User:Egm6321.f10.team4/HW7

= Problem 1 Plot Legendre Polynomials and Functions = From Meeting 37, p 37-1.

Given
The general form of the Legendre Equation:

And the solutions have the following format.

Definitions of the Legendre Polynomials:


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

Find
part.a Plot the given Legendre Polynomials and Infinite Series

With observing $$\displaystyle P_n(\mu)$$ and $$\displaystyle Q_n(\mu)$$ as $$\displaystyle \mu \rightarrow |\mu|=1 $$

part.b Plot each pairs of Legendre Polynomials and Functions

Observe even-ness and odd-ness of $$\displaystyle {P_i,Q_i} $$ where i=0 to 3 and make predictions

Solution for the part a


Solution for the part.b
Observation and prediction for Determining the function inside of the integral,



Egm6321.f10.team4.Yoon 00:00, 25 November 2010 (UTC) - Primary Author

Given
We have the following function f such that
 * {| style="width:100%" border="0"

\begin{align} f = \sum_{i}g_{i} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

Find
Prove that

if $$ \begin{align} \{g_i\} \end{align}$$ is odd, then function $$ \begin{align} f \end{align}$$ is odd.

if $$ \begin{align} \{g_i\} \end{align}$$ is even, then function $$ \begin{align} f \end{align}$$ is even.

Solution
A function $$ \begin{align} f(x) \end{align}$$ is odd, if and only if $$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$

Hence,

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(x) \end{align}$$ = $$ \begin{align} \sum_{i}-g_{i}(-x) \end{align}$$

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -\sum_{i}g_{i}(-x) \end{align}$$

when i = 3 $$ \begin{align} -\sum_{i=3}g_{i}(-x) \end{align}$$ = $$ \begin{align} -[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$

let

$$ \begin{align} -[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$

which then makes

$$ \begin{align} -f(-x) \end{align}$$ = $$ \begin{align} -\sum_{i}g_{i}(-x) \end{align}$$

then leads us to

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$ being an odd function

Now lets go to the even case:

A function $$ \begin{align} f(x) \end{align}$$ is even, if and only if $$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$

Hence,

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

when i = 3 $$ \begin{align} \sum_{i=3}g_{i}(-x) \end{align}$$ = $$ \begin{align} +[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$

let

$$ \begin{align} +[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$

which then makes

$$ \begin{align} f(-x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

then leads us to

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$ being an even function

Egm6321.f10.team4.Corella.RL 18:39, 6 December 2010 (UTC) - Primary Author

= Problem 3* - Evenness and Oddness of Legendre Polynomials = Note: * The asterisk implies that this problem has been solved or partially solved in homework assignment for this course in a previous semester. Any sources used in developing these solutions will be cited accordingly.

3.1

 * {| style="width:100%" border="0" align="left"


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


 * where [n/2] is treated like floor(n/2), meaning that only the integer portion is used and the remainder is dropped. For example, [5/2] = [2.5] = 2.

3.2

 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle q \in P_5 \Rightarrow q(x)=\sum_{i=0}^{5}c_i x^i $$
 * $$\displaystyle (Eq. 3-2) $$
 * }
 * }
 * }

where $$\displaystyle \begin{Bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{Bmatrix} = \begin{Bmatrix} 2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{Bmatrix} $$

3.1
Show that $$\displaystyle P_{2k}(x) $$ is even and that $$\displaystyle P_{2k+1}(x) $$ is odd, for $$\displaystyle k=0,1,2,\dots $$.

3.2
Find $$\displaystyle {a_i} $$ such that $$\displaystyle q(x)= \sum_{i=0}^5 a_i P_i(x) $$.

Also, plot $$\displaystyle q=\sum_i c_i x^i = \sum_i a_i P_i $$.

3.1
Let us begin by finding the first several values of $$\displaystyle P_{2k}(x) $$ and $$\displaystyle P_{2k+1}(x) $$:

Case: k=0


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle P_{2 \cdot 0}(x)=P_{n=0}(x)=\frac{1 \cdot (0-0-1)}{2^0 \cdot 1 \cdot 1} (-1)^0 x^0 =1 $$
 * $$\displaystyle (Eq. 3-3) $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle P_{2 \cdot 0+1}(x)=P_{n=1}(x)= \frac{1 \cdot (2-0-1)}{2^0 \cdot 1 \cdot (1-0)!} (-1)^0 x^1 = x $$
 * $$\displaystyle (Eq. 3-4) $$
 * }
 * }
 * }

Note that when $$\displaystyle n $$ is odd in the summation in Eq. 3-1, the maximum value for $$\displaystyle i $$ in the summation in Eq. 3-1 is the same as it is when $$\displaystyle n=n-1 $$. For example, for:


 * $$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} \underbrace{\frac{1 \cdot 3 \cdot \dots \cdot \left(2n - 2i - 1\right)}{2^i \cdot i! (n-2i)!} (-1)^i x^{n-2i}}_{ \left( \cdot \right)} $$

When n=1:
 * $$\displaystyle P_1(x) = \sum_{i=0}^{[1/2]} \left( \cdot \right) = \sum_{i=0}^{[0.5]} \left( \cdot \right) = \sum_{i=0}^{0} \left( \cdot \right) $$

Case: k=1


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle P_{2 \cdot 1}(x)=P_{n=2}(x)=\frac{1 \cdot (4-0-1)}{2^0 \cdot 1 \cdot (2-0)!} (-1)^0 x^2 + \frac{1 \cdot (4-2-1)}{2^1 \cdot 1 \cdot (2-2)!} (-1)^1 x^0 =\frac{3}{2}x^2-\frac{1}{2} $$
 * $$\displaystyle (Eq. 3-5) $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle P_{2 \cdot 1+1}(x)=P_{n=3}(x)=\frac{1 \cdot 3 \cdot 5}{2^0 \cdot 1 \cdot (3)!} (-1)^0 x^3 + \frac{1 \cdot 3}{2^1 \cdot 1 \cdot (3-2)!} (-1)^1 x^1 =\frac{5}{2}x^3-\frac{3}{2}x $$
 * $$\displaystyle (Eq. 3-6) $$
 * }
 * }
 * }

For the sake of brevity, results of several cases are tabulated below:

Table 3-1: Even and Odd Legendre Polynomials

It can readily be observed that the polynomials $$\displaystyle P_{2k}(x)$$ have only even exponents on $$\displaystyle x$$, thus $$\displaystyle P_{2k}(x)=P_{2k}(-x)$$, so these are all even functions. Likewise, the polynomials $$\displaystyle P_{2k+1}(x)$$ have only odd exponents on $$\displaystyle x$$, so $$\displaystyle P_{2k+1}(x)=-P_{2k+1}(-x)$$, and these are all odd functions.

3.2
Let us expand the equation:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle q(x)= \sum_{i=0}^5 a_i P_i(x) $$
 * $$\displaystyle = a_0 P_0(x) + a_1 P_1(x) +a_2 P_2(x) +a_3 P_3(x) +a_4 P_4(x) +a_5 P_5(x) $$
 * $$\displaystyle = a_0 P_0(x) + a_1 P_1(x) +a_2 P_2(x) +a_3 P_3(x) +a_4 P_4(x) +a_5 P_5(x) $$


 * $$\displaystyle (Eq. 3-7) $$
 * }
 * }

Substituting the Legendre Polynomials in Table 3-1:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle q(x) = a_0 (1) + a_1 (x) +a_2 \left(\frac{3}{2}x^2-\frac{1}{2} \right) +a_3 \left(\frac{5}{2}x^3-\frac{3}{2}x \right) +a_4 \left( \frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \right) +a_5 \left(\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x \right) $$
 * $$\displaystyle = \sum_{i=0}^5c_ix^i= \underbrace{\left(a_0-\frac{1}{2}a_2+\frac{3}{8}a_4 \right)}_{=c_0=2}x^0
 * $$\displaystyle = \sum_{i=0}^5c_ix^i= \underbrace{\left(a_0-\frac{1}{2}a_2+\frac{3}{8}a_4 \right)}_{=c_0=2}x^0

+ \underbrace{\left(a_1-\frac{3}{2}a_3+\frac{15}{8}a_5 \right)}_{=c_1=-5}x^1 + \underbrace{\left(\frac{3}{2}a_2-\frac{15}{4}a_4 \right)}_{=c_2=-3}x^2 + \underbrace{\left(\frac{5}{2}a_3-\frac{35}{4}a_5 \right)}_{=c_3=11}x^3 + \underbrace{\left(\frac{35}{8}a_4 \right)}_{=c_4=7}x^4 + \underbrace{\left(\frac{63}{8}a_5 \right)}_{=c_5=6}x^5 $$
 * $$\displaystyle (Eq. 3-8) $$
 * }
 * }

From Eq. 3-8 we can deduce the following:


 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_4 = \frac{56}{35}=\frac{8}{5} $$ $$\displaystyle a_3 = \frac{2}{5} \left[11+\frac{35}{4}\left(\frac{16}{21}\right) \right] =\frac{2}{5}\left(\frac{33}{3}+\frac{20}{3}\right)=\frac{106}{15} $$ $$\displaystyle a_2 = \frac{2}{3} \left[-3+\frac{5}{4}\left(\frac{8}{5} \right)\right]=\frac{2}{3}\left(3\right)=2 $$ $$\displaystyle a_1 = 5+\frac{3}{2}\left(\frac{106}{15}\right)-\frac{15}{8}\left(\frac{16}{21}\right) =5+\frac{53}{5}-\frac{10}{7}=\frac{175}{35}+\frac{371}{35}-\frac{50}{35}=\frac{496}{35} $$ $$\displaystyle a_0 =2+ \frac{1}{2}\left(2\right)-\frac{3}{8}\left(\frac{8}{5}\right) =3-\frac{3}{5}=\frac{12}{5} $$
 * $$\displaystyle a_5 = \frac{48}{63}=\frac{16}{21} $$
 * $$\displaystyle a_5 = \frac{48}{63}=\frac{16}{21} $$
 * $$\displaystyle (Eq. 3-7) $$
 * }
 * }

Thus: $$\displaystyle \begin{Bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{Bmatrix} = \begin{Bmatrix} \frac{12}{5} \\ \frac{496}{35} \\ 2 \\ \frac{106}{15} \\ \frac{8}{5} \\ \frac{16}{21} \end{Bmatrix} $$

Plots
The two definitions of q(x) are shown below and compared in plots, with both individual values and the summed values:



Figure 3-1: q(x), showing each individual $$ \displaystyle c_i x^i $$, for i=0,1,2,3,4, and 5. The sum $$ \displaystyle \sum_i c_i x^i $$ is shown in red.

Figure 3-1 was made in MATLAB using the following code:
 * clear all; close all; clc; hold on;
 * C=[2,-5,-3,11,7,6]
 * x=-1:0.1:1;
 * for i=0:5
 * plot(x,C(i+1)*x.^i)
 * end
 * plot(x,C(1)*x.^0+C(2)*x.^1+C(3)*x.^2+C(4)*x.^3+C(5)*x.^4+C(6)*x.^5,'r-')
 * xlabel('x'); ylabel('c(i)*x^i');



Figure 3-2: q(x), showing each individual $$ \displaystyle a_i P_i(x) $$, for i=0,1,2,3,4, and 5. The sum $$ \displaystyle \sum_i a_i P_i(x)$$ is shown in red.

Figure 3-2 was made in MATLAB using the following code: clear all; close all; clc; hold on; a=[12/5,496/35,2,106/15,8/5,16/21] x=-1:0.1:1; plot(x,a(1)*1) %a0*P0 plot(x,a(2)*x) %a1*P1 plot(x,a(3)*(1.5*x.^2-.5)) %a2*P2 plot(x,a(4)*(2.5*x.^3-1.5*x)) %a3*P3 plot(x,a(5)*((35/8)*x.^4-(15/4)*x.^2+3/8)) %a4*P4 plot(x,a(6)*((63/8)*x.^5-(35/4)*x.^3+(15/8)*x)) %a5*P5 % sum of the above in red: plot(x,a(1)*1+a(2)*x+a(3)*(1.5*x.^2-.5)+a(4)*(2.5*x.^3-1.5*x)+... a(5)*((35/8)*x.^4-(15/4)*x.^2+3/8)+a(6)*((63/8)*x.^5-(35/4)*x.^3+(15/8)*x),'r-') xlabel('x'); ylabel('a_i * P_i(x)');

Note that the blue curves in Figures 3-1 and 3-2 differ, because $$ \displaystyle a_i P_i(x) \neq c_ix^i $$. However, the red curves (the corresponding sums from i=0 to i=5) are identical.

Egm6321.f10.team4.ejm 15:17, 6 December 2010 (UTC) - Author

= Problem 4 =

From Meeting 38, p 37-5.

Given
And

Find
Part 1 Without calculation, find property of $$\displaystyle A_n(\mu)$$

i.e., $$\displaystyle A_{2k}=0?(\mu)$$, $$\displaystyle A_{2k+1}=0?(\mu)$$ for $$\displaystyle k=?(\mu)$$

Part 2 Evaluate non-zero coefficients

Solution for the Part 1
We know

$$\displaystyle f(\mu)$$ is an even function. i.e. for equation 4-1:

and we know $$\displaystyle P_n$$ is even for $$\displaystyle n=2k$$ as well as odd for $$\displaystyle n=2k+1$$

An integral of an even function is an odd function so for $$\displaystyle n=2k$$

and an integral of an odd function is an even function so for $$\displaystyle n=2k+1$$

Solution for the Part 2
For equation 4-1, start off by assuming

and we know

so

where

For $$\displaystyle n=0$$

using Wolfram Alpha with an input of: Integrate[(1-x^2)^3, x,0,1]

For $$\displaystyle n=2$$

using Wolfram Alpha with an input of: 5*Integrate[(1-x^2)^3*(3/2*x^2-1/2), x,0,1]

For $$\displaystyle n=4$$

using Wolfram Alpha with an input of: 9*Integrate[(1-x^2)^3*(35/8*x^4-15/4*x^2+3/8), x,0,1]

For $$\displaystyle n=6$$

using Wolfram Alpha with an input of: 13*Integrate[(1-x^2)^3*(231/16*x^6-315/16*x^4+105/16*x^2-5/16), x,0,1]

For equation 4-2, start off by assuming

so

For $$\displaystyle n=0$$

using Wolfram Alpha with an input of: Integrate[exp(-(2x/pi)^2), x,0,pi/2]

For $$\displaystyle n=2$$

using Wolfram Alpha with an input of: 5*Integrate[exp(-(2x/pi)^2)*(3/2*sin(x)^2-1/2), x,0,pi/2]

For $$\displaystyle n=4$$

using Wolfram Alpha with an input of: 9*Integrate[exp(-(2x/pi)^2)*(35/8*sin(x)^4-15/4*sin(x)^2+3/8), x,0,pi/2]

For $$\displaystyle n=6$$

using Wolfram Alpha with an input of: 13*Integrate[exp(-(2x/pi)^2)*(231/16*sin(x)^6-315/16*sin(x)^4+105/16*sin(x)^2-5/16),x,0,pi/2]

Egm6321.f10.team4.Auerbach 06:17, 7 December 2010 (UTC)Author

= Problem 5. Relationship Between Artanh(x) function And Legendre Functions $$\displaystyle {{Q}_{0}}(x),{{Q}_{1}}(x)$$ =

Given
Legendre functions are given as below,

Find
Reveal the relationship between Artanh(x) and Legendre Functions.

Solution for part.a
To find out the relationship between Legendre equation and artanh function, let's derivate artanh function from tanh fuction first.

let

Then, moving the denominator of the right hand side to the left hand side.

Multiplying e^y on the both side of the equation

Collecting the term of x yields

Deviding by (x-1) NoteSince the left land side is greater than 0, the condition of rigthand side is limited to |x|<1. Taking log on the both side of the equation,

Deviding by 2

In conclusion, artanh fuction is identical with Legendre series(Q0).

Solution for part.b
Multiplying x and substract 1 on both side of equation yield,

Egm6321.f10.team4.Yoon 16:00, 6 December 2010 (UTC) - Primary Author

= Problem 6 : Eveness and Oddness of the second solution to the Legendre Equation=

Given
The second solution to the Legendre Differential Equation

where

and

Find
Show that $$Q(x)$$ will be an even or odd function based on the integer value of $$ n $$

Solution
Noting that $$ acrtanh(x)$$ is an odd function and taking into account that $$j$$ can only take odd values we can create two tables.

The first shows that $$ Q(x)$$ is an odd function when $$n$$ takes even values



and the second shows that $$Q(x)$$ is an even function when $$n$$ takes odd values.



Egm6321.f10.team4.petralanda.n 21:06, 6 December 2010 (UTC)

= Problem 7 – Distance Between Two Points = From Meeting 40, p 40-2.

Given
To convert from rectangular to spherical coordinates (by the astronomic convention)

$$\displaystyle x_1 = x = r cos \phi cos \theta $$

$$\displaystyle x_2 = y = r sin \phi cos \theta $$

$$\displaystyle x_3 = z = r sin \theta $$

The expression for the distance between two points in space is given by

$$\displaystyle r_{PQ}^2 = \sum^{3}_{i = 1} ( x_Q^i - x_P^i )^2 $$ $$\displaystyle (Eq. 7-1) $$

Show
that the resulting expression for the distance between the points P and Q is

$$\displaystyle r_{PQ}^2 = r_Q^2 (1 + \rho^2 - 2 \rho cos \Gamma) $$ $$\displaystyle (Eq. 7-2) $$

where

$$\displaystyle \rho = \frac {r_P}{r_Q} $$

$$\displaystyle cos \Gamma = cos \theta_Q cos \theta_P cos (\phi_Q + \phi_P) + sin \theta_Q sin \theta_P $$

Solution
Starting from

$$\displaystyle r_{PQ}^2 = \sum^{3}_{i = 1} (x_Q^i - x_P^i)^2 $$ $$\displaystyle (Eq. 7-1) $$

$$\displaystyle = (x_{Q3} - x_{P3})^2 + (x_{Q2} - x_{P2})^2 + (x_{Q1} - x_{P1})^2 $$ $$\displaystyle (Eq. 7-3) $$

$$\displaystyle = (r_Q sin \theta_Q - r_P sin \theta_P)^2 + (r_Q sin \phi_Q cos \theta_Q - r_P sin\phi_P cos \theta_P)^2 + (r_Q cos \phi_Q cos \theta_Q - r_P cos \phi_P cos \theta_P)^2 $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-4) $$

$$\displaystyle = r_Q^2 sin^2 \theta_Q + r_P^2 sin^2 \theta_P - 2 r_Q r_P sin \theta_Q sin \theta_P + r_Q^2 sin^2 \phi_Q cos^2 \theta_Q + r_P^2 sin^2 \phi_P cos^2 \theta_P - 2 r_Q r_P sin \phi_Q sin \phi_P cos \theta_Q cos \theta_P + r_Q^2 cos^2 \phi_Q cos^2 \theta_Q + r_P^2 cos^2 \phi_P cos^2 \theta_P - 2 r_Q r_P cos \phi_Q cos \phi_P cos \theta_Q cos \theta_P $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-5) $$

Using the trigonometric identity

$$\displaystyle sin^2 \alpha + cos^2 \alpha = 1 $$

The expression simplifies to

$$\displaystyle = r_Q^2 + r_P^2 - 2 r_Q r_P (sin \theta_Q sin \theta_P + sin \phi_Q sin \phi_P cos \theta_Q cos \theta_P + cos \Phi_Q cos \phi_P cos \theta_Q cos \theta_P) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-6) $$

Which can be regrouped

$$\displaystyle = r_Q^2 + r_P^2 - 2 r_Q r_P (sin \theta_Q sin \theta_P + cos \theta_Q cos \theta_P (sin \phi_Q sin \phi_P + cos \phi_Q cos \phi_P)) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-7) $$

Dividing through the expression by $$\displaystyle r_{Q}^2 $$, yields

$$\displaystyle = r_Q^2 ( 1 + \frac {r_P^2}{r_Q^2} - 2 \frac {r_P}{r_Q} (sin \theta_Q sin \theta_P + cos \theta_Q cos \theta_P cos( \phi_Q - \phi_P))) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-8) $$

Lastly, using our defintions of $$\displaystyle cos \Gamma $$ and $$\displaystyle \rho $$, our expression becomes

$$\displaystyle r_{PQ}^2 = r_Q^2 ( 1 + \rho^2 - 2 \rho cos \Gamma ) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7-2) $$

50.10.0.112 07:39, 5 December 2010 (UTC)Egm6321.f10.team4.osentowski - Author

Note: While this problem was assigned during the Fall 2009 term, it should be noted that the above solution was completed without reference to previously posted solutions.

= Problem 8 - Binomial Theorem Expansion= From Meeting 40, p 40-4.

Given
We have the following equation for the binomial theorem represented as
 * {| style="width:100%" border="0"

\begin{align} (x+y)^{r} = \sum_{k=0}^{\infty }\binom{r}{k}(x^{r-k})(y^{k}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-1)
 * }
 * }

where
 * {| style="width:100%" border="0"

\begin{align} \binom{r}{k}=\frac{(r)(r-1)\cdots (r-k+1)}{k!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-2)
 * }
 * }

Find
Use Eq. 8-1 and Eq. 8-2 to obtain the following:


 * {| style="width:100%" border="0"

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{k=0}^{\infty }(\alpha _{i})(x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-3)
 * }
 * }

and


 * {| style="width:100%" border="0"

\begin{align} \alpha _{i} = \frac{1\cdot 3\cdots (2i-1)}{2\cdot 4\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * }
 * }

Solution
Comparing Eq. 8-1 to Eq. 8-3 we can suggest that:

$$\displaystyle \begin{align} x=1, y=-x, r=-\frac{1}{2}, k=i \end{align} $$

Now plug-in the numbers from Eq. 8-3 to Eq. 8-1, we have


 * {| style="width:100%" border="0"

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{i=0}^{\infty }\binom{-\frac{1}{2}}{i}(1^{-\frac{1}{2}-i})(-x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-5)
 * }
 * }

Since $$\displaystyle \begin{align} (1)^{-\frac{1}{2}}=1 \end{align} $$ we can cancel it out and rearrange it to be


 * {| style="width:100%" border="0"

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{i=0}^{\infty }\binom{-\frac{1}{2}}{i}(-1)^{i}(x)^{i} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-6)
 * }
 * }

To obtain Eq. 8-4 we must set $$\displaystyle \begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} \end{align} $$ and expand

By using Eq. 8-2 and doing the expansion we obtain


 * {| style="width:100%" border="0"

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =(-1)^{i} \frac{-\frac{1}{2}(-\frac{1}{2}-1)\cdots (-\frac{1}{2}-i+1)}{i!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-7)
 * }
 * }

Simplifying other terms we then get


 * {| style="width:100%" border="0"

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =(-1)^{i} \frac{-\frac{1}{2}(-\frac{3}{2})\cdots (\frac{1}{2}-i)}{i!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-8)
 * }
 * }

Redistributing $$\displaystyle \begin{align} (-1)^{i} \end{align} $$ and factoring out $$\displaystyle \begin{align}(-\frac{1}{2}) \end{align} $$we have our final equation


 * {| style="width:100%" border="0"

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =\frac{(1)\cdot (3)\cdots (2i-1)}{(2)\cdot (4)\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * }
 * }

We have now proven that we can derive Eq. 8-3 and Eq. 8-4 by using Eq. 8-1 and Eq. 8-2:


 * {| style="width:100%" border="0"

\begin{align} (x+y)^{r} = \sum_{k=0}^{\infty }\binom{r}{k}(x^{r-k})(y^{k})= \sum_{k=0}^{\infty }(\alpha _{i})(x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-1)
 * }
 * }

where we showed from above that it is indeed


 * {| style="width:100%" border="0"

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =\frac{(1)\cdot (3)\cdots (2i-1)}{(2)\cdot (4)\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * }
 * }

Egm6321.f10.team4.Corella.RL 18:36, 6 December 2010 (UTC) - Primary Author

= Problem 9 – Generating Polynomials = From Meeting 41, p 41-2

Given
$$\displaystyle \mu P_n' - P_{n-1}' =  n P_n $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-1) $$

$$\displaystyle (n + 1) P_{n + 1} - (2n + 1) \mu P_n + n P_{n-1} =  0 $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-2) $$

$$\displaystyle P_0 =  1 $$

$$\displaystyle P_1 =  x $$

Find
$$\displaystyle P_2 $$ through $$\displaystyle P_6 $$, and compare the results to those derived using the formula

$$\displaystyle P_n (x) = \sum^{\frac {n}{2}}_{i =0} -1^i \frac {(2n - 2i)! x^{n-2i}}{2^n i! (n-1)! (n-2i)!} $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-3) $$

from Meeting 36 p. 36-2.

Solution
First, we must find $$\displaystyle \mu $$ using Equation 9-1, and the values for $$\displaystyle P_0 $$ and $$\displaystyle P_1 $$

$$\displaystyle \mu P_1' - P_0' = 1 P_1 $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-1) $$

$$\displaystyle \mu x' - 1' = x $$

$$\displaystyle \mu = x $$

Using this value for \mu, and Equation 9-2, the polynomials can be obtained.

$$\displaystyle P_2 $$ is found using $$\displaystyle n =  1 $$

$$\displaystyle (1+1) P_2 - (2*1 + 1) x P_1 + P_0 = 0 $$

$$\displaystyle 2 P_2 - 3 x (x) + 1 = 0 $$

$$\displaystyle 2 P_2 = 3 x^2 - 1 $$

$$\displaystyle P_2 = \frac {1}{2} (3x^2 - 1) $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-4) $$

$$\displaystyle P_3 $$ is found using $$\displaystyle n = 2 $$

$$\displaystyle (2 + 1) P_3 - (2*2 + 1) x P_2 + 2 P_1 = 0 $$

$$\displaystyle 3 P_3 - 5 x (\frac {1}{2} (3x^2 - 1)) + 2 x = 0 $$

$$\displaystyle 3 P_3 = 5 x (\frac {1}{2} (3x^2 - 1)) - 2 x $$

$$\displaystyle 3 P_3 = \frac {15}{2} x^3 - \frac{5}{2}x - 2x $$

$$\displaystyle P_3 =  \frac {1}{2} (5 x^3 - 3 x) $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-5) $$

$$\displaystyle P_4 $$ is found using $$\displaystyle n = 3 $$

$$\displaystyle (3 + 1) P_4 - (2*3 + 1) x P_3 + 3 P_2 =  0 $$

$$\displaystyle 4 P_4 - 7 x (\frac {1}{2} (5 x^3 - 3 x)) + 3 (\frac {1}{2} (3 x^2 - 1)) =  0 $$

$$\displaystyle 4 P_4 =  \frac {7}{2} x (5 x^3 - 3x) - \frac {3}{2} (3 x^2 - 1) $$

$$\displaystyle P_4 =  \frac {1}{8} (35 x^4 - 21 x^2 - 9 x^2 + 3) $$

$$\displaystyle P_4 =  \frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8} $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-6) $$

$$\displaystyle P_5 $$ is found using $$\displaystyle n = 4 $$

$$\displaystyle (4+1) P_5 - (2*4+1) P_4 + 4 P_3 =  0 $$

$$\displaystyle 5 P_5 - 9x P_4 + 4 P_3 =  0 $$

$$\displaystyle 5 P_5 - 9x (\frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8}) + 4 (\frac {1}{2} (5 x^3 - 3 x)) =  0 $$

$$\displaystyle 5 P_5 = \frac {315}{8} x^5 - \frac {135}{4} x^3 + \frac {27}{8} x - 10 x^3 + 6x $$

$$\displaystyle P_5 = \frac {63}{8} x^5 - \frac {35}{4} x^3 + \frac {15}{8} x $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-7) $$

$$\displaystyle P_6 $$ is found using $$\displaystyle n = 5 $$

$$\displaystyle (5 + 1) P_6 - (2*5 + 1) x P_5 - 5 P_4 =  0 $$

$$\displaystyle 6 P_6 - 11 x (\frac {63}{8} x^5 - \frac {35}{4} x^3 + \frac {15}{8} x) + 5 (\frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8}) $$

$$\displaystyle 6 P_6 =  \frac {693}{8} x^6 - \frac {385}{4} x^4 + \frac {165}{8} x^2 - \frac {175}{8} x^4 + \frac {75}{4} x^2 - \frac {15}{8} $$

$$\displaystyle 6 P_6 =  \frac {693}{8} x^6 - \frac {945}{8} x^4 + \frac {315}{8} x^2 - \frac {15}{8} $$

$$\displaystyle P_6 =  \frac {231}{16} x^6 - \frac {315}{16} x^4 + \frac {105}{16} x^2 - \frac {5}{16} $$ <p style="text-align:right;">$$\displaystyle (Eq. 9-8) $$

The polynomials, represented in Equations 9-4 through 9-8, do indeed agree with the results presented in the Meeting 36 notes, developed using Equation 9-3. It can therefore be assumed that this is an acceptable method for generating Legendre Polynomials.

50.10.0.112 06:32, 6 December 2010 (UTC)Egm6321.f10.team4.osentowski - Author

128.227.11.220 19:00, 7 December 2010 (UTC) Egm6321.f10.team4.petralanda.n - Editor

= Problem 10: First solution to the Legendre Equation using Polynomial Expansion and Recurrence Method =

Find
$$P_2(x)$$ through $$P_6(x)$$ and compare the values with those obtained in Problem 7.9 using the second recurrence relation.

Solution
First we start by computing $$ P_5(x)$$ and $$P_6(x)$$.

$$n=5$$ and Eq (10-1) takes the following form


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle
 * $$\displaystyle

P_{5}(x)=\sum\limits_{i=0}^{5/2}{\frac{(-1)^{i}(2(5)-2i)!x^{5-2i}}{2^{5}i!(5-i)!(5-2i)!}}=\frac{(2(5))!x^{5}}{2^{5}(5)!(5)!}+\frac{(-1)^{1}(2(5)-2)!x^{5-2}}{2^{5}1!(5-1)!(5-2)!}+\frac{(-1)^{2}(2(5)-4)!x^{5-4}}{2^{5}2!(5-2)!(5-4)!}

$$


 * }
 * }
 * }

Which will yield


 * {| style="width:100%" border="0" align="left"

P_{5}(x)=\frac{63}{8}x^{5}-\frac{35}{4}x^{3}+\frac{15}{8}x $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10-2) $$
 * }
 * }

$$n=6$$ and Eq (10-1) will take the following form


 * {| style="width:100%" border="0" align="left"

P_{6}(x)=\sum\limits_{i=0}^{6/2}{\frac{(-1)^{i}(2(6)-2i)!x^{6-2i}}{2^{6}i!(6-i)!(6-2i)!}}=\frac{(2(6))!x^{6}}{2^{6}(6)!(6)!}-\frac{(2(6)-2)!x^{6-2}}{2^{6}(6-1)!(6-2)!}+\frac{(2(6)-4)!x^{6-4}}{2^{6}2!(6-2)!(6-4)!}-\frac{(2(6)-6)!}{2^{6}3!(6-3)!} $$
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }
 * }

Which will result in


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle
 * $$\displaystyle

P_{6}(x)=\frac{231}{16}x^{6}-\frac{315}{16}x^{4}+\frac{105}{16}x^{2}-\frac{5}{16}$$


 * <p style="text-align:right;">$$\displaystyle (Eq. 10-3) $$
 * }
 * }

The following table compiles all the values generated using the polynomial expansion and the recurrence method.



Egm6321.f10.team4.petralanda.n 21:10, 6 December 2010 (UTC)