# R 9.1 - Coefficients for heat conduction on a sphere

## Given

 $\displaystyle f(\theta) = {T}_{0}cos(\theta)$ (9.1.1)

 $\displaystyle f(\theta) = {T}_{0}cos^{6}(\theta)$ (9.1.2)

 $\displaystyle f(\theta) = {T}_{0}exp[-\frac{2\theta}{\pi}^2]$ (9.1.3)

 $\displaystyle f(\mu) = {T}_{0}sinh(1-\mu^2)$ (9.1.4)

## Find

1. Find $A_{i}$
2. Compute the first 5 non-zero coefficients
3. Plot Exact Function, Fourier-Legendre series with 3 Terms and 5 Terms

## Solution

Problem 1.

 $\displaystyle \int_{-1}^{1}({p}_{2k}\times Even) \ne 0, \int_{-1}^{1}({p}_{2k}\times Odd) = 0, \int_{-1}^{1}({p}_{2k+1}\times Even) = 0, \int_{-1}^{1}({p}_{2k+1}\times Odd) \ne 0$ (9.1.5)
 $\displaystyle \mu = cos(\theta)$ (9.1.6)

Converting to $\theta$

 $\displaystyle f(\mu) = {T}_{0}cos(\mu)$ (9.1.7)

 $\displaystyle f(\mu) = {T}_{0}(\mu)^{6}$ (9.1.8)
 $\displaystyle f(\mu) = {T}_{0}exp[-(\frac{2{cos}^{-1}(\mu)}{\pi})^2]$ (9.1.9)
 $\displaystyle f(\mu) = \sum_{j=0}^\infin {A}_{j}{P}_{j}(\mu)$ (9.1.10)

 $\displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)F(\mu)d(\mu)$ (9.1.11)

Therefore,

 $\displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)T_{0}\mu d(\mu)$ (9.1.12)
 $\displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu)T_{0}(\mu)^{6}d(\mu)$ (9.1.13)
 $\displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu){T}_{0}exp[-\frac{2{cos}^{-1}(\mu)}{\pi}^2]d(\mu)$ (9.1.14)
 $\displaystyle A_{j}=\frac{2j+1}{2}\int_{-1}^{1}{P}_{j}(\mu){T}_{0}sinh(1-\mu^2)d(\mu)$ (9.1.15)

Problem 2.
Substitute j into Eq 9.1.11 to Eq 9.1.15
$A_{1}= T_{0},A_{3}=0,A_{5}=0,A_{7}=0, A_{9}=0$
$A_{0}= T_{0}, A_{2}=\frac{1}{7}T_{0}, A_{4}=\frac{10}{21}T_{0}, A_{6}=\frac{24}{77}T_{0}, A_{8}=0$
$A_{0}= 0.417T_{0}, A_{1}=0.478T_{0}, A_{2}=0.097T_{0}, A_{3}=0.0076T_{0}, A_{4}=-0.00157T_{0}$
$A_{0}= 0.7459T_{0}, A_{2}=-0.7997T_{0}, A_{4}=0.06671T_{0}, A_{6}=-0.01341T_{0}, A_{8}=0.00046T_{0}$
Problem 3.
Use Equation(9.1.10),Expand the fourier-legendre function

Egm6322.s12.sungsik 01:54, 7 February 2012 (UTC) Egm6322.s12.sungsik 02:05, 10 February 2012 (UTC)

# R 9.2 - Coefficients for general solution of heat conduction on a sphere

## Given

Boundary condition on the surface of the sphere:

 $\displaystyle f(\mu) = T_0(1-\mu ^2)^2$ (9.2.1)

Fourier-Legendre series for the boundary condition in 9.2.1:

 $\displaystyle f(\mu) = T_0(1-\mu ^2)^2 = \sum_{j=0,2,4}A_jP_j(\mu)$ (9.2.2)

General solution for heat conduction on a sphere:

 $\displaystyle u(r,\theta) = \sum_{j=0}^\infty \bar{A_j}r^jP_j(\mu)$ (9.2.3)

## Find

Show that

 $\displaystyle \sum_{j=0}^\infty \bar{A_j}P_j(\mu) = \sum_{j=0}^\infty A_jP_j(\mu) \Rightarrow \bar{A_j} = A_j, \forall j$ (9.2.4)

## Solution

 Solved on my own

The purpose of this problem is to show that the coefficients of the general formulation of the heat conduction problem on a sphere (obtained through solving the Laplace equation via separation of variables) are equal to the coefficients obtained from expanding the Fourier-Legendre series subject to the boundary condition of Eq. 9.2.1. As this condition is valid when r=1, Eq. 9.2.3 becomes:

 $\displaystyle u(r=1,\theta) = \sum_{j=0}^\infty \bar{A_j}P_j(\mu)$ (9.2.5)

Applying the boundary conditions and comparing to the Fourier-Legendre expansion of 9.2.2:

 \displaystyle \begin{align} \sum_{j=0}^\infty \bar{A_j}P_j(\mu) &= f(\mu) = T_0(1-\mu^2)^2 \\ &= \sum_{j=0}^\infty A_jP_j(\mu) \end{align} (9.2.6)

Recall from class notes that the only nonzero indices of $A_j$ are j=0,2,4. This is due to the orthogonality and oddness/eveness of the Legendre polynomials. This is covered in the lecture notes but will be reviewed here for clarity.

We know that the Legendre polynomials, by definition, are a subset of all polynomials:

 $\displaystyle P_j\in\mathcal P_j$ (9.2.7)

Any set of polynomials of degree n are spanned by the same number of orthogonal polynomials (for our purposes, Legendre polynomials will suffice). This also indicates that a polynomial of degree m > n is orthogonal to the subspace $\mathcal{P_n}$ as no linear combination of degree n polynomials can produce a new polynomial of a higher order. This is written as:

 \displaystyle \begin{align} P_m &\bot \{P_j,j=0,1,2,\cdots,n\}\forall m \ge n+1\\ P_m &\bot \mathcal P_n \forall m \ge n+1 \end{align} (9.2.8)

The boundary condition of the problem, given in Eq. 9.2.1, is a 4th order polynomial; thus, when computing the coefficients of the Fourier-Legendre expansion in Eq. 9.2.2 according to the formula

 $\displaystyle A_j = \frac{\langle P_j,f(\mu) \rangle}{\langle P_j,P_j \rangle}$ (9.2.9)

the numerator on right hand side will be zero for j>4 due to the orthogonality of the polynomials (the inner product of orthogonal functions is zero). Thus,

 $\displaystyle \langle P_j,f(\mu) \rangle = 0, \forall j \ge 5$ (9.2.10)

It should also be noted that the boundary condition $f(\mu)$ is an even function whereas Legendre polynomials of order 2k+1 for k = 0,1,2,... are odd functions; thus, since the inner product of odd and even functions are zero,

 $\displaystyle A_{2k+1} \sim \langle P_{2k+1},f(\mu) \rangle = 0 \text{ for } k=0,1,2,\cdots$ (9.2.11)

From Eq. 9.2.10 and 9.2.11, the only surviving coefficients are $A_0, A_2 \text{ and } A_4$. Thus, after evaluating these coefficients using Eq. 9.2.9, Eq. 9.2.6 becomes:

 \displaystyle \begin{align} \sum_{j=0}^\infty \bar{A_j}P_j(\mu) &= \sum_{j=0,2,4} A_jP_j(\mu) \\ &= \frac{8T_0}{15}P_0(\mu) - \frac{16T_0}{21}P_2(\mu) + \frac{8T_0}{35}P_4(\mu) \end{align} (9.2.12)

As the expansion of the left hand side of Eq. 9.2.12 is in terms of the same Legendre polynomials as the right hand side, it is clear that the coefficients must be identical. For $j\neq 0,2,4$, the values of $\bar{A_j}$ are zero and for $j=0,2,4$ the coefficients must be the same as $A_0, A_2, \text{and} A_4$ given in the class notes. Thus,

 $\displaystyle \bar{A_j}=A_j,\forall j$ (9.2.13)

# R*9.3 Identities of Legendre Equation

## Given

We are given the 1st and 2nd homogeneous solutions of the Legendre Equation from p.41-4, 41-5, 7-1, and 37-4.

$\{P_j, j=0,1,2,...\}$

$\{Q_j, j=0,1,2,...\}$

## Find

Derive the following identities for the 2nd homogeneous solutions of the Legendre Equation.

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

$Q_1(x) = \frac{x}{2} \log\left(\frac{1+x}{1-x} \right) - 1 = x \tanh^{-1}x - 1$

Also find the expression for the $Q_2(x)$ in terms of the inverse hyperbolic tangent of x

$Q_2(x) = \frac{1}{4}(3x^2 - 1) \log \left(\frac{1+x}{1-x} \right) - \frac{3}{2}x$
From lecture notes Mtg 43b

## Solution

 Solved on our own

The Legendre differential equation was defined in lecture page 7-1 for EGM 6321.

$L_2(y) := (1-x^2)y'' - 2xy' + n(n+1)y = 0$

King (2003) very clearly derives the solutions for these equations in pages 5-6 and 31-33. First assert the method of Frobenius. Choose a power series solution of the form below.

$y = \sum_{i=0}^{\infty}a_i x^{i+c}$

Substitute the term above into Legendre's equation to get a series representation.

$(1-x^2)\sum_{i=0}^{\infty}a_i(i+c)(i+c-1)x^{i+c-2} - 2x \sum_{i=0}^{\infty} a_i(i+c)x^{i+c-1} + n(n+1) \sum_{i=0}^{\infty}a_ix^{i+c} = 0$

Rearrange the equation to simplify the grouping of terms below.

$\sum_{i=0}^{\infty}a_i(i+c)(i+c-1)x^{i+c-2} +\sum_{i=0}^{\infty} a_i(n(n+1) - (i + c)(i + c + 1))x^{i+c}$

Further simplification gives a form that allows derivation of homogeneous solutions.

$a_0c(c-1)x^{c-2} + a_1c(c+1)x^{c-1} + \sum_{i=2}^{\infty}(a_i(i+c)(i+c-1) + a_{i-2}(n(n+1) - (i+c-2)(i+c-1))) x^{i+c-2} = 0$

The series definitions gives an indicial equation of the form c(c-1) = 0. Apply Frobenius General Rule 1 on p. 14 of King (2003) - "If the indicial quation has two distinct roots, $c = \alpha , \beta (\alpha < \beta) ,$whose difference is an integer, and one of the coefficients of $x^k$ becomes indeterminate on putting $c = \alpha$, both solutions can be generated by putting $c = \alpha$ in the recurrence relation." Therefore, the solution of Legendre's equation is given in expanded form below.

$y = a_0[1 - \frac{n(n+1)}{2!}x^2 - {3*2 - n(n+1))(n(n+1)}{4!}x^4 + ...] + a_1[x + \frac{(2*1 - n(n+1))}{3!}x^3$

$+ \frac{(4*3 - n(n+1))(2*1 - n(n+1))}{5!}x^5 + ...]$

The simplified form of the soltuion is given below. This equation gives rise to the Legendre polynomials.

$y = AP_n(x) + BQ_n(x)$ In the solution, the Legendre polynomials are given by the following summation.

$P_n(x) = \sum_{r=0}^m(-1)^r \frac{(2n-r)!x^{n-24}}{x^nr!(n-r)!(n-24)!}$

The first three terms are listed as $P_0, P_1, and P_2$.

$P_0(x) = 1, P_1(x) = x, P_2(x) = \frac{1}{2} (3x^2 - 1)$

These terms provide the basis for finding the Legendre functions: $Q_0(x), Q_1(x), and Q_2(x)$. The key is to utilize the reduction of order formula. For a homogeneous differential equation, we have the definition below.

$y'' + a_1(x)y' + a_0(x)y = 0$

If a solution $u_1(x)$ exists, then a 2nd solution is given by this integral.

$u_2(x) = u_1(x)\int^x \frac{1}{u_1^2(t)} \exp{-\int^t a_1(s)ds} dt$

If we apply this equation to Legendre polynomials, then the following form is given for the differential equation.

$y'' - \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0$

The astute mathematician must note that the integral for the inner integral must be derived first in the reduction of order formula.

$\int^t a_1(s)ds = \int^t \frac{2s}{1-s^2} ds = \log(1 - t^2)$

So the Legendre function can be derived by using substitution variables of t. In general, the equation below gives the nth Legendre function for the nth Legendre polynomial.

$Q_n(x) = P_n(x)\int^x{\frac{1}{{P_n(t)}^2}} \exp{-\log{(1-t^2)}} dt$

For n=0, the following derivation occurs for $Q_0(x)$.

\begin{align} Q_0(x) & = P_0(x)\int^x {\frac{1}{{P_0(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = \int^x{\frac{dt}{(1-t^2)}} \\ & = \int^x{(\frac{1}{2(1+t)} + \frac{1}{2(1-t)})}dt \\ & = [\frac{1}{2}\log{\frac{1+t}{1-t}}]^x \\ & = \frac{1}{2} \log{\frac{(1+x)}{(1-x)}} \end{align}

Similarly, for n=1, $Q_1(x)$ can be derived.

\begin{align} Q_1(x) & = P_1(x)\int^x {\frac{1}{{P_1(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = x\int^x{\frac{dt}{t^2(1-t^2)}} \\ & = x\int^x{(\frac{1}{t^2} + \frac{1}{2(1+t)} + \frac{1}{2(1-t)})}dt \\ & = x[-\frac{1}{t} + \frac{1}{2}\log{\frac{1+t}{1-t}}]^x \\ & = x\frac{1}{2} \log{\frac{(1+x)}{(1-x)}} - 1 \end{align}

Lastly, for n=2, the integration gives $Q_2(x)$.

\begin{align} Q_2(x) & = P_2(x)\int^x {\frac{1}{{P_2(x)}^2}} \exp{-\log(1-t^2)} dt \\ & = \frac{1}{4}(3x^2-1) \log{\frac{1+x}{1-x}} - \frac{3}{2}x \end{align}

Note that in each Legendre function there exists a term that can be simplified into the inverse hyperbolic tangent. Apply the definition from Wolfram for inverse hyperbolic tangent at http://mathworld.wolfram.com/InverseHyperbolicTangent.html.

\begin{align} \frac{1}{2} \log{(1+x)}{(1-x)} & = \frac{1}{2} \log{(1+x)} - \log{(1-x)} \\ & = \frac{1}{2} \ln{(1+x)} - \ln{(1-x)} \\ & = \tanh^{-1}{\frac{(1+x)}{(1-x)}} \end{align}

Therefore, the following definitions are derived for 9.3.

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

$Q_1(x) = x\frac{1}{2} \log{\frac{(1+x)}{(1-x)}} - 1 = x\tanh^{-1}{\frac{(1+x)}{(1-x)}}-1$

$Q_2(x) = \frac{1}{4}(3x^2-1) \log{\frac{1+x}{1-x}} - \frac{3}{2}x = \frac{1}{2}(3x^2-1) \tanh^{-1}{\frac{1+x}{1-x}} - \frac{3}{2}x$

Note 4 tilde command by Manuel Steele below.

Egm6322.s12.team2.steele.m2 08:24, 10 February 2012 (UTC)

# R 9.4 -Verify the second homogeneous solutions of Legendre equation

## Given

Given the following equations $\displaystyle Q_{n}(x)=P_{n}(x)\tanh^{-1}x-2\sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)$

(9.4.1)

$\displaystyle J\,:=1+2\left[\frac{n-1}{2}\right]=1+2\mathrm{Int}(\frac{n-1}{2})$

(9.4.2)

## Find

Verify the expressions for $\displaystyle Q_{0},\, Q_{1},\, Q_{2}$ as shown on the following equations

$\displaystyle Q_{0}(x)=\frac{1}{2}\log(\frac{1+x}{1-x})=\tanh^{-1}x$

(9.4.3)

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

(9.4.4)

$\displaystyle Q_{2}(x)=\frac{1}{2}(3x^{2}-1)\log(\frac{1+x}{1-x})-\frac{3}{2}x= ?$

(9.4.5)

## Solution

 Solved on my own

The Legendre polynomial is as follows:

$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)!$

(9.4.6)

where $\displaystyle m = \frac{n}{2}$.

Thus

$\displaystyle {P}_{-1}(x) = 0$

$\displaystyle {P}_{0}(x) = 1$

$\displaystyle {P}_{1}(x) = x$

$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1)$

Then from 9.4.1 we have

$\displaystyle Q_{0}(x)=\underbrace{P_{0}(x)}_{=1}\tanh^{-1}x-2\left(\frac{-2+1}{-1+1}\right)^{\dagger}\underbrace{P_{0-1}(x)}_{=0}=\tanh^{-1}x$

$\displaystyle Q_{1}(x)=\underbrace{P_{1}(x)}_{=x}\tanh^{-1}x-2\frac{2-2+1}{2-1+1}\underbrace{P_{1-1}(x)}_{=1}=x\tanh^{-1}x-1$

$\displaystyle Q_{2}(x)=\underbrace{P_{2}(x)}_{(3x^{2}-1)/2}\tanh^{-1}x-2\sum_{j=1,3}^{3}\frac{4-2j+1}{(4-j+1)j}P_{2-j}(x)$

$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-2\left(\frac{3}{4}\underbrace{P_{1}(x)}_{=x}+\frac{4-6+1}{(4-3+1)\cdot3}\underbrace{P_{-1}(x)}_{=0}\right)$

$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-\frac{3}{2}x$

$\displaystyle \dagger$

Although $\displaystyle P_{-1}(0)$, the fact that 1-1=0 makes this term meaningless. We'd better define $\displaystyle \sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}=0$, if $\displaystyle n=0$

# R 9.5 - Legendre Function

## Given

Show that

 $\displaystyle Q_n(x) = P_{n}(x)\tanh^{-1}x-2\sum_{j=1,3,5,...}^J\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)$ (9.5.1)

Show that

 $\displaystyle J := 1 + 2\left[\frac{n-1}{2}\right] = 1+2\text{Int}\left(\frac{n-1}{2}\right)$ (9.5.2)

## Find

$Q_{n}$ is odd or even depending on the index n

## Solution

 $\displaystyle \tan^{-1}(x) = \text{Odd}$ (9.5.3)

 $\displaystyle P_{2k} = \text{Even}$ (9.5.4)

 $\displaystyle P_{2k+1} = \text{Odd}$ (9.5.4)

When n=even
From Eq(9.5.1)
$\text{Even}\times\text{Odd}-\text{Odd}= \text{Odd}-\text{Odd}$
when n=Odd
$\text{Odd}\times\text{Odd} = \text{Even}-\text{Even}$
Hence,
$Q_{0},Q_{2},...,Q_{2k} = \text{Odd Function}, k=1,2,3,...$
$Q_{1},Q_{3},...,Q_{2k+1} = \text{Odd Function}, k=1,2,3,...$
Egm6322.s12.sungsik 02:22, 7 February 2012 (UTC)

# R 9.6 - Equalities in spherical coordinates

## Given

Definition of $\rho$: Boundary condition on the surface of the sphere:

 $\displaystyle \rho = \frac{r_P}{r_Q}$ (9.6.1)

Cosine law:

 $\displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2r_Pr_Q\text{cos}\gamma$ (9.6.2)

## Find

Show the following equalities:

 \displaystyle \begin{align} (r_{PQ})^2 &= \left[\left(\frac{r_P}{r_Q}\right)^2 + 1 - 2\frac{r_P}{r_Q}\text{cos}\gamma\right]\\ &= (r_Q)^2[\rho^2 + 1 -2\rho\text{cos}\gamma] \end{align} (9.6.3)
 $\displaystyle \text{cos}\gamma = \text{cos}\theta_Q\text{cos}\theta_P\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_Q\text{sin}\theta_P$ (9.6.4)

## Solution

 Solved on my own

The first inequality in Eq. 9.6.3 can be shown trivially simply by substituting the definition of $\rho$ from Eq. 9.6.1 into the equation. This immediately yields the solution.

 $\displaystyle (r_{PQ})^2 = [(\rho)^2 + 1 - 2(\rho)\text{cos}\gamma]$ (9.6.5)

Showing the equality in Eq. 9.6.4 requires somewhat more effort. It is first necessary to define the coordinates of points P and Q in spherical coordinates, as shown in the class notes:

 \displaystyle \begin{align} \vec{x_p} &= (r_P\text{cos}\theta_P\text{cos}\phi_P,r_P\text{cos}\theta_P\text{sin}\phi_P,r_P\text{sin}\theta_P)\\ \vec{x_q} &= (r_Q\text{cos}\theta_P\text{cos}\phi_Q,r_Q\text{cos}\theta_Q\text{sin}\phi_Q,r_Q\text{sin}\theta_Q) \end{align} (9.6.6)

The distance d between two points $\vec{x_1}, \vec{x_2}$ in Cartesian space is given by the three-dimensional Pythagorean theorem:

 $\displaystyle d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$ (9.6.7)

Substituting the spherical coordinates in Eq. 9.6.6 into Eq. 9.6.7 yields an expression for $r_{PQ}$:

 $\displaystyle r_{PQ} = \sqrt{A + B + C}$ (9.6.8)

where

 \displaystyle \begin{align} A &= (r_P\text{cos}\theta_P\text{cos}\phi_P - r_Q\text{cos}\theta_P\text{cos}\phi_Q)^2 = (r_P\text{cos}\theta_P\text{cos}\phi_P)^2 - 2r_Pr_Q\text{cos}\theta_P\text{cos}\phi_P\text{cos}\theta_Q\text{cos}\phi_Q + (r_Q\text{cos}\theta_Q\text{cos}\phi_Q)^2\\ B &= (r_P\text{cos}\theta_P\text{sin}\phi_P - r_Q\text{cos}\theta_P\text{sin}\phi_Q)^2 = (r_P\text{cos}\theta_P\text{sin}\phi_P)^2 - 2r_Pr_Q\text{cos}\theta_P\text{sin}\phi_P\text{cos}\theta_Q\text{sin}\phi_Q + (r_Q\text{cos}\theta_Q\text{sin}\phi_Q)^2\\ C &= (r_P\text{sin}\theta_P - r_Q\text{sin}\theta_Q)^2 = (r_P\text{sin}\theta_P)^2 - 2r_Pr_Q\text{sin}\theta_P\text{sin}\theta_Q + (r_Q\text{sin}\theta_Q)^2 \end{align} (9.6.9)

Expanding and collecting like terms yields:

 $\displaystyle r_{PQ}^2 = r_P^2(\underbrace{\text{cos}^2\theta_P[\underbrace{\text{cos}^2\phi_P+\text{sin}^2\phi_P}_{=1}] + \text{sin}^2\theta_P}_{=1}) + r_Q^2(\underbrace{\text{cos}^2\theta_Q[\underbrace{\text{cos}^2\phi_Q+\text{sin}^2\phi_Q}_{=1}] + \text{sin}^2\theta_Q}_{=1}) - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\phi_P\text{cos}\theta_Q\text{cos}\phi_Q + \text{cos}\theta_P\text{sin}\phi_P\text{cos}\theta_Q\text{sin}\phi_Q + \text{sin}\theta_P\text{sin}\theta_Q]$ (9.6.10)

Again collecting like terms in the term on the far right yields:

 $\displaystyle r_{PQ}^2 = r_P^2 + r_Q^2 - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\theta_Q(\text{cos}\phi_P\text{cos}\phi_Q + \text{sin}\phi_P\text{sin}\phi_Q) + \text{sin}\theta_P\text{sin}\theta_Q]$ (9.6.11)

Recognizing the term inside the parenthesis as the sum of cosines identity shown in Eq. 9.6.12:

 $\displaystyle \text{cos}(u\pm v) = \text{cos}u\text{cos}v\mp\text{sin}u\text{sin}v$ (9.6.12)

Eq. 9.6.11 can now be written as:

 $\displaystyle r_{PQ}^2 = r_P^2 + r_Q^2 - 2r_Pr_Q[\text{cos}\theta_P\text{cos}\theta_Q\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_P\text{sin}\theta_Q]$ (9.6.13)

Comparing this expression for $r_{PQ}$ with the one given by the law of cosines in Eq. 9.6.2 indicates that the term in the brackets must be equal to cos$\gamma$ for the equality to hold true:

 $\displaystyle \text{cos}\gamma = \text{cos}\theta_P\text{cos}\theta_Q\text{cos}(\phi_Q-\phi_P) + \text{sin}\theta_P\text{sin}\theta_Q$ (9.6.14)

# R9.7 Identities of Binomial Theorem

## Given

We are given the following Binomial Equation from 43-21.

$(1+x)^r = \sum^{\infty}_{k=0} \binom{r}{k} \, x^k, \text{for} |x| <1$

## Find

Derive the following identities for the Binomial Theorem.

$(1-x)^{-\frac{1}{2}} = \sum^{\infty}_{i=0} \alpha_i x^i$

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

From lecture notes Mtg 43c

## Solution

 Solved on our own

The first step is to write the general equation of the Binomial Theorem.

$(x+y)^n = \sum^n_{k=0} \binom{n}{k} x^{n-k}y^k$

Note that the Binomial Theorem can be extended to infinity when $n \equiv r \in \mathbb{R}$.

$(x+y)^r = \sum^{\infty}_{k=0} \binom{r}{k} x^{r-k} \, y^k$

The biniomial coefficent can be expanded as follows.

$\binom{r}{k} = \frac{r}{(r-1) \cdots(r-k+1)}{k!}$

Note that if we set y=1 then the equation can be simplified.

$(1+x)^r = \sum{\infty}_{k=0} \binom{r}{k} \, x^k, \text{for} |x| < 1$

The Binomial Equation was adapted by Isaac Newton for real numbers beyond non-negative integers. The key is to factor (n-k)! from the binomial coefficient. Then if r=-s, we get the following form of the Binomial Equation.

$\frac{1}{{(1-x)}^s} = \sum^{\infty}_{k=0} \binom{s+k-1}{k} x^k$

Note that if s = 1/2, then the equation can be further simplified.

$\frac{1}{(1-x)^{\frac{1}{2}}} = \sum^{\infty}_{k=0} \binom{k-{\frac{1}{2}}}{k} x^k$

Realize that the binomial coefficient is a fraction, so the numerator and denominator can be multiplied by 2. This will give us the form for the solution.

$\frac{1}{(1-x)^{\frac{1}{2}}} = \sum^{\infty}_{k=0} \binom{2k-1}{2k} x^k$

Therefore, the given steps derive the desired form below where we let i=k.

$(1-x)^{-\frac{1}{2}} = \sum^{\infty}_{i=0} \alpha_i x^i$

$\alpha_i = \frac{1 \cdot 3 \cdots(2i-1)}{2 \cdot 4 \cdots(2i)} = \binom{2k-1}{2k}$

Manuel Steele 4 tilde command signature

Egm6322.s12.team2.steele.m2 08:26, 10 February 2012 (UTC)

# R* 9.8 -Generating Legendre polynomials

## Given

Part 1

Recurrence relation RR2:

$\displaystyle (n+1)P_{n+1}-(2n+1) \mu P_n+nP_{n-1}=0$

(9.8.1)

with $\displaystyle P_0(\mu)=1$ and $\displaystyle P_1(\mu)= \mu$

Part 2

Define:

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

$\displaystyle -x:=-2\mu\rho+\rho^2$

(9.8.2)

and the binomial theorem:

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

(9.8.3)

$\displaystyle \alpha_{i}=\frac{1\cdot3\cdots(2i-1)}{2\cdot4\cdots(2i)}$

(9.8.4)

## Find

Finding

$\displaystyle \{P_{2},\cdots,P_{6}\}$

in terms of condition 1, and 2 respectively. and compare the result

## Solution

 Solved on my own

Part 1

From 9.8.1, we have

$\displaystyle P_{n+1}=\frac{2n+1}{n+1}\mu P_{n}-\frac{n}{n+1}P_{n-1}$

(9.8.5)

with $\displaystyle P_0(\mu)=1$ and $\displaystyle P_1(\mu)= \mu$ then

$\displaystyle \begin{array}{l} \\ P_{2}=\frac{3}{2}\mu P_{1}-\frac{1}{2}P_{0}=\frac{1}{2}(3\mu^{2}-1)\\ \\ P_{3}=\frac{5}{3}\mu P_{2}-\frac{2}{3}P_{1}=\frac{1}{2}(5\mu^{3}-3\mu)\\ \\ P_{4}=\frac{7}{4}\mu P_{3}-\frac{3}{4}P_{2}=\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\\ \\ P_{5}=\frac{9}{5}\mu P_{4}-\frac{4}{5}P_{3}=\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\\ \\ P_{6}=\frac{11}{6}\mu P_{5}-\frac{5}{6}P_{4}=\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5) \end{array}$

Part 2

Since

$\displaystyle x:=2\mu\rho-\rho^{2}=\rho(2\mu-\rho)$

From the 9.8.3 and 9.8.4 we have

$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{\infty}\alpha_{i}x^{i}=\sum_{i=0}^{\infty}\alpha_{i}\rho^{i}(2\mu-\rho)^{i}$

when computing for $\displaystyle P_{6}$ , we can see that the highest order is $\displaystyle i=6$，thus

$\displaystyle \mathcal{G}_{L}(\mu,\rho):=\sum_{i=0}^{6}\alpha_{6}x^{6}=\sum_{i=0}^{6}\alpha_{6}\rho^{6}(2\mu-\rho)^{6}$

$\displaystyle =\frac{231{\left(2\mu\rho-{\rho}^{2}\right)}^{6}}{1024}+\frac{63{\left(2\mu\rho-{\rho}^{2}\right)}^{5}}{256}+\frac{35{\left(2\mu\rho-{\rho}^{2}\right)}^{4}}{128}+\frac{5{\left(2\mu\rho-{\rho}^{2}\right)}^{3}}{16}+\frac{3{\left(2\mu\rho-{\rho}^{2}\right)}^{2}}{8}+\frac{2\mu\rho-{\rho}^{2}}{2}+1$

$\displaystyle =\frac{231\,{\rho}^{12}}{1024}-\frac{693\,\mu\,{\rho}^{11}}{256}+\frac{3465\,{\mu}^{2}\,{\rho}^{10}}{256}-\frac{63\,{\rho}^{10}}{256}-\frac{1155\,{\mu}^{3}\,{\rho}^{9}}{32}+\frac{315\,\mu\,{\rho}^{9}}{128}$

$\displaystyle +\frac{3465\,{\mu}^{4}\,{\rho}^{8}}{64}-\frac{315\,{\mu}^{2}\,{\rho}^{8}}{32}+\frac{35\,{\rho}^{8}}{128}-\frac{693\,{\mu}^{5}\,{\rho}^{7}}{16}+\frac{315\,{\mu}^{3}\,{\rho}^{7}}{16}-\frac{35\,\mu\,{\rho}^{7}}{16}$

$\displaystyle +\frac{231\,{\mu}^{6}\,{\rho}^{6}}{16}-\frac{315\,{\mu}^{4}\,{\rho}^{6}}{16}+\frac{105\,{\mu}^{2}\,{\rho}^{6}}{16}-\frac{5\,{\rho}^{6}}{16}+\frac{63\,{\mu}^{5}\,{\rho}^{5}}{8}-\frac{35\,{\mu}^{3}\,{\rho}^{5}}{4}+\frac{15\,\mu\,{\rho}^{5}}{8}$

$\displaystyle +\frac{35\,{\mu}^{4}\,{\rho}^{4}}{8}-\frac{15\,{\mu}^{2}\,{\rho}^{4}}{4}+\frac{3\,{\rho}^{4}}{8}+\frac{5\,{\mu}^{3}\,{\rho}^{3}}{2}-\frac{3\,\mu\,{\rho}^{3}}{2}+\frac{3\,{\mu}^{2}\,{\rho}^{2}}{2}-\frac{{\rho}^{2}}{2}+\mu\,\rho+1$

Since the highest order we consider is 6，then

$\displaystyle \mathcal{G}_{L}(\mu,\rho):=1+\mu\rho+\frac{1}{2}(3\mu^{2}-1)\rho^{2}+\frac{1}{2}(5\mu^{3}-3\mu)\rho^{3}+\frac{1}{8}(35\mu^{4}-30\mu^{2}+3)\rho^{4}$

$\displaystyle +\frac{1}{8}(63\mu^{5}-70\mu^{3}+15\mu)\rho^{5}+\frac{1}{16}(231\mu^{6}-315\mu^{4}+105\mu^{2}-5)\rho^{6}$

Therefore

\displaystyle \begin{align} P_0 &= 1\\ P_1 &= \mu \\ P_2 &= \frac{1}{2}(3\mu^2-1) \\ P_3 &= \frac{1}{2}(5\mu^3-3\mu) \\ P_4 &= \frac{1}{8}(35\mu^4-30\mu^2+3) \\ P_5 &= \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \\ P_6 &= \frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5) \end{align}

So, using two different methods, we obtained the same result.

# Contributing Members

 Team Contribution Table Problem Number Assigned To Solved By Typed By Proofread By 9.1 sungsik sungsik sungsik Matt Shields 9.2 Matt Shields Matt Shields Matt Shields Lang Xia 9.3 Manuel Steele Manuel Steele Manuel Steele 9.4 Lang Xia Lang Xia Lang Xia Manuel Steele 9.5 sungsik sungsik sungsik Manuel Steele 9.6 Matt Shields Matt Shields Matt Shields Lang Xia 9.7 Manuel Steele Manuel Steele Manuel Steele 9.8 Lang Xia Lang Xia Lang Xia

## Members Signatures

Egm6322.s12.team2.Xia, I solved R9.4 and R*9.8 without others assistance and proofread R9.2 and R9.6. 10:55, 10 February 2012 (UTC)
Egm6322.s12.sungsik 02:07, 10 February 2012 (UTC)

# References for Report 9

King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) Mtg 43a 24 Jan 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) Mtg 43b 24 Jan 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43c) Mtg 43c 31 Jan 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 43a) Mtg 43d 31 Jan 2011.

http://en.wikipedia.org/wiki/Binomial_theorem

http://mathworld.wolfram.com/InverseHyperbolicTangent.html