User:Egm6321.f10.team5/hw7

=Problem 1 - Plot and Observe = From (meeting 37 page 1)

Given
Legendre equation $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$

Find
Plot {$$P_0$$, $$P_1$$, $$P_2$$, $$P_3$$} and {$$Q_0$$, $$Q_1$$, $$Q_2$$, $$Q_3$$}. Observe $$P_n(\mu)$$ and $$Q_n(\mu)$$ as $$\mu \rightarrow \pm 1$$. Then plot {$$P_0$$, $$Q_0$$}, {$$P_1$$, $$Q_1$$}, {$$P_2$$, $$Q_2$$}, {$$P_3$$, $$Q_3$$}, and {$$P_4$$, $$Q_4$$}. Observe evenness and oddness of {$$P_i$$,$$Q_i$$} where $$i=0,\cdots,4$$ and guess $$\int \limits_{-1}^{+1} P_i(\mu) Q_i(\mu) d\mu$$.

Solution
For $$n=0$$, Legendre equation is $$(1-x^2)y'' - 2xy' = 0$$ $$y'' - \frac {2x} {1-x^2} y' = 0$$ $$P_0(x) = 1$$ $$Q_0(x) = P_0(x) \int^x \frac 1 {P_0^2(t)} \exp\left[ - \int^t - \frac {2s} {1-s^2} ds \right] dt = 1 \int^x \frac 1 {1^2} \exp[-\log(1-t^2)] dt = \int^x \frac 1 {1-t^2} dt$$ $$= \frac 1 2 [\log (-x-1) - \log(x-1)] = \frac 1 2 \log \left( \frac {-x-1} {x-1} \right) = \frac 1 2 \log \left( \frac {1+x} {1-x} \right)$$ $$P_0(x) = 1$$ and $$Q_0(x)=\pm \infty$$ as $$x \rightarrow \pm 1$$ $$\int \limits_{-1}^{+1} P_0(x) Q_0(x) dx \approx 0$$ by guess.

For $$n=1$$, Legendre equation is $$(1-x^2)y'' - 2xy' + 2y = 0$$ $$y'' - \frac {2x} {1-x^2} y' + \frac 2 {1-x^2} y = 0$$ $$P_1(x) = x$$ $$Q_1(x) = P_1(x) \int^x \frac 1 {P_1^2(t)} \exp\left[ - \int^t - \frac {2s} {1-s^2} ds \right] dt = x \int^x \frac 1 {t^2} \exp[-\log(1-t^2)] dt = x \int^x \frac 1 {t^2} \frac 1 {1-t^2} dt$$ $$= \frac x 2 \log \left( \frac {1+x} {1-x} \right) - 1$$ $$P_1(x) = \pm 1$$ and $$Q_1(x)=\infty$$ as $$x \rightarrow \pm 1$$ $$\int \limits_{-1}^{+1} P_1(x) Q_1(x) dx \approx 0$$ by guess.

For $$n=2$$, Legendre equation is $$(1-x^2)y'' - 2xy' + 6y = 0$$ $$y'' - \frac {2x} {1-x^2} y' + \frac 6 {1-x^2} y = 0$$ $$P_2(x) = \frac 1 2 (3x^2-1)$$ $$Q_2(x) = P_2(x) \int^x \frac 1 {P_2^2(t)} \exp\left[ - \int^t - \frac {2s} {1-s^2} ds \right] dt = \frac 1 2 (3x^2-1) \int^x \frac 1 {[\frac 1 2 (3t^2-1)]^2} \exp[-\log(1-t^2)] dt $$ $$= \frac 1 2 (3x^2-1) \int^x \frac 1 {[\frac 1 2 (3t^2-1)]^2} \frac 1 {1-t^2} dt = \frac 1 2 (3 x^2 - 1) \frac 1 2 \left( \log \frac {1+x} {1-x} - \frac {6 x} {3x^2-1} \right) = \frac 1 4 (3 x^2 - 1) \log \frac {1+x} {1-x} - \frac 3 2 x$$ $$P_2(x) = 1$$ and $$Q_2(x)=?$$ as $$x \rightarrow \pm 1$$ $$\int \limits_{-1}^{+1} P_2(x) Q_2(x) dx \approx -1$$ by guess.

For $$n=3$$, Legendre equation is $$(1-x^2)y'' - 2xy' + 12y = 0$$ $$y'' - \frac {2x} {1-x^2} y' + \frac {12} {1-x^2} y = 0$$ $$P_3(x) = \frac 1 2 (5x^3-3x)$$ $$Q_3(x) = P_3(x) \int^x \frac 1 {P_3^2(t)} \exp\left[ - \int^t - \frac {2s} {1-s^2} ds \right] dt = \frac 1 2 (5x^3-3x) \int^x \frac 1 {[\frac 1 2 (5t^3-3t)]^2} \exp[-\log(1-t^2)] dt $$ $$= \frac 1 2 (5x^3-3x) \int^x \frac 1 {[\frac 1 2 (5t^3-3t)]^2} \frac 1 {1-t^2} dt = \frac 1 {36} (5x^3-3x) \left[ \frac {50x} {3-5x^2} - \frac 8 x + 9 \log (-x-1) - 9 \log(x-1) \right]$$ $$P_3(x) = \pm 1$$ and $$Q_3(x)=\infty$$ as $$x \rightarrow \pm 1$$ $$\int \limits_{-1}^{+1} P_3(x) Q_3(x) dx \approx 0$$ by guess.

For $$n=4$$, Legendre equation is $$(1-x^2)y'' - 2xy' + 20y = 0$$ $$y'' - \frac {2x} {1-x^2} y' + \frac {20} {1-x^2} y = 0$$ $$P_4(x) = \frac {35} 8 x^4 - \frac {15} 4 x^2 + \frac 3 8$$ $$Q_4(x) = P_4(x) \int^x \frac 1 {P_4^2(t)} \exp\left[ - \int^t - \frac {2s} {1-s^2} ds \right] dt = \left(\frac {35} 8 x^4 - \frac {15} 4 x^2 + \frac 3 8\right) \int^x \frac 1 {\left(\frac {35} 8 t^4 - \frac {15} 4 t^2 + \frac 3 8 \right)^2} \exp[-\log(1-t^2)] dt $$ $$= \left(\frac {35} 8 x^4 - \frac {15} 4 x^2 + \frac 3 8\right) \int^x \frac 1 {\left(\frac {35} 8 t^4 - \frac {15} 4 t^2 + \frac 3 8 \right)^2} \frac 1 {1-t^2} dt = \frac 1 6 \left(\frac {35} 8 x^4 - \frac {15} 4 x^2 + \frac 3 8\right) \left[3 \log \left(\frac {1+x} {1-x}\right) - \frac {10x (21x^2-11)} {35x^4-30x^2+3} \right]$$ $$P_4(x) = 1$$ and $$Q_4(x)=?$$ as $$x \rightarrow \pm 1$$ $$\int \limits_{-1}^{+1} P_0(x) Q_0(x) dx \approx -1$$ by guess.

The plot of {$$P_0$$, $$P_1$$, $$P_2$$, $$P_3$$, $$P_4$$} is given below. The plot of {$$Q_0$$, $$Q_1$$, $$Q_2$$, $$Q_3$$, $$Q_4$$} is given below.

Author and proof-reader
[Author] Jo

[Proof-reader] Mike Faraone

=Problem 2 - Sums of Even and Odd Functions= From (meeting 38 page 1)

Given

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f = \sum_i g_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (2-1)
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Find
Show that: 1. If $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd.

2. If $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
 We referred and used mediawiki code of f09.Team1 but we discussed and improved about not only basic properties of even and odd functions but verification of $$\displaystyle  = 0 $$ using this problem also.

[Background Knowledge] Even and odd functions

With above background knowledge, we can verify straightly forward.

'''Part 1. If $$\displaystyle \{g_i\}$$ is odd'''

According to the character of an odd function, such that $$\displaystyle f(x) = -f(-x)$$


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$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$
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'''Part 2. If $$\displaystyle \{g_i\}$$ is even'''

According to the character of an even function, such that $$\displaystyle f(x) = f(-x)$$


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$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$
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[Improvement] : We discussed about verification of $$\displaystyle  = 0$$

Basic properties of even and odd functions

1. The sum of two even functions is even, and any constant multiple of an even function is even. 2. The sum of two odd functions is odd, and any constant multiple of an odd function is odd. 3. The product of two even functions is an even function. 4. The product of two odd functions is an even function. 5. The product of an even function and an odd function is an odd function. 6. The integral of an odd function from −A to +A is zero (where A is finite, and the function has no vertical asymptotes between −A and A). 7. The integral of an even function from −A to +A is twice the integral from 0 to +A (where A is finite, and the function has no vertical asymptotes between −A and A).

 According to 5 and 6 of basic properties, 


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 = \int_{-1}^{+1} \underbrace{P_n(x)}_{even \ func.} \cdot \underbrace{Q_n(x)}_{odd \ func.} dx \ = \ \int_{-1}^{+1} (odd \ func.(x)) dx \ = \ 0, $$
 * $$\displaystyle
 * $$\displaystyle
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Author and proof-reader
[Author] Oh, Sang Min

[Proof-reader]

=Problem 3 - 2 questions= From (meeting 38 page 1 and 3)

Given
'''Part 1. Legendre Polynomials' Evenness and Oddness'''


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


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

'''Part 2. Find $$ \{a_i\} $$ and plot $$ q $$'''
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q(x) = \sum_{i=0}^5 c_i x^i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (3-2-1)
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with the coefficients
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c_0 = 2, \quad c_1 = -5, \quad c_2 = -3, \quad c_3 = 11, \quad c_4 = 7, \quad c_5 = 6 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (3-2-2)
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Find
'''Part 1. Legendre Polynomials' Evenness and Oddness'''

1. Show $$ P_{2k}(x)\qquad \qquad $$     is even k=0,1,2...

2. Show $$ P_{2k+1}(x) \qquad \qquad $$    is odd

'''Part 2. Find $$ \{a_i\} $$ and plot $$ q $$'''

1. Find $$ \{a_i\} $$

2. Plot $$ q $$

Solution
 We referred and used mediawiki code of f09.Team5 but we solved again by ourselves .

Part 1. Legendre Polynomials' Evenness and Oddness
1. In case of $$\displaystyle n=2k$$


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$$
 * $$ P_{2k}(x) = \sum_{i=0}^{[k]}\frac{(-1)^{i} \cdot (4k-2i)! \cdot \color{blue}{(x)^{2k-2i}}}{2^{2k} \cdot i! \cdot (2k-i)! \cdot (2k-2i)!} $$
 * $$\displaystyle (3-1-3)
 * $$\displaystyle (3-1-3)
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$$
 * $$ P_{2k}(-x) = \sum_{i=0}^{[k]}\frac{(-1)^{i} \cdot (4k-2i)! \cdot \underbrace{\color{blue}{(-x)^{2k-2i}}}_{(x)^{2k-2i}}}{2^{2k} \cdot i! \cdot (2k-i)! \cdot (2k-2i)!} $$
 * $$\displaystyle (3-1-4)
 * $$\displaystyle (3-1-4)
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According to the character of an even function, such that $$\displaystyle f(x) = f(-x)$$

$$\displaystyle (3-1-3), \ (3-1-4)$$ -->


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$$\displaystyle P_{2k}(x) =P_{2k}(-x) $$ Therefore, clearly $$\displaystyle P_{2k}(x) $$ is even
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2. In case of $$\displaystyle n=2k+1$$


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$$
 * $$ P_{2k+1}(x) = \sum_{i=0}^{[\frac{2k+1}{2}]}\frac{(-1)^{i} \cdot (4k+2-2i)! \cdot \color{red}{(x)^{2k+1-2i}}}{2^{2k+1} \cdot i! \cdot (2k+1-i)! \cdot (2k+1-2i)!} $$
 * $$\displaystyle (3-1-5)
 * $$\displaystyle (3-1-5)
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$$
 * $$ P_{2k+1}(-x) = \sum_{i=0}^{[\frac{2k+1}{2}]}\frac{(-1)^{i} \cdot (4k+2-2i)! \cdot \underbrace{\color{red}{(-x)^{2k+1-2i}}}_{-(x)^{2k+1-2i}}}{2^{2k+1} \cdot i! \cdot (2k+1-i)! \cdot (2k+1-2i)!} $$
 * $$\displaystyle (3-1-6)
 * $$\displaystyle (3-1-6)
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 * }


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 * }
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According to the character of an even function, such that $$\displaystyle f(x) = -f(-x)$$

$$\displaystyle (3-1-5), \ (3-1-6)$$ -->


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$$\displaystyle P_{2k+1}(x) = \ - \ P_{2k+1}(-x) $$ Therefore, clearly $$\displaystyle P_{2k+1}(x) $$ is odd
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Part 2. Find $$ \{a_i\} $$ and plot $$ q $$
1. Find $$ \{a_i\} $$

From HW6 problem 8


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P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-3)
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P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-4)
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 * }


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P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-5)
 * }
 * }


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P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-6)
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P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-7)
 * }
 * }


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P_5(x) = \frac{63}{8} x ^ 5 - \frac{35}{4} x ^ 3 + \frac{15}{8}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3-2-8)
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q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) + a_{5}\left(\frac{63}{8} x ^ 5 - \frac{35}{4} x ^ 3 + \frac{15}{8}x\right)\\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-\frac{3}{2}a_{3} + \frac{15}{8}a_5\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3} - \frac{35}{4}a_5\right) + x^{4}\left(\frac{35}{8}a_{4}\right) + x^{5}\left(\frac{63}{8}a_5\right) \end{align} $$ $$
 * $$ \begin{align}
 * $$ \begin{align}
 * <p style="text-align:right;">$$\displaystyle (3-2-9)
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 * $$ \begin{align}

\sum_{i=0}^{5}\ a_{i} &P_{i} = \sum_{i=0}^{5}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 2 \\ &\text{(2) }a_{1}-\frac{3}{2}a_{3} + \frac{15}{8}a_5 &= -5 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= -3 \\ &\text{(4) }\frac{5}{2}a_{3} - \frac{35}{4}a_5 &= 11 \\ &\text{(5) }\frac{35}{8}a_{4} &= 7 \\ &\text{(6) }\frac{63}{8}a_{5} &= 6 \end{align} $$

Clearly,

$$\displaystyle a_5 = \frac{16}{21}, \ a_4 = \frac{8}{5} $$

$$\displaystyle \frac{5}{2}a_{3} = 11 + \frac{35}{4}a_5 $$

$$\displaystyle a_{3} = (11 + \frac{35}{4} \cdot \frac{16}{21}) \cdot \frac{2}{5} \ = \ \frac{106}{15}$$

$$\displaystyle \frac{3}{2}a_{2} = -3 + \frac{15}{4}a_{4} $$

$$\displaystyle a_{2} = (-3 + 6) \cdot \frac{2}{3} \ = \ 2$$

$$\displaystyle a_{1} = -5 + \frac{3}{2}a_{3} - \frac{15}{8}a_5 $$

$$\displaystyle a_{1} = -5 + \frac{3}{2} \cdot \frac{106}{5} - \frac{10}{7} \ = \ \frac{496}{35}$$

$$\displaystyle a_{0} = 2 + \frac{1}{2}a_{2} + \frac{3}{8}a_{4} $$

$$\displaystyle a_{0} = 2 + 1 - \frac{3}{5} \ = \ \frac{12}{5}$$

Therefore


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$$\displaystyle a_{0} = \frac{12}{5}, \ a_{1} = \frac{496}{35}, \ a_{2} = 2, \ a_{3} = \frac{106}{15}, \ a_4 = \frac{8}{5}, \ a_5 = \frac{16}{21} $$
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 2. Plot $$\displaystyle q$$

Author and proof-reader
[Author]
 * Oh, Sang Min

[Proof-reader]
 * Raul Riveros

=Problem 4 - Find property and evaluate= From (meeting 38 page 5)

Given
Consider

a) $$ f(\theta) = T_0cos^6\theta \! $$

b) $$ f(\theta) = T_0exp \Bigg[\frac{2\theta}{\pi}\Bigg] $$

Find
1) Without calculation, find property of $$ A_n \! $$ i.e. $$ A_{2k} = ? \, A_{2k+1} = ? \mbox{ for } k = ? $$.

2) Evaluate non-zero coefficients (use WA).

Part 1
The Legendre polynomials $$ P_n \! $$ are even when n is even and odd when n is odd (see Problem 7.3 above). That is to say that

$$ P_{2k} \! $$ is EVEN

and

$$ P_{2k+1} \! $$ is ODD for $$ k = 0, 1, 2, 3 ... $$

Mathematically, the product of 2 odd functions is an even function, the product of 2 even functions is an even function and the product of an even function with an odd function is an odd function.

If the integrand of $$ A_n $$ is odd, the integral will be zero since our limits of integration equally straddle $$ \mu = 0 $$. However, if the integrand is even, $$ A_n $$ will be non-zero. We have already examined the even and odd properties of the Legendre polynomials $$ P_n $$. Thus, we must now look at $$ f(\mu) $$ to determine the even and odd properties of $$ A_n $$.

Part 1a
where we recall that $$ \mu := sin(\theta) \! $$

Equation (4.3) is a 6th order polynomial in $$ \mu $$, therefore it is an even function.

Therefore, with this $$ f(\mu) $$ we may conclude the following:

The reason k stops at 3 is that for $$ n \ge 7 $$, $$ A_n $$ is zero, since all Legendre polynomials at or beyond that order are orthogonal to $$ f $$.

Part 1b
For this part, we will examine $$ A_n $$ as a function of $$ \theta $$ since we cannot easily make f an explicit function of $$ \mu $$.

By merely looking at the behavior of $$ f(\theta) $$ at the limits of integration and at zero, it is apparent that it is an even function.

Therefore, we observe that

Part 2a
We now evaluate the non-zero values of $$ A_n $$ using Wolfram Alpha.

Part 2b
We now evaluate the few non-zero values of $$ A_n $$ using Wolfram Alpha.

Author and proof-reader
[Author] Michael Steinberg

[Proof-reader]

=Problem 5 - Second Solution to the 0th Legendre Polynomial= From (meeting 39 page 1)

Given
Part 1


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P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5-1)
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\ Q_0(x) = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) = \tanh^{-1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5-2)
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Part 2


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P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5-3)
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\ Q_1(x) = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) - 1 = x \tanh^{-1}(x) - 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5-4)
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Find
'''Part 1. Verify $$\displaystyle (5-2)$$'''

'''Part 2. Verify $$\displaystyle (5-4)$$'''

Solution
 We referred and used mediawiki code of f09.Team1 but we solved again by ourselves then discussed and found about Inverse and Hyperbolic functions and relationship with lecture.

Part 1. Verification of $$\displaystyle (5-2) $$
[Background Knowledge] Inverse function Hyperbolic function

According to above background knowledge,


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$$
 * $$\displaystyle \tanh(Q_0) = x $$
 * <p style="text-align:right;">$$\displaystyle (5-5)
 * <p style="text-align:right;">$$\displaystyle (5-5)
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$$
 * $$\displaystyle \tanh(x) = \frac{e^{2x} - 1}{e^{2x} + 1} $$
 * <p style="text-align:right;">$$\displaystyle (5-6)
 * <p style="text-align:right;">$$\displaystyle (5-6)
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Using $$\displaystyle (5-5) $$ and $$\displaystyle (5-6) $$
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\tanh(Q_0) = \frac{e^{2Q_0} - 1}{e^{2Q_0} + 1} $$
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\ = \ \frac{e^{2 \cdot \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} - 1}{e^{2 \cdot \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} + 1}
 * $$\displaystyle
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\ = \ \frac{\frac{1+x}{1-x} - 1}{\frac{1+x}{1-x} +1}

\ = \ \frac{\frac{1+x}{1-x} - \frac{1-x}{1-x}}{\frac{1+x}{1-x} + \frac{1-x}{1-x}}

\ = \ \frac{\frac{2x}{1-x}}{\frac{2}{1-x}} $$
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$$\displaystyle \ = \ x $$
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'''Therefore, $$\displaystyle \ Q_0(x) \ = \ tanh^{-1}(x) $$'''
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Part 2. Verification of $$\displaystyle (5-4) $$
Using $$\displaystyle \ (1) \ p.39-1$$ such that


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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), $$ $$
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In this case, $$ n=1, j=1$$


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Q_1(x) = P_1(x) \tanh^{-1}(x) - 2 \sum_{j=1}^{1} \frac{2-2+1}{(2-1+1)} P_{1-1}(x) \ = \ x \tanh^{-1}(x) - 2 \cdot \frac{1}{2} $$
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\ = \ x \tanh^{-1}(x) - 1 $$
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'''Therefore, $$\displaystyle Q_1(x) \ = \ x \tanh^{-1}(x) - 1 $$ '''
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Author and proof-reader
[Author] Oh, Sang Min

[Proof-reader]

=Problem 6 - Odd and Even Solutions of Legendre= From (meeting 39 page 2)

Given
$$ Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.1) $$

$$ J = 1+2[\frac {n-1} {2}] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.2) $$

Find
Use Eq. 7.1 to show $$Q_n(x)$$ is even or odd depending on $$n$$

Solution
In order to get a preliminary solution we must review when a function is even or odd. A function is said to be odd if $$-f(x) = f(-x)$$ and even if $$f(x) = f(-x)$$.

Since Eq. 7.1 is a combination of functions we must also review the properties of what happens when even/odd functions are combined either through multiplication or summation.

Even/Odd function Basic Properties

1) The sum of a number of odd functions is odd

2) The sum of a number of even functions is even

3) The sum of an even and an odd function is not even or odd, unless one of the functions is equal to zero

4) The product of an even (or odd) function and a constant multiple is even (odd)

5) The product of two even functions is even

6) The product of two odd functions is even

7) the product of an even and odd function is odd

Reference: Even and Odd functions

Now, to use these properties in eq. 7.1 we must determine whether $$P_n$$ is even or odd based on $$n$$. To do that we simply recall that $$P_n$$ is odd if $$n$$ is odd and $$P_n$$ is even if $$n$$ is even.

Reference: Legendre Polynomials

So, we can now look at Eq. 7.1 in terms of evenness and oddness. For the first case we assume $$ n = even $$

$$ Q_n(x) = (even) tanh^{-1}(x) - 2 \sum_{j=1,3,5,..}^J \frac {2n-2j+1} {(2n-j+1)j} (odd) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.3) $$

Eq. 7.3 must be further simplified in terms of it evenness/oddness. By simple visual analysis of a graph of $$tanh^{-1}(x)$$ or by plugging in random numbers and then their negative counterparts we can see that

$$ tanh^{-1}(x) = odd \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.4) $$

Also, we can apply Basic Properties (1) and (4) above to see that the summation term is the sum of a number of odd functions after each has been multiplied by a constant. Therefore the entire second term of Eq. 7.3 is an odd function. This allows us to reduce Eq. 7.3 to

$$ Q_n(x) = (even) (odd) - (odd) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.5) $$

Then using Basic Properties (1) and (7) we see that when

$$ n = even \,$$

$$ Q_n(x) = odd \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.6) $$

Now we use the same process to determine if $$Q_n(x)$$ is even or odd if $$n$$ is odd

If $$n = odd$$, then $$P_n(x) = odd$$ and $$P_{n-j}(x) = even$$. Also keeping in mind that $$tanh^{-1}(x)$$ is always odd. So Eq. 7.1 reduces to

$$ Q_n(x) = (odd) (odd) - 2 \sum_{j=1,3,5,..}^J (constant) (even) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.7) $$

Using Basic Properties (2), (4) and (5) it is clear that when

$$ n = odd \,$$

Eq. 7.7 reduces to

$$ Q_n(x) = even \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.8) $$

To prove that $$Q_n(x)$$ is even when $$n$$ is odd and vice versa we will look at the specific examples when $$n = 1, 2, 3, 4$$

To do this we must first get the Legendre Polynomials for $$n = 0, 1, 2, 3, 4$$. The quickest way to do this is to use the Rodriquez formula.

$$ P_n(x) = \frac {1} {2^n n!} \frac {d^n} {dx^n} (x^2-1)^n \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.9) $$

Reference: Rodrigues Formula

From Eq. 7.9 we can get

$$ P_0(x) = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.10) $$

$$ P_1(x) = x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.11) $$

$$ P_2(x) = \frac {1} {2} (3x^2-1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.12) $$

$$ P_3(x) = \frac {1} {2} (5x^3-3x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.13) $$

$$ P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.14) $$

For $$n=1$$

First, we see for $$n=1$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {1-1} {2}] = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.15) $$

Then plug Eq. 7.15, 7.11 into Eq 7.1 to get

$$ Q_1(x) = (x) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^1 \frac {2(1)-2(j)+1} {(2(1)-(j)+1)(j)} P_{1-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.16) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.10 for $$P_0(x)$$ Eq. 7.16 reduces to

$$ Q_1(x) = (x) tanh^{-1}(x) - (1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.17) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_1(\frac 1 2) = -0.725347... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.18) $$

and

$$ Q_1(\frac {-1} {2}) = -0.725347... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.19) $$

Therefore we can see that $$Q_1(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=2$$

First, we see for $$n=2$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {2-1} {2}] = 2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.20) $$

Then plug Eq. 7.12, and 7.20 into Eq 7.1 to get

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.21) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.11 for $$P_1(x)$$ Eq. 7.21 reduces to

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.22) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_2(\frac 1 2) = -0.818663... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.23) $$

and

$$ Q_2(\frac {-1} {2}) = 0.818663... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.24) $$

Therefore we can see that $$Q_2(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

For $$n=3$$

First, we see for $$n=3$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {3-1} {2}] = 3 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.25) $$

Then plug Eq. 7.13, and 7.25 into Eq 7.1 to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^3 \frac {2(3)-2(j)+1} {(2(3)-(j)+1)(j)} P_{3-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.26) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.26 reduces to

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 [\frac 5 6 (\frac 1 2 (3x^2-1)) + \frac 1 12 (1)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.27) $$

Eq. 7.27 can then be further simplified to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - \frac 5 6 (3x^2-1) - \frac 1 6 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.28) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_3(\frac 1 2) = -0.198655... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.29) $$

and

$$ Q_3(\frac {-1} {2}) = -0.198655... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.30) $$

Therefore we can see that $$Q_3(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=4$$

First, we see for $$n=4$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {4-1} {2}] = 4 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.31) $$

Then plug Eq. 7.14, and 7.31 into Eq 7.1 to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^4 \frac {2(4)-2(j)+1} {(2(4)-(j)+1)(j)} P_{4-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.32) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.32 reduces to

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 [\frac 7 8 (\frac 1 2 (5x^3-3x)) + \frac 3 18 (x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.33) $$

Eq. 7.33 can then be further simplified to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - \frac 7 8 (5x^3-3x)) - \frac 1 3 (x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.34) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_4 (frac 1 2) = 0.440175... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.35) $$

and

$$ Q_4(\frac {-1} {2}) = -0.440175... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.36) $$

Therefore we can see that $$Q_4(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

From the generalization of evenness and oddness made in the first part of the problem along with the examples provided it has been proven that Eq. 7.1 is an odd function when "n" is an even integer and even when "n" is an odd integer.

Author and proof-reader
[Author] Rob Carroll [Proof-reader]

=Problem 7 - Attraction of Spheres= From (meeting 40 page 3)

Given


Image from lecture 40 of EGM6321

$$ r_{PQ}^2= \sum_{i=1}^3(x_Q^i-x_P^i)^2 $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.1) $$ $$\displaystyle x^1=r \cos(\theta) \cos(\rho) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.2) $$ $$\displaystyle x^2=r \cos(\theta) \sin(\rho) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.3) $$ $$\displaystyle x^3=r \sin(\theta) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.4) $$ $$\displaystyle \cos(\gamma)=\cos(\theta_Q)\cos(\theta_P)\cos(\rho_Q-\rho_P)+\sin(\theta_Q)\sin(\theta_P) $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.5) $$

Find
Show the following. $$\displaystyle r_{PQ}^2=r_Q^2 \left [ \left ( \frac{r_P}{r_Q} \right )^2+1-2 \left ( \frac{r_P}{r_Q} \right ) \cos(\gamma) \right ] $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.6) $$

Solution
 Solution was compared to F09.TEAM1.HW7.P5 

$$\displaystyle r_{PQ}^2= \left ( r_Q \cos(\theta_Q)\cos(\rho_Q)-r_P \cos(\theta_P)\cos(\rho_P) \right )^2 + \left ( r_Q \cos(\theta_Q)\sin(\rho_Q)-r_P \cos(\theta_P)\sin(\rho_P) \right )^2 + \left ( r_Q \sin(\theta_Q)-r_P \sin(\theta_P) \right )^2 $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.7) $$

$$\displaystyle \begin{align} r_{PQ}^2= &r_Q^2 \cos^2(\theta_Q)\cos^2(\rho_Q)-2r_Qr_P \cos(\theta_Q)\cos(\rho_Q) \cos(\theta_P)\cos(\rho_P) + r_P^2 \cos^2(\theta_P)\cos^2(\rho_P) + \\ &r_Q^2 \cos^2(\theta_Q)\sin^2(\rho_Q)-2r_Q r_P\cos(\theta_Q)\sin(\rho_Q) \cos(\theta_P)\sin(\rho_P) + r_P^2 \cos^2(\theta_P)\sin^2(\rho_P) + \\ &r_Q^2 \sin^2(\theta_Q)-2r_Q r_P\sin(\theta_Q) \sin(\theta_P) + r_P^2 \sin^2(\theta_P) \\ \end{align} $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.8) $$

$$\displaystyle \begin{align} r_{PQ}^2= &r_Q^2 \cos^2(\theta_Q) \underbrace{( \cos^2(\rho_Q)+\sin^2(\theta_Q))}_{1} + r_P^2 \cos^2(\theta_P) \underbrace{( \cos^2(\rho_P)+\sin^2(\theta_P))}_{1} - \\ &2r_Qr_P \cos(\theta_Q) \cos(\theta_P) \cos(\rho_Q)\cos(\rho_P) - 2r_Q r_P\cos(\theta_Q)\cos(\theta_P) \sin(\rho_Q) \sin(\rho_P) + \\ &r_Q^2 \sin^2(\theta_Q)-2r_Q r_P\sin(\theta_Q) \sin(\theta_P) + r_P^2 \sin^2(\theta_P) \\ \end{align} $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.9) $$

$$\displaystyle \begin{align} r_{PQ}^2= & r_Q^2 \underbrace{(\cos^2(\theta_Q)+ \sin^2(\theta_Q))}_{1} + r_P^2 \underbrace{(\cos^2(\theta_P)+ \sin^2(\theta_P))}_{1} - \\ &2r_Qr_P \cos(\theta_Q) \cos(\theta_P) \underbrace{(\cos(\rho_Q)\cos(\rho_P)+\sin(\rho_Q) \sin(\rho_P))}_{\cos(\rho_Q-\rho_P)} - \\ &2r_Q r_P\sin(\theta_Q) \sin(\theta_P) \\ \end{align} $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.10) $$

$$\displaystyle r_{PQ}^2= r_Q^2 + r_P^2 - 2r_Qr_P \underbrace{\left [ \cos(\theta_Q) \cos(\theta_P) \cos(\rho_Q-\rho_P) + \sin(\theta_Q) \sin(\theta_P) \right ]}_{\cos(\gamma)} $$ <p style="text-align:right;">$$\displaystyle (Eq. 7.11) $$

Divide Eq. 7.11 by $$\displaystyle r_Q^2$$ results in Eq. 7.6

Author and proof-reader
[Author] Mike Faraone

[Proof-reader]

=Problem 8 - Binomial Series= From (meeting 40 page 4)

Given

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(x+y)^r = \sum_{k=0}^\infty \begin{pmatrix} r \\ k \end{pmatrix} x^{r-k} y^k $$ $$ where
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\begin{pmatrix} r \\ k \end{pmatrix} = \frac{r(r-1) \cdot\cdot\cdot (r-k+1)}{k!} $$ $$
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(1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$
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where
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\alpha_i = \frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ $$
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Find
Use $$\displaystyle (8-1), \ (8-2)$$ to obtain $$\displaystyle (8-3), \ (8-4)$$

Solution
 We referred and used mediawiki code of f09.Team1 but we solved again by ourselves then discussed and found about background knowledge(factorial) and relationship with lecture.

[Background knowledge] Factorial 

$$\displaystyle x=1, \ y=-x, \ r=-\frac{1}{2}$$ plug into $$\displaystyle (8-1)$$


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(1-x)^{-1/2} = \sum_{k=0}^\infty \begin{pmatrix} -1/2 \\ k \end{pmatrix} 1^{-1/2-k} (-x)^k $$ $$
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We can expand using $$\displaystyle (8-2)$$
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(1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(-1/2)(-3/2)\cdot\cdot\cdot(1/2-k)}{k!} (-x)^k $$ $$
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(1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2k-1)}{(-2)^k k!} (-x)^k = \sum_{k=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2k-1)}{\underbrace{2^k k!}_{2 \cdot 4 \cdot \cdot \cdot (2k)}} x^k $$ $$
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Change $$\displaystyle k \rightarrow i$$
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(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 $$ $$
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Therefore, we obtained $$\displaystyle (8-3), \ (8-4)$$ using $$\displaystyle (8-1), \ (8-2)$$
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Author and proof-reader
[Author] Oh, Sang Min

[Proof-reader]

=Problem 9 -  = From (meeting 41 page 1)

Given
Using $$(n+1) P_{n+1}(\mu) - (2n+1) \mu P_n(\mu) + n P_{n-1}(\mu) 0$$ starting from $$P_0(\mu)=1$$ and $$P_1(\mu)=x$$

Find
Generate {$$P_2$$, $$P_3$$, $$P_4$$, $$P_5$$, $$P_6$$}

Solution
We can modify the equation as $$P_{n+1}(\mu) - \frac {2n+1} {n+1} \mu P_n(\mu) + \frac n {n+1} P_{n-1}(\mu) = 0$$ $$P_{n+1}(\mu) = \frac {2n+1} {n+1} \mu P_n(\mu) - \frac n {n+1} P_{n-1}(\mu)$$ $$P_2(\mu) = \frac 3 2 \mu P_1(\mu) - \frac 1 2 P_0(\mu) = \frac 3 2 \mu \mu - \frac 1 2 1 = \frac 1 2 (3\mu^2 - 1)$$ $$P_3(\mu) = \frac 5 3 \mu P_2(\mu) - \frac 2 3 P_1(\mu) = \frac 5 3 \mu \frac 1 2 (3\mu^2-1) - \frac 2 3 \mu = \frac 1 2 (5\mu^3-3\mu)$$ $$P_4(\mu) = \frac 7 4 \mu P_3(\mu) - \frac 3 4 P_2(\mu) = \frac 7 4 \mu \frac 1 2 (5\mu^3-3\mu) - \frac 3 4 \frac 1 2 (3\mu^2 - 1) = \frac {35} 8 \mu^4 - \frac {15} 4 \mu^2 + \frac 3 8 = \frac 1 8 (35\mu^4-30\mu^2+3)$$ $$P_5(\mu) = \frac 9 5 \mu P_n(\mu) - \frac 4 5 P_3(\mu) = \frac 9 5 \mu \frac 1 8 (35\mu^4-30\mu^2+3) - \frac 4 5 \frac 1 2 (5\mu^3-3\mu) = \frac 1 8 (63\mu^5-70\mu^3+15\mu)$$ $$P_6(\mu) = \frac {11} 6 \mu P_5(\mu) - \frac 5 6 P_4(\mu) = \frac {11} 6 \mu \frac 1 8 (63\mu^5-70\mu^3+15\mu) - \frac 5 6 \frac 1 8 (35\mu^4-30\mu^2+3) = \frac 1 {16} (231\mu^6-315\mu^4+105\mu^2-5)$$

Author and proof-reader
[Author] Jo

[Proof-reader]

=Problem 10 - Legendre Polynomial Expansion and Recurrence Method= From lecture 41-3.

Given
We have the Legendre polynomial,
 * $$\displaystyle P_{n}(x)=\sum\limits_{i=0}^{n/2}\left[(-1)^{i}\cdot\frac{(2n-2i)!\cdot x^{n-2i}}{2^{n}\cdot i!\cdot(n-i)!\cdot(n-2i)!}\right]$$

and their recurrence relation,
 * $$\displaystyle (n+1)\cdot P_{n+1}-(2n+1)\cdot x\cdot P_n+n\cdot P_{n-1}=0$$

From lecture 41-3 we have,
 * $$\displaystyle P_0(x)=1$$
 * $$\displaystyle P_1(x)=x$$
 * $$\displaystyle P_2(x)=3/2\cdot x^2-1/2$$

Find
We are to find $$\left\{P_3,\ldots,P_6\right\}$$ using the given power series expansion and compare results with those obtained using the second recurrence relation.

Polynomial Expansion
We first find $$\left\{P_3,\ldots,P_6\right\}$$ by using the given polynomial expansion, wherein the subscripts 3 through 6 are substituted for $$n$$. This yields,
 * $$\displaystyle P_3(x)=5/2\cdot x^3-3/2\cdot x$$
 * $$\displaystyle P_4(x)=35/8\cdot x^4-15/4\cdot x^2+3/8$$
 * $$\displaystyle P_5(x)=63/8\cdot x^5-35/4\cdot x^3+15/8$$
 * $$\displaystyle P_6(x)=231/16\cdot x^6-315/16\cdot x^4+105/16\cdot x^2-5/16$$

Recurrence Relation
We know that $$\displaystyle P_0(x)=1$$ and $$\displaystyle P_1(x)=x$$. We then must find $$\left\{P_2,\ldots,P_6\right\}$$ by using the recurrence relation, which may be rewritten as,
 * $$\displaystyle P_n=\frac{(n+1)\cdot P_{n+1}+n\cdot P_{n-1}}{(2n+1)\cdot x}$$

We must solve for the following,
 * $$\displaystyle P_2=\frac{3\cdot P_{3}+2\cdot P_{1}}{5\cdot x}$$
 * $$\displaystyle P_3=\frac{4\cdot P_{4}+3\cdot P_{2}}{7\cdot x}$$
 * $$\displaystyle P_4=\frac{5\cdot P_{5}+4\cdot P_{3}}{9\cdot x}$$
 * $$\displaystyle P_5=\frac{6\cdot P_{6}+5\cdot P_{4}}{11\cdot x}$$

We now have enough equations to solve for the polynomials by substitution. The results yield,
 * $$\displaystyle P_0(x)=1$$
 * $$\displaystyle P_1(x)=x$$
 * $$\displaystyle P_2(x)=3/2\cdot x^2-1/2$$
 * $$\displaystyle P_3(x)=5/2\cdot x^3-3/2\cdot x$$
 * $$\displaystyle P_4(x)=35/8\cdot x^4-15/4\cdot x^2+3/8$$
 * $$\displaystyle P_5(x)=63/8\cdot x^5-35/4\cdot x^3+15/8$$
 * $$\displaystyle P_6(x)=231/16\cdot x^6-315/16\cdot x^4+105/16\cdot x^2-5/16$$

Comparison
The polynomials, computed by both methods as shown above yield the same solutions.

Author and proof-reader
[Author]
 * Raul Riveros

[Proof-reader] =References=

=Contributing members= Oh, Sang min Author of problem 2, 3, 5, 8

Jo Authored Problem 1 and Problem 9.

Mike Faraone Authored Problem 7 and Proofed Problem 1.

Raul Riveros Solved problem 10, reviewed Problem 3.