left [\cos(\psi_Q) \cos(\psi_P) \right] + \\ &\ r_P^2 \cos^2(\theta_P) \cancelto{1}{\left[ \cos^2(\psi_P) + \sin^2(\psi_P) \right]} + r_P^2 \sin^2(\theta_P) - 2 r_P r_Q \cos(\theta_Q)\cos(\theta_P) \left [\sin(\psi_Q) \sin(\psi_P)\right] - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P) \end{align}
 \displaystyle \begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 \cancelto{1}{\left[\cos^2(\theta_Q) + \sin^2(\theta_Q) \right]} - 2 r_P r_Q \cos(\theta_Q) \cos(\theta_P) \underbrace{\left [\cos(\psi_Q) \cos(\psi_P) + \sin(\psi_Q) \sin(\psi_P)\right]}_{=\ \cos(\psi_Q-\psi_P)} + \\ &\ r_P^2 \cancelto{1}{\left[ \cos^2(\theta_P) + \sin^2(\theta_P) \right]} - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P) \end{align}
 \displaystyle \begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 + r_P^2 - 2 r_P r_Q \left[\underbrace{\cos(\theta_Q) \cos(\theta_P) {\cos(\psi_Q-\psi_P)} + \sin(\theta_Q)\sin(\theta_P)}_{=\ \cos\gamma} \right] \end{align}

And so,

 $\displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) \cos\gamma$

# Problem 6: Binomial Series

From lecture note slide 38-4.

## Given

The binomial series expansion is

 $\displaystyle (x+y)^r = \sum_{k=0}^\infty \begin{pmatrix} r \\ k \end{pmatrix} x^{r-k} y^k$ $\displaystyle (Eq. 1)$

where

 $\displaystyle \begin{pmatrix} r \\ k \end{pmatrix} = \frac{r(r-1) \cdot\cdot\cdot (r-k+1)}{k!}$ $\displaystyle (Eq. 2)$

## Find

Use Eqs. 1 and 2 to show that

 $\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i$ $\displaystyle (Eq. 3)$

where

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

## Solution

Using Eq. 3, we can rewrite the LHS of Eq. 1 as

 $\displaystyle (1-x)^{-1/2} = \sum_{k=0}^\infty \begin{pmatrix} -1/2 \\ k \end{pmatrix} 1^{-1/2-k} (-x)^k$ $\displaystyle (Eq. 5)$

We can expand this

 $\displaystyle (1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(-1/2)(-3/2)\cdot\cdot\cdot(1/2-k)}{k!} (-x)^k$ $\displaystyle (Eq. 6)$

The number of factors in the numerator is $\displaystyle k$, hence we can rewrite this as

 $\displaystyle (1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2k-1)}{(-2)^k k!} (-x)^k$ $\displaystyle (Eq. 7)$

Change $\displaystyle k \rightarrow i$, and cancel two minus signs

 $\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2i-1)}{2^i i!} x^i$ $\displaystyle (Eq. 8)$

The denominator can then be expanded

 $\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \underbrace{\frac{1 \cdot 3\cdot\cdot\cdot(2i-1)}{2 \cdot 4 \cdot \cdot \cdot (2i)}}_{\alpha_i} x^i$ $\displaystyle (Eq. 9)$

So that

 $\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i$ $\displaystyle (Eq. 10)$

where

 $\displaystyle \alpha_i = \frac{1 \cdot 3\cdot\cdot\cdot(2i-1)}{2 \cdot 4 \cdot \cdot \cdot (2i)}$ $\displaystyle (Eq. 11)$

# Problem 7: Generating Functon for Legendre Polynomials

From lecture note slide 39-2.

## Given

We have that

 $\displaystyle A(\mu, \rho) := 1 - 2 \mu \rho + \rho^2$ $\displaystyle (Eq. 1)$

and the binomial expansion

 $\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i$ $\displaystyle (Eq. 2)$

where

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

Using Eqs. 1 and 2, we can expand the expression

 $\displaystyle [A(\mu,\rho)]^{-1/2} = \alpha_0 + \alpha_1(2\mu \rho- \rho^2) + \alpha_2(2\mu \rho- \rho^2)^2 + ...$
 $\displaystyle = \underbrace{\alpha_0}_{P_0(\mu)} + \underbrace{2 \mu \alpha_1}_{P_1(\mu)} \rho + \underbrace{(- \alpha_1 + 4 \mu^2 \alpha_2)}_{P_2(\mu)} \rho^2 + ...$ $\displaystyle (Eq. 4)$

## Find

Continue the expansion given by Eq. 4 to yield $\displaystyle P_3$, $\displaystyle P_4$, and $\displaystyle P_5$, and compare the result to that obtained by Eq. 7 on lecture note slide 31-3

 $\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} = \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$ $\displaystyle (Eq. 5)$

## Solution

We need the first 6 terms of the expansion

 $\displaystyle [A(\mu,\rho)]^{-1/2} = \alpha_0 + \alpha_1(2\mu \rho- \rho^2) + \alpha_2(2\mu \rho- \rho^2)^2 + \alpha_3(2\mu \rho- \rho^2)^3 + \alpha_4(2\mu \rho- \rho^2)^4 + \alpha_5(2\mu \rho- \rho^2)^5 + ...$
 $\displaystyle = \alpha_0$
 $\displaystyle + \alpha_1(2\mu \rho - \rho^2)$
 $\displaystyle + \alpha_2(4\mu^2 \rho^2 - 4\mu \rho^3 + \rho^4)$
 $\displaystyle + \alpha_3 (8\mu^3 \rho^3 - 12\mu^2 \rho^4 + 6\mu \rho^5 - \rho^6)$
 $\displaystyle + \alpha_4 (16\mu^4 \rho^4 - 32\mu^3 \rho^5 + 24\mu^2 \rho^6 - 8 \mu \rho^7 + \rho^8)$
 $\displaystyle + \alpha_5 (32\mu^5 \rho^5 - 80 \mu^4 \rho^6 + 80 \mu^3 \rho^7 - 40 \mu^2 \rho^8 + 10 \mu \rho^9 - \rho^{10}) + ...$
 $\displaystyle = \alpha_0 + \alpha_1 \rho$
 $\displaystyle + (-\alpha_1 + 4 \alpha_2 \mu^2) \rho^2$
 $\displaystyle + (- 4 \alpha_2 \mu + 8 \alpha_3 \mu^3) \rho^3$
 $\displaystyle + (\alpha_2 - 12 \alpha_3 \mu^2 + 16 \alpha_4 \mu^4) \rho^4$
 $\displaystyle + (6 \alpha_3 \mu - 32 \alpha_4 \mu^3 + 32 \alpha_5 \mu^5) \rho^5 + ...$ $\displaystyle (Eq. 6)$

Using Eq. 3, we have that

 $\displaystyle \alpha_0 = 1 \quad \alpha_1 = \frac{1}{2} \quad \alpha_2 = \frac{3}{8} \quad \alpha_3 = \frac{15}{48} \quad \alpha_4 = \frac{105}{384} \quad \alpha_5 = \frac{945}{3840} \quad$

we can substitute this into Eq. 6

 $\displaystyle = 1 + \frac{1}{2} \rho$
 $\displaystyle + \left(-\frac{1}{2} + \frac{3}{2} \mu^2\right) \rho^2$
 $\displaystyle + \left(- \frac{3}{2} \mu + \frac{5}{2} \mu^3\right) \rho^3$
 $\displaystyle + \left(\frac{3}{8} - \frac{15}{4} \mu^2 + \frac{35}{8} \mu^4\right) \rho^4$
 $\displaystyle + \left(\frac{15}{8} \mu - \frac{35}{4} \mu^3 + \frac{63}{8} \mu^5\right) \rho^5 + ...$ $\displaystyle (Eq. 7)$

From the above, we can extract

 $\displaystyle P_3(\mu) = - \frac{3}{2} \mu + \frac{5}{2} \mu^3$ $\displaystyle (Eq. 8)$
 $\displaystyle P_4(\mu) = \frac{3}{8} - \frac{15}{4} \mu^2 + \frac{35}{8} \mu^4$ $\displaystyle (Eq. 9)$
 $\displaystyle P_5(\mu) = \frac{15}{8} \mu - \frac{35}{4} \mu^3 + \frac{63}{8} \mu^5$ $\displaystyle (Eq. 10)$

Using Eq. 5, we get

 $\displaystyle P_3(x) = \sum_{i=0}^1 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$ $\displaystyle (Eq. 11)$
 $\displaystyle P_3(x) = \frac{1 \cdot 3 \cdot 5}{ 3!} x^3 - \frac{1 \cdot 3 }{2} x$ $\displaystyle (Eq. 12)$
 $\displaystyle P_3(x) = \frac{5}{ 2} x^3 - \frac{3}{2} x$ $\displaystyle (Eq. 13)$
 $\displaystyle P_4(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$ $\displaystyle (Eq. 14)$
 $\displaystyle P_4(x) = \frac{1 \cdot 3 \cdot ... \cdot 7}{4!} x^4 - \frac{1 \cdot 3 \cdot 5}{4} x^2 + \frac{1 \cdot 3 }{8}$ $\displaystyle (Eq. 15)$
 $\displaystyle P_4(x) = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}$ $\displaystyle (Eq. 16)$
 $\displaystyle P_5(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$ $\displaystyle (Eq. 17)$
 $\displaystyle P_5(x) = \frac{1 \cdot 3 \cdot ... \cdot 9}{5!} x^5 - \frac{1 \cdot 3 \cdot ... \cdot 7}{2 \cdot 3!} x^3 + \frac{1 \cdot 3 \cdot 5}{8} x$ $\displaystyle (Eq. 18)$
 $\displaystyle P_5(x) = \frac{63}{8} x^5 - \frac{35}{4} x^3 + \frac{15}{8} x$ $\displaystyle (Eq. 19)$

We have confirmed that the methods are equivalent, at least for terms up to $\displaystyle P_5$.

# Notes and References

1. King et al., "Differential Equations (Linear, Nonlinear, Ordinary, Partial)", Cambridge University Press, 2003

# Contributing Team Members

Egm6321.f09.Team1.andy 12:03, 29 November 2009 (UTC)

Egm6321.f09.Team1.sallstrom 18:48, 2 December 2009 (UTC)

Egm6321.f09.Team1.vasquez 05:09, 5 December 2009 (UTC)

Egm6321.f09.Team1.AH 15:59, 8 December 2009 (UTC)