text:
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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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
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$$\displaystyle (3-2-3)
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P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (3-2-4)
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P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (3-2-5)
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P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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
 *
$$\displaystyle (3-2-7)
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P_5(x) = \frac{63}{8} x ^ 5 - \frac{35}{4} x ^ 3 + \frac{15}{8}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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}
 *
$$\displaystyle (3-2-9)
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 * $$ \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
 *
$$\displaystyle (5-1)
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\ Q_0(x) = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) = \tanh^{-1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (5-2)
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Part 2

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P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (5-3)
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\ Q_1(x) = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) - 1 = x \tanh^{-1}(x) - 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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 $$
 *
$$\displaystyle (5-5)
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$$\displaystyle (5-5)
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$$
 * $$\displaystyle \tanh(x) = \frac{e^{2x} - 1}{e^{2x} + 1} $$
 *
$$\displaystyle (5-6)
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$$\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} $$
 * $$\displaystyle
 * $$\displaystyle
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\ = \ \frac{e^{2 \cdot \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} - 1}{e^{2 \cdot \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} + 1}
 * $$\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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$$\displaystyle (5-7)
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In this case, $$ 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} $$
 * $$\displaystyle
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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
$$ Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x) \,$$
$$\displaystyle (Eq. 7.1) $$

$$ J = 1+2[\frac {n-1} {2}] \,$$
$$\displaystyle (Eq. 7.2) $$

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 $$ n = even $$

$$ Q_n(x) = (even) tanh^{-1}(x) - 2 \sum_{j=1,3,5,..}^J \frac {2n-2j+1} {(2n-j+1)j} (odd) \,$$
$$\displaystyle (Eq. 7.3) $$

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

$$ tanh^{-1}(x) = odd \,$$
$$\displaystyle (Eq. 7.4) $$

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

$$ Q_n(x) = (even) (odd) - (odd) \,$$
$$\displaystyle (Eq. 7.5) $$

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

$$ n = even \,$$

$$ Q_n(x) = odd \,$$
$$\displaystyle (Eq. 7.6) $$

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

$$ Q_n(x) = (odd) (odd) - 2 \sum_{j=1,3,5,..}^J (constant) (even) \,$$
$$\displaystyle (Eq. 7.7) $$

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

$$ n = odd \,$$

Eq. 7.7 reduces to

$$ Q_n(x) = even \,$$
$$\displaystyle (Eq. 7.8) $$

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.

$$ P_n(x) = \frac {1} {2^n n!} \frac {d^n} {dx^n} (x^2-1)^n \,$$
$$\displaystyle (Eq. 7.9) $$

Reference: Rodrigues Formula

From Eq. 7.9 we can get

$$ P_0(x) = 1 \,$$
$$\displaystyle (Eq. 7.10) $$

$$ P_1(x) = x \,$$
$$\displaystyle (Eq. 7.11) $$

$$ P_2(x) = \frac {1} {2} (3x^2-1) \,$$
$$\displaystyle (Eq. 7.12) $$

$$ P_3(x) = \frac {1} {2} (5x^3-3x) \,$$
$$\displaystyle (Eq. 7.13) $$

$$ P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3) \,$$
$$\displaystyle (Eq. 7.14) $$

For $$n=1$$

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

$$ J = 1+2[\frac {1-1} {2}] = 1 \,$$
$$\displaystyle (Eq. 7.15) $$

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

$$ Q_1(x) = (x) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^1 \frac {2(1)-2(j)+1} {(2(1)-(j)+1)(j)} P_{1-j}(x) \,$$
$$\displaystyle (Eq. 7.16) $$

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

$$ Q_1(x) = (x) tanh^{-1}(x) - (1) \,$$
$$\displaystyle (Eq. 7.17) $$

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

$$ Q_1(\frac 1 2) = -0.725347... \,$$
$$\displaystyle (Eq. 7.18) $$

and

$$ Q_1(\frac {-1} {2}) = -0.725347... \,$$
$$\displaystyle (Eq. 7.19) $$

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

$$ J = 1+2[\frac {2-1} {2}] = 2 \,$$
$$\displaystyle (Eq. 7.20) $$

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

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x) \,$$
$$\displaystyle (Eq. 7.21) $$

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

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) \,$$
$$\displaystyle (Eq. 7.22) $$

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

$$ Q_2(\frac 1 2) = -0.818663... \,$$
$$\displaystyle (Eq. 7.23) $$

and

$$ Q_2(\frac {-1} {2}) = 0.818663... \,$$
$$\displaystyle (Eq. 7.24) $$

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

$$ J = 1+2[\frac {3-1} {2}] = 3 \,$$
$$\displaystyle (Eq. 7.25) $$

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

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^3 \frac {2(3)-2(j)+1} {(2(3)-(j)+1)(j)} P_{3-j}(x) \,$$
$$\displaystyle (Eq. 7.26) $$

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

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 [\frac 5 6 (\frac 1 2 (3x^2-1)) + \frac 1 12 (1)] \,$$
$$\displaystyle (Eq. 7.27) $$

Eq. 7.27 can then be further simplified to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - \frac 5 6 (3x^2-1) - \frac 1 6 \,$$
$$\displaystyle (Eq. 7.28) $$

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

$$ Q_3(\frac 1 2) = -0.198655... \,$$
$$\displaystyle (Eq. 7.29) $$

and

$$ Q_3(\frac {-1} {2}) = -0.198655... \,$$
$$\displaystyle (Eq. 7.30) $$

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

$$ J = 1+2[\frac {4-1} {2}] = 4 \,$$
$$\displaystyle (Eq. 7.31) $$

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

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^4 \frac {2(4)-2(j)+1} {(2(4)-(j)+1)(j)} P_{4-j}(x) \,$$
$$\displaystyle (Eq. 7.32) $$

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

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 [\frac 7 8 (\frac 1 2 (5x^3-3x)) + \frac 3 18 (x)] \,$$
$$\displaystyle (Eq. 7.33) $$

Eq. 7.33 can then be further simplified to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - \frac 7 8 (5x^3-3x)) - \frac 1 3 (x)] \,$$
$$\displaystyle (Eq. 7.34) $$

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

$$ Q_4 (frac 1 2) = 0.440175... \,$$
$$\displaystyle (Eq. 7.35) $$

and

$$ Q_4(\frac {-1} {2}) = -0.440175... \,$$
$$\displaystyle (Eq. 7.36) $$

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

$$ r_{PQ}^2= \sum_{i=1}^3(x_Q^i-x_P^i)^2 $$
$$\displaystyle (Eq. 7.1) $$ $$\displaystyle x^1=r \cos(\theta) \cos(\rho) $$
$$\displaystyle (Eq. 7.2) $$ $$\displaystyle x^2=r \cos(\theta) \sin(\rho) $$
$$\displaystyle (Eq. 7.3) $$ $$\displaystyle x^3=r \sin(\theta) $$
$$\displaystyle (Eq. 7.4) $$ $$\displaystyle \cos(\gamma)=\cos(\theta_Q)\cos(\theta_P)\cos(\rho_Q-\rho_P)+\sin(\theta_Q)\sin(\theta_P) $$
$$\displaystyle (Eq. 7.5) $$

Find
Show the following. $$\displaystyle r_{PQ}^2=r_Q^2 \left [ \left ( \frac{r_P}{r_Q} \right )^2+1-2 \left ( \frac{r_P}{r_Q} \right ) \cos(\gamma) \right ] $$
$$\displaystyle (Eq. 7.6) $$

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

$$\displaystyle r_{PQ}^2= \left ( r_Q \cos(\theta_Q)\cos(\rho_Q)-r_P \cos(\theta_P)\cos(\rho_P) \right )^2 + \left ( r_Q \cos(\theta_Q)\sin(\rho_Q)-r_P \cos(\theta_P)\sin(\rho_P) \right )^2 + \left ( r_Q \sin(\theta_Q)-r_P \sin(\theta_P) \right )^2 $$
$$\displaystyle (Eq. 7.7) $$

$$\displaystyle \begin{align} r_{PQ}^2= &r_Q^2 \cos^2(\theta_Q)\cos^2(\rho_Q)-2r_Qr_P \cos(\theta_Q)\cos(\rho_Q) \cos(\theta_P)\cos(\rho_P) + r_P^2 \cos^2(\theta_P)\cos^2(\rho_P) + \\ &r_Q^2 \cos^2(\theta_Q)\sin^2(\rho_Q)-2r_Q r_P\cos(\theta_Q)\sin(\rho_Q) \cos(\theta_P)\sin(\rho_P) + r_P^2 \cos^2(\theta_P)\sin^2(\rho_P) + \\ &r_Q^2 \sin^2(\theta_Q)-2r_Q r_P\sin(\theta_Q) \sin(\theta_P) + r_P^2 \sin^2(\theta_P) \\ \end{align} $$
$$\displaystyle (Eq. 7.8) $$

$$\displaystyle \begin{align} r_{PQ}^2= &r_Q^2 \cos^2(\theta_Q) \underbrace{( \cos^2(\rho_Q)+\sin^2(\theta_Q))}_{1} + r_P^2 \cos^2(\theta_P) \underbrace{( \cos^2(\rho_P)+\sin^2(\theta_P))}_{1} - \\ &2r_Qr_P \cos(\theta_Q) \cos(\theta_P) \cos(\rho_Q)\cos(\rho_P) - 2r_Q r_P\cos(\theta_Q)\cos(\theta_P) \sin(\rho_Q) \sin(\rho_P) + \\ &r_Q^2 \sin^2(\theta_Q)-2r_Q r_P\sin(\theta_Q) \sin(\theta_P) + r_P^2 \sin^2(\theta_P) \\ \end{align} $$
$$\displaystyle (Eq. 7.9) $$

$$\displaystyle \begin{align} r_{PQ}^2= & r_Q^2 \underbrace{(\cos^2(\theta_Q)+ \sin^2(\theta_Q))}_{1} + r_P^2 \underbrace{(\cos^2(\theta_P)+ \sin^2(\theta_P))}_{1} - \\ &2r_Qr_P \cos(\theta_Q) \cos(\theta_P) \underbrace{(\cos(\rho_Q)\cos(\rho_P)+\sin(\rho_Q) \sin(\rho_P))}_{\cos(\rho_Q-\rho_P)} - \\ &2r_Q r_P\sin(\theta_Q) \sin(\theta_P) \\ \end{align} $$
$$\displaystyle (Eq. 7.10) $$

$$\displaystyle r_{PQ}^2= r_Q^2 + r_P^2 - 2r_Qr_P \underbrace{\left [ \cos(\theta_Q) \cos(\theta_P) \cos(\rho_Q-\rho_P) + \sin(\theta_Q) \sin(\theta_P) \right ]}_{\cos(\gamma)} $$
$$\displaystyle (Eq. 7.11) $$

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
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-1)
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\begin{pmatrix} r \\ k \end{pmatrix} = \frac{r(r-1) \cdot\cdot\cdot (r-k+1)}{k!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-2)
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(1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-3)
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where
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\alpha_i = \frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ $$
 * $$\displaystyle
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$$\displaystyle (8-4)
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Find
Use $$\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 $$ $$
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-5)
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We can expand using $$\displaystyle (8-2)$$
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(1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(-1/2)(-3/2)\cdot\cdot\cdot(1/2-k)}{k!} (-x)^k $$ $$
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-6)
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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 $$ $$
 * $$\displaystyle
 * $$\displaystyle
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$$\displaystyle (8-7)
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Change $$\displaystyle k \rightarrow i$$
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(1-x)^{-1/2} = \sum_{i=0}^\infty \underbrace{\frac{1 \cdot 3\cdot\cdot\cdot(2i-1)}{2 \cdot 4 \cdot \cdot \cdot (2i)}}_{\alpha_i}x^i $$ $$
 * $$\displaystyle
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$$\displaystyle (8-8)
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Therefore, we obtained $$\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.

title: User:Egm6321.f10.team5/hw7


text:
User:Egm6321.f2010.team5.riveros/hw6p6

= Problem 6 - Spherical Coordinate Problem = From (meeting 35 page 4)

Given
We are given the following relations,
 * $$\displaystyle x_1 = r \sin \bar{\theta} \cos \phi =  \xi_1 \sin \xi_2 \cos \xi_3 $$
 * $$\displaystyle x_2 = r \sin \bar{\theta} \sin \phi =  \xi_1 \sin \xi_2 \sin \xi_3 $$
 * $$\displaystyle x_3 = r \cos \bar{\theta} =  \xi_1 \cos \xi_2$$

Noting that $$\displaystyle \xi_1=r$$, $$\displaystyle \xi_2=\bar{\theta}$$, and $$\displaystyle \xi_3=\phi$$

Find
We are to derive the laplacian, $$\Delta\Psi$$, in spherical coordinates using the math/physics convention in which,
 * $$\bar\theta=\frac{\pi}{2}-\theta$$

Solution
We first calculate the total derivatives, $$dx_i$$, in spherical coordinates using the given relations. We have,
 * $$\displaystyle dx_1=\frac{\partial x_1}{\partial \xi_1}d \xi_1 + \frac {\partial x_1}{\partial \xi_2} d \xi_2 + \frac{\partial x_1}{\partial\xi_3}d\xi_3=sin\xi_2 cos \xi_3 d\xi_1 + \xi_1 cos \xi_2 cos \xi_3 d \xi_2 - \xi_1 sin \xi_2 sin \xi_3 d \xi_3 $$
 * $$\displaystyle dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3=dx_2=sin\xi_2 sin\xi_3 d\xi_1 + \xi_1 cos\xi_2 sin\xi_3 d\xi_2 + \xi_1 sin\xi_2 cos\xi_3 d\xi_3 $$
 * $$\displaystyle dx_3=\frac{\partial x_3}{\partial\xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3=dx_3=cos\xi_2 d\xi_1 - \xi_1 sin\xi_2 d\xi_2 + 0 $$

Next, $$(ds)^2$$ is found by similarly. We have,
 * $$\displaystyle (ds)^2=\sum (dx_i)^2 $$

By substituting the given relations and using the first Pythagorean identity, $$(ds)^2$$ simplifies to,
 * $$\displaystyle ds^2=d\xi_1^2+\xi_1^2d\xi_2^2+\xi_1^2sin^2\xi_2d\xi_3^2$$

We note that $$h_i^2$$ is the coefficient of $$\xi^2_i$$. Listing the coefficients,
 * $$\displaystyle h_1=1$$
 * $$\displaystyle h_2=\xi_1$$
 * $$\displaystyle h_3=\xi_1 sin\xi_2$$

The Laplacian for orthogonal curvilinear coordinates can be expressed as
 * $$\displaystyle \Delta\Psi=\frac{1}{h_1h_2h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left(\frac {h_1h_2h_3}{(h_i)^2}\cdot\frac{\partial\Psi}{\partial\xi_i}\right)$$

We substitute the values for $$h_i$$ found above to yield
 * $$\displaystyle \Delta\Psi=\frac{1}{\xi^2_1sin\xi_2}\left[\frac{\partial}{\partial\xi_1}\left(\xi^2_1sin\xi_2\frac{\partial\Psi}{\partial \xi_1}\right)

+\frac{\partial}{\partial\xi_2}\left(sin\xi_2\frac{\partial\Psi}{\partial \xi_2}\right) +\frac{\partial}{\partial\xi_3}\left(\frac{1}{sin\xi_2}\frac{\partial\Psi}{\partial \xi_3}\right)\right]$$ We expand the above relation to yield the Laplacian in spherical coordinates,
 * $$\Delta\Psi =\frac {1}{r^2 sin \bar\theta} \left[ \frac {\partial}{\partial r} \left(r^2 sin \bar\theta \frac {\partial\Psi}{\partial r} \right)

+ \frac {\partial}{\partial \bar\theta} \left(sin \bar\theta \frac {\partial \Psi}{\partial \bar\theta} \right) + \frac {\partial}{\partial \phi} \left(\frac {1}{sin \bar\theta} \frac {\partial \Psi}{\partial \phi}\right) \right] $$

Author and proof-reader
[Author]
 * Raul Riveros

[Proof-reader]

=Problem 5: Derive the Laplacian in Cylindrical Coordinates=

Given
From the lecture notes p 35-3
 * {| style="width:100%" border="0"

{x}_{1}=x= { \xi}_{1} \cos { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.1)
 * style= |
 * }

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

{x}_{2}=y= { \xi}_{1} \sin { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.2)
 * style= |
 * }

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

{x}_{3}=z= { \xi}_{3} $$ $$
 * $$\displaystyle (Eq 5.3)
 * style= |
 * }

Find
$$ 1) $$ Find $$ \left\{  {dx}_{i}\right\}=  \left\{  {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$ in terms of $$  \left\{  {\xi}_{j}\right\}=  \left\{  {\xi}_{1}, {\xi}_{2}, {\xi}_{3}\right\} $$ and $$ \left\{  {d\xi}_{k}\right\} $$

$$ 2) $$ Find $${ds}^{2} = \sum_{i} { \left(d {x}_{i} \right)}^{2}=  \sum_{i}  {\left(  {h}_{i}\right)}^{2}  { \left( {d\xi}_{i} \right)}^{a}$$ Identify $$ \left\{  {h}_{i}\right\} $$ in terms of $$ \left\{  {\xi}_{i}\right\} $$

$$ 3) $$ Find $$ \Delta \Psi $$ in cylindrical coordinates

$$ \left\{ {dx}_{i}\right\} $$
To solve we find the total derivatives of $$ \left\{ {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$

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

{dx}_{1} = \frac{\partial  {x}_{1}}{\partial  {\xi}_{1}} {d\xi}_{1}+\frac{\partial  {x}_{1}}{\partial  {\xi}_{2}} {d\xi}_{2}+\frac{\partial  {x}_{1}}{\partial  {\xi}_{3}} {d\xi}_{3} $$ $$ Substituting $$\displaystyle (Eq 5.1) $$ into $$\displaystyle (Eq 5.4) $$ we get
 * $$\displaystyle (Eq 5.4)
 * style= |
 * }

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

{dx}_{1} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \cos {\xi}_{2}\right){d\xi}_{3} $$
 *
 * style= |
 * }

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

$$ {dx}_{1} =  \cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.5)
 * style= |
 * }

Similarly,

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

{dx}_{2} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \sin {\xi}_{2}\right){d\xi}_{3} $$
 *
 * style= |
 * }

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

$$ {dx}_{2} =  \sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.6)
 * style= |
 * }

Since there is no projection on $$ {x}_{3}$$ from $$\displaystyle (Eq 5.4) $$

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

$$ {dx}_{3} =  {d\xi}_{3} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.7)
 * style= |
 * }

$$ {\left(ds\right)}^{2} $$
To solve we use the relationship

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

{ds}^{2} = { \left(d {x}_{1} \right)}^{2}+ { \left(d {x}_{2} \right)}^{2}+ { \left(d {x}_{3} \right)}^{2} $$
 *
 * style= |
 * }

And substituting equations 5.5 through 5.7 we obtain the following

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

{ds}^{2} = { \left(\cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(\sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(d {\xi}_{3} \right)}^{2} $$ $$
 * $$\displaystyle (Eq 5.8)
 * style= |
 * }

Expanding equation 5.8 and combining like terms gives us

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

{ds}^{2} = \cancelto{1}{\left( {\cos {\xi}_{2}}^{2}+ { \sin {\xi}_{2}}^{2} \right)} {\left( {d\xi}_{1} \right)}^{2}+ \cancelto{{{\xi}_{1}}^{2}}{\left({{\xi}_{1}}^{2}{\cos {\xi}_{2}}^{2}+ {  {{\xi}_{1}}^{2}\sin {\xi}_{2}}^{2} \right)}  {\left( {d\xi}_{2} \right)}^{2}+  {\left( {d\xi}_{3} \right)}^{2} $$ $$
 * $$\displaystyle (Eq 5.9)
 * style= |
 * }

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

$$ {ds}^{2}={\left( {d\xi}_{1} \right)}^{2}+ {{\xi}_{1}}^{2}{\left( {d\xi}_{2} \right)}^{2}+ {\left( {d\xi}_{3} \right)}^{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.10)
 * style= |
 * }

Now by inspection we see that

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

$$ {h}_{1}= 1 ; {h}_{2} =  {\xi}_{1} ; {h}_{3}= 1 $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.11)
 * style= |
 * }

$$\Delta \Psi $$
To obtain the Laplacian in cylindrical coordinates we use the following equation

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

\Delta \Psi = \frac{1}{ {h}_{1}{h}_{2} {h}_{3}} \sum_{i=1}^{3}  \frac{\partial }{\partial  {\xi}_{i}} \left(  \frac{{h}_{1}{h}_{2} {h}_{3}}{ {\left( {h}_{i} \right)}^{2}} \frac{\partial  \Psi}{\partial {\xi}_{i}}\right) $$ $$
 * $$\displaystyle (Eq 5.12)
 * style= |
 * }

Now we plug in our values for $$ {h}_{i}$$ and expand the summation

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

\Delta \Psi =  \frac{1}{ {\xi}_{1}}  \frac{\partial }{\partial  {\xi}_{1}} \left(  \frac{1}{{\xi}_{1}} \frac{\partial  \Psi}{\partial {\xi}_{1}}\right)+ \cancelto{1}\frac{{\xi}_{1}}{ {\xi}_{1}} \frac{\partial {}^{2} \Psi}{\partial  {\xi}_{2}^{2}} + \cancelto{1}\frac{{\xi}_{1}}{ {\xi}_{1}} \frac{\partial {}^{2} \Psi}{\partial  {\xi}_{3}^{2}} $$ $$
 * $$\displaystyle (Eq 5.13)
 * style= |
 * }

Finally we replace the $$ \xi s $$ with the cylindrical coordinate convention to obtain

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

$$ \Delta \Psi =  \frac{1}{r}  \frac{\partial }{\partial  r} \left(  \frac{1}{r} \frac{\partial  \Psi}{\partial r}\right)+ \frac{\partial {}^{2} \Psi}{\partial  { \theta}^{2}} + \frac{\partial {}^{2} \Psi}{\partial  {z}^{2}} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq 5.14)
 * style= |
 * }

Contributing Members
Solved and Posted by--Egm6321.f10.team3.cook 02:45, 16 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:58, 16 November 2010 (UTC)

= Problem 6 The Laplace Equation in Cylindrical Coordinates =

Given
Converting from rectangular to spherical coordinates:

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

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

$$\displaystyle x_3 = z = r cos\bar\theta $$

$$\displaystyle \xi_1 = r $$

$$\displaystyle \xi_2 = \bar\theta $$

$$\displaystyle \xi_3 = \phi $$

where $$\displaystyle \bar\theta $$ is equal to $$\displaystyle \pi / 2 - \theta $$ by the math/physics convention, as opposed to the astronomic convention

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of spherical coordinates.

Solution
$$\displaystyle dx_1 = \frac {\partial x_1}{\partial xi_1} d \xi_1 + \frac {\partial x_1}{\partial xi_2} d \xi_2 + \frac {\partial x_1}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_1 = sin \xi_2 cos \xi_3 d \xi_1 + \xi_1 cos \xi_2 cos \xi_3 d \xi_2 - \xi_1 sin \xi_2 sin \xi_3 d \xi_3 $$

$$\displaystyle dx_2 = \frac {\partial x_2}{\partial xi_1} d \xi_1 + \frac {\partial x_2}{\partial xi_2} d \xi_2 + \frac {\partial x_2}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_2 = sin\xi_2 sin\xi_3 d\xi_1 + \xi_1 cos\xi_2 sin\xi_3 d\xi_2 + \xi_1 sin\xi_2 cos\xi_3 d\xi_3 $$

$$\displaystyle dx_3 = \frac {\partial x_3}{\partial xi_1} d \xi_1 + \frac {\partial x_3}{\partial xi_2} d \xi_2 + \frac {\partial x_3}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_3 = cos\xi_2 d\xi_1 - \xi_1 sin\xi_2 d\xi_2 + 0 $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

$$\displaystyle $$

Using the trigonometric identity

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

and regrouping terms yields

$$\displaystyle ds^2 = d\xi_1^2 + \xi_1^2 d\xi_2^2 + \xi_1^2 sin^2 \xi_2 d\xi_3^2 $$

where $$\displaystyle h_i^2 $$ are the coefficients of each term in the above expression.

$$\displaystyle h_1 = 1 $$

$$\displaystyle h_2 = \xi_1 = r $$

$$\displaystyle h_3 = \xi_1 sin\xi_2 = r sin\bar\theta $$

For any set of orthogonal curvilinear coordinates, the Laplacian is given by

$$\displaystyle \Delta \Psi = \frac {1}{h_1h_2h_3} \sum \frac {\partial}{\partial \xi_i} (\frac {h_1h_2h_3}{h_i^2} \frac {\partial \Psi}{\partial \xi_i}) $$

for $$\displaystyle i = 1 $$

$$\displaystyle = \frac {1}{\xi_1} ( \frac {\partial}{\partial \xi_1} \frac {\xi_1}{h_1^2} \frac {\partial \Psi}{\partial \xi_1} ) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac {\partial}{\partial r} (r^2 sin\bar\theta \frac {\partial \Psi}{\partial r} $$

for $$\displaystyle i = 2 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_2} \frac {\xi_1}{h_2^2} \frac {\partial \Psi}{\partial \xi_2}) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac{\partial}{\partial \bar\theta} (\frac {r^2 sin \bar\theta}{r^2} \frac {partial \Psi}{\partial \bar\theta}) $$

for $$\displaystyle i = 3 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_3} \frac {\xi_1}{h_3^2} \frac {\partial \Psi}{\partial \xi_3}) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac{\partial}{\partial phi} (\frac {r^2 sin\bar\theta}{r^2 sin^2 \bar\theta} \frac {\partial \Psi}{\partial \phi}) $$

Therefore, the expression for the Laplacian in terms of spherical coordinates becomes

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6321.f2010.team5.riveros/hw6p6


text:
User talk:Egm6321.f10.team5.steinberg/hw6

Part 1
From the Wikipedia article, the elliptic coordinates are defined as shown in (7.1) and (7.2).

Part 3
For orthogonal curvilinear coordinates (2D), the Laplacian operator may be defined as

Since $$ h_1 $$ and $$ h_2 $$ are equal, (7.10) becomes


text:
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User talk:Binyaswaitmail

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title: MediaWiki:Listusers-summary


title: File:Convolution.1.A.pdf


text:
User:Egm6321.f10.team6.Kim.MK/hw6

=Problem 3 - Solving the Legendre Equation Using the Direct Method =

Find
The complete solution $$y(x)$$ using the direct method

Solution
We can make (3.1) as like (3.3) form,

1st homogeneous solutions is (3.2) $$\begin{align}{{u}_{1}}(x)=x\end{align}$$ with $${{a}_{1}}(x)=\frac{-2x}{1-{{x}^{2}}}$$

The integrating factor is

2nd homogeneous solution can be obtained as

Then

Thus,

The particular solution can be obtained as

with $$\begin{align}f(x)=\frac{1}\end{align}$$

Then,

$$\begin{align} \end{align}$$

Therefore, the complete solution can be written as

Note
The integral of $$\begin{align}y_{p}\end{align}$$ can be evaluated via Wolfram Alpha

Given
The equations for spherical coordinates with Math/Phys. convention are given by:

Solution of Problem 4 and definition of $$\theta $$ measured clockwisely from the North Pole

Find
The expression of Lapalcian

Solution
Eq. (6.2) can be written as

Replace (6.4)in the Laplacian obtained in Problem 4,

Thus,

So, we can get

title: User:Egm6321.f10.team6.Kim.MK/hw6


text:
User:Ottava Rima/To Autumn source analysis

SandyGeorgia has asked for a Source analysis of To Autumn, a FA. I shall base the analysis off of this version of the page, the one listed in the history as the first FA version. I shall be placing excerpts from the text with quotations from the books. I shall do so in a way that may violate "fair use" but should not since it is for an educational reason - verifying if a work has plagiarism or not.

Lead
I shall skip over the lead, as it is summary from the body of the text and contains no original concepts. It was also substantially reworked.

Background
1. Article: "During the spring of 1819, Keats wrote many of his major odes: "Ode on a Grecian Urn", "Ode on Indolence", "Ode on Melancholy", "Ode to a Nightingale", and "Ode to Psyche". After the month of May, he began to pursue other forms of poetry, including the verse tragedy Otho the Great in collaboration with friend and roommate Charles Brown, the second half of Lamia, and a return to his unfinished epic Hyperion.[1]"

Source: Bate 1963 pp. 526–562 This is a summary of a large section that describes all of the poems listed above.

2. His efforts from spring until autumn were dedicated completely to a career in poetry; he switched between writing long and short poems, and his goal for each day was to compose more than fifty lines of verse. He devoted his free time to studying works such as Robert Burton's Anatomy of Melancholy to further his own ideas.[2]

Source: Gittings 1968 pp. 269–270

A. Page 269 talks about Hyperion and smaller poems. The "goal for each day" comes from: "In his new mood of release, Keats was not content with reaching his old standard of fifty lines of poetry a day" and "He did not stop there, but wrote a longer poem in the same metre."

B. Page 270 has the statement about Burton: "he borrowed from the 'Dialogue between Pleasure and Pain', which he found at the beginning of The Anatomy of Melancholy by Robert Burton". The emphasis on "career in poetry" is related to the paragraph (emphasis on first sentence) beginning: "Once more he began to feel himself fully a poetry, with poetry as his main concern."

3. "Although Keats managed to write many poems in 1819, he was suffering from a multitude of financial troubles throughout the year. These troubles were compounded with his concerns over his brother, George, who, after emigrating to America, was badly in need of money. Keats was distracted by his and his brother's fiscal problems, but on 19 September 1819 he set aside time to write "To Autumn". The poem marks the final moment of his poetry career. He could no longer afford to devote his time to the composition of poems and began working on more lucrative projects.[1]"

Source: See 1 about Bate 1963 pp. 526–562. The statement on George comes from p. 526: "George, who had so often helped him in the past, now depended on him. A large reservoir of guilt, accumulating since Keats had left Guy's Hospital, was also suddenly tapped." Page 562 talks about letters from George mentioning money problems. - "but Keats's dispatch shows his feeling of urgency. On September 10 bad news and appeals for help arrived from George; and from then until September 21, when he gave up the new version of Hyperion, the days were distracted by anxiety, by fruitless efforts to help George, and by he determination to turn to some other kind of work."

The page numbers should probably have added 580-581, where it is made explicit that he worked on "To Autumn" on 19 September (it is mentioned partly before, but for redundancy of the date).

4. "In addition to his monetary problems, Keats's declining health and personal responsibilities provided more obstacles to his poetic efforts.[3]"

Source: Motion 1999 p. 461 "Keats wrote the poem when his precarious freelance life was finally coming to an end, when his poor health was becoming unignorable, when he realised that he ould not continue to postpone some sort of resolution with Fanny, when he felt gloomy about the reliability of his 'set', and when his worries about his brother and sister-in-law were acute."

5. "On 19 September 1819, Keats walked near Winchester along the River Itchen. In a letter to his friend Joshua Reynolds written on 21 September, Keats described the impression the scene had made upon him and its influence on the composition of "To Autumn":[4] "

Source: Bate 1963 p. 580 "The Sunday afer he returned to Winchester from London, he took the same walk out to the St. Cross meadows along the small clear River Itchen (September 19). He mentioned the walk in a letter to Reynolds two days later... The poem is the last of the great odes, 'To Autumn.'"

6. "How beautiful the season is now – How fine the air. A temperate sharpness about it [...] I never lik'd stubble fields so much as now [...] Somehow a stubble plain looks warm – in the same way that some pictures look warm – this struck me so much in my sunday's walk that I composed upon it."[5]

Source: Quoted material that can be found in Bate 1963 p. 580 as attributed. However, the quote is attributed to a book on Keats's letters. Spelling is the same in both.

7. "Not everything on Keats's mind at the time was bright; the poet knew in September that he would have to finally abandon Hyperion. Thus, in the letter that he wrote to Reynolds, Keats also included a note saying that he abandoned his long poem.[6]"

Source: Bate 1963 p. 585 Derived from "No further delay was possible. He had probably been thinking since he returned from London on September 15 that he would have to abandon Hyperion--this effort that symbolized so much in his hope to be 'among the English poets.' Three days after the ode 'To Autumn,' he wrote to Reynolds (in the same letter where he described the warm stubble fields and his walk just before he wrote 'To Autumn') that he had 'given up Hyperion.'"

8. "Keats did not send "To Autumn" to Reynolds, but did include the poem within a letter to Richard Woodhouse, Keats's publisher and friend, and dated it on the same day.[7]"

Source: Evert 1965 pp. 296–297 Derived from this: "On September 21, 1819, Keats announced in a letter to Reynolds that he had given up The Fall of Hyperion... What he had composed was, of course, the ode 'To Autumn,' which he did not include in the letter to Reynolds but sent to Woodhouse in a letter of the same date."

Similar language used in part but only from the compactness of facts.

9. "The poem was revised and included in Keats's 1820 collection of poetry titled Lamia, Isabella, the Eve of St. Agnes, and Other Poems. Although the publishers Taylor and Hessey feared the kind of bad reviews that had plagued Keats's 1818 edition of Endymion, they were willing to publish the collection after the removal of any potentially controversial poems to ensure that there would be no politically motivated reviews that could give the volume a bad reputation.[8]"

Source: McGann 1979 pp. 988–989 I am going to quote A Routledge literary Sourcebook on the poems of John Keats p. 58 copy of the text (since it is visible on googlebooks for others to check). "'To Autumn' was first published in Lamia, Isabella, The Even of St. Agnes, and other Poems, the so called 1820 volume" and "The publishers of the 1820 volume were Taylor and Hessey, who also published Endymion in 1818... that had been the target of hostile reviews of Keats, and the poet was not the only person who suffered in that literary whirl-wind. consequently, when Keats approached Taylor and Hessey again, in the later part of 1819, about publishing the new book of poems he had been planning, they were interested but wary. They had no intention of bringing out a volume that would all down again the sort of hostility and ridicule which greeted Endymion.... The key fact in the pre-publication history of the 1820 poems is the insistence by Keats's publishers that the book not contain anything that would provoke the reviewers to attack (they were especially oncerned about charges of indecency and political radicalism). Keats sruggled with them over these issues, but he was eventually persuaded to follow their line. The two poems published in Leigh Hunt's Indicator did not find a place in the 1820 volume, and the reason for this is that Keats and his publishers did not want to give the reviewers any occasion for linking Keats's new work with the politically sensitive name of Leigh Hunt."

Structure
1. "Like many of Keats's 1819 odes, the structure of the poem is that of an odal hymn.[9]"

Source: Bate 1963 p. 499 "One was the odal hymn, of which he 'Ode on Melancholy' and the later 'To Autumn' are triumphant examples."

2. "While the earlier 1819 odes perfected techniques and allowed for variations that appear within "To Autumn", Keats dispenses with some aspects of the previous poems (such as the narrator) and ensures that the poem deals only with concrete concepts. There is no dramatic movement in "To Autumn" as there is in the earlier poems, and the poem attempts to discuss the poetic process without a progression of the temporal scene, an idea that Keats termed as "stationing".[10] "

Source: Bate 1963 pp. 581–582 Derived from this on p. 581: "Most of what Keats had developed in the structure of the ode stanza the previous April and May reappears effortlessly now... There is only one new variation, simple but altogether appropriate: the ode stanza is given a more prolonged effect." and "The poet himself is completely absent; there is no 'I,' no suggestion of the discursive language that we find in the other odes; the poem is entirely concrete, and self-sufficient in and through its concreteness. But if dramatic debate, protest, and qualification are absent, it is not because any premises from which they might proceed are disregarded but because these premises are being anticipated and absorbed at each step."

The rest is derived from this on p. 582: "These resolutions are attained partly through still another one to which Keats's poetry has so often aspired: a union of process and stasis (or what Keats has called 'station')."

3. "Some of the language of the poem resembles phrases found in earlier poems Keats had written and there are similarities between the lines of "To Autumn" and lines in poems such as Endymion, Sleep and Poetry, and Calidore.[11]"

Source: Ridley 1933 pp. 283–285 These pages lists lines and how they are similar to previous poems. An example on p. 284: "Then there is ' Sleep quiet with his poppy coronet' in Sleep and Poetry (348), and a passage in Endymion whih is interesting for a probably associative link"

4. "Keats relies heavily on monosyllabic words and consonantal sounds – especially bilabial consonants – along with an emphasis on long vowels to control the flow of the poem. His syntax lacks hiatus and there is only a single instance medial inversion of an accent within the poem. However, he does incorporate the Augustan inversion (a reversal of an accent at the beginning of a line) approximately 4.2% of the time. Within his measure, Keats incorporates spondees in approximately 13.9% of his verses. The rhyme follows a pattern of starting with a Shakespearian ABAB pattern which is followed by CDEDCCE rhyme scheme. The verse differentiates itself from his previous odes through use of 11 line stanzas, instead of 10, with a couplet placed before the concluding line of each stanza.[12]"

Source: Bate 1962 pp. 182–184 The first part is derived from p. 182: "The diction, like that of the other odes, is almost monosyllabic, strong in consonantal body, and English in origin" and "Bilabial consonants, which Keats had so commonly employed in Hyperion, the Eve of St. Agnes, and other odes, are equally abundant now: 'Drows'd with the fumme of poppies'; 'Or by a cider-press with patient look'..."

The next is derived from p. 183: "'Long' vowels are dominant, and spondai feet are drawn upon as never before (13.9%), except in the Grecian Urn and the Ode on Melancholy. Senses other that that of sight are once again appealed to, as in the inspired alteration from 'Drows'd with red poppies' (17) to 'Drows'd with the fume of poppies'; and Keats's former happy preference for the passive verbal participle as epithet is once again given free play, as in the translation from While a gold cloud (25) to 'While barred'' clouds'..."

More pp. 183-184: "Rigorous structural care is once again apparent at every hand: hiatus is non-existent; medial inversion of accent occurs only one; the strict Augustan device of initial inversion of accent alone is relied upon for variety, and is employed more frequently (4.2%) than in any other lyric of Keats except Ode to a Nightingale; and an even more severely orthodox distribution of pause is employed than in the previous odes. It is enough to add that the stanza of the ode differs from that of the earlier ones in consisting of eleven rather than ten lines, and in introducing a couplet before the concluding line. The former stanza, it will be remebered, consisted of what amounted to a quatrain from the Shakespearian octave, abab, follwed, in the main, by a strictly Petrarchan sestet, cdecde. The rhyme-scheme of Autumn is abab and, in the first stanza, cde dcce; in the other two cde cdde."

The line with "lacks hiatus" is derived from [the quote "hiatus is non-existent" which has been altered by Amandajm to suggest that there is hiatus in the poem.

5. "Between the manuscript version and the published version of "To Autumn" Keats tightened the langague of the poem. One of Keats's changes emphasized by critics is the change in line 17 of "Drows'd with red poppies" to "Drows'd with the fume of poppies", which emphasizes the sense of smell instead of sight. The later edition relies more on passive, past participles, as apparent in the change of "While a gold cloud" in line 25 to "While barred clouds".[13"

Source: Bate 1962 p. 183 Derived from: "Senses other that that of sight are once again appealed to, as in the inspired alteration from 'Drows'd with red poppies' (17) to 'Drows'd with the fume of poppies'; and Keats's former happy preference for the passive verbal participle as epithet is once again given free play, as in the translation from While a gold cloud (25) to 'While barred'' clouds'..." "

6. "Other changes involve the strengthening of phrases, especially within the transformation of the phrase in line 13 "whoever seeks for thee may find" into "whoever seeks abroad may find". Many of the lines within the second stanza were completely rewritten, especially those which did not fit into a rhyme scheme. Some of the minor changes involved adding punctuation missing from the original manuscript copy and altering capitalisation changes between the versions.[14]"

Source: Ridley 1933 pp. 285–287 Ridley analyzes how the editions differ. p. 285 has the first example and says: "when he reaches the end of the third line, Keats alters, feeling also no doubt a kind of thin abruptness in the half-line question, and a certain feebleness both of sound and sense in for thee". The second is commented on 286: "this has at least achieved a rhyme; but if the line about the sun is to disappear altogether the rhyme is in the wrong place... So Keats cancels the whole passage with some vigorous cross-hatching, and begins all over again using the re-written sixth line as the fifth".

The rest is derived from p. 287: "The copy in the Woodhouse letter omits to notice the cancellation of the s of stores; corrects some spellings, but writes Stready for Steady; does some punctuating; reads a brook for the brook, and Dased for Dos'd, either an easy misreading of a word so written that it might be either, or a deliberate alteration; and greatly acentuates the opiate z sound of the last line by reading oozings for oozing."

Poem
1. "The first stanza of the poem describes natural processes, unlike the following which deal more with sensual observations, as it presents a harvest in its final stages."

Source: Bloom 1971 p. 432 "The first stanza is natural process; the remaining two stanzas are sensuous observations of the consequences of that process: first, sights of the harvest in its final stages; then, post-harvest sounds, heralding the coming-on of winter. The sequence of the three stanzas then is pre-harvest ripeness, late-harvest repletion, and post-harvest natural music." Any similarity is in the tight statement of fact.

2. "The Stanza provides a union of maturation and growth, two oppositional forces within the work, and this union instills an idea within nature that the season will not end:[16]"

Source: Bate 1963 p. 582 Derived from: "Each of the three stanzas concentrates on a dominant, even archetypal, aspect of autumn, but, while doing so, admits and absorbs its opposite. The theme of the first is ripeness, of growth now reaching its climax beneath the 'maturing sun,' as the strain of the weighty fruit bends the apple tree and loads the vines.... Yet growth is still surprisingly going on, as autumn and the sun conspire 'to set budding more...' and as the bees are deceived into feeling that summer will never end."

3. "The second stanza reverses the images of the first stanza and describes the process of harvesting. Autumn, a harvester, is not actually harvesting but exists in a stasis. Only near the end of the stanza is there movement:[16]"

Source: Bate 1963 p. 582 Derived from: "If, in the first stanza, we find process continuing with a context of stillness and attained fulfillment, in the second--which is something of a reverse or mirror image of the first--we find stillness where we expect process. For now autumn is conceived as a reaper or harvester. Yet it is a harvest that is not harvesting. This benevolent deity is at first motionless... Movement begins only in the latter part of the stanza. Even then it is only suggested int he momentary glimpses of the figure of the gleaner..."

4. "Within the final moments of the poem, there is an introduction of the harvest and Autumn is manifested in the role of a harvester. The end approaches within the final moments of the song and death is slowly approaching alongside of the end of the year. However, Autumn is replaced by an image of life in general, and the songs of autumn becomes a song about life in general:[17]"

Source: Bate 1963 pp. 582–583 "in what follows is the withdrawal of autumn, the coming death of the year, and of course the familiar archetypal relevance of the association of our feeling of sequence in our own lives." and "the procedure now is almost completely indirect and left solely to inference... autumn is replaced by the concrete images of life, and of life unafflicted by any thought of death: the gnats, the hedge crickets, the redbreast. Moreover, it is life that an exist in much the same way at other times than autumn. Only two images are peculiar to the season--the 'stubble plains,' and the 'full-grown lambs.' The mind is free to assoiate the wailful mourning of the gnats with a funeral dirge for the dying year, but the sound is no more confined to autumn alone than is the 'soft-dying' of any day"

Themes
1. ""To Autumn" is thematically connected to many of Keats's 1819 odes. For example, his "Ode to Melancholy" introduces the acceptance of the process of life, and the concept is taken up again within "To Autumn".[18]"

Source: Bate 1962 p. 522 (Source is really Bate 1963, this was later corrected). Derived from a page following analysis of "Ode to Melancholy": "Anticipating the ode 'To Autumn' of four months later, the second stanza then turns directly to the vivid acceptance of process. In the very springing of the flowers and the new green of the hill..."

2. "There is a union between the ideal and the real which leads to fulfillment. Of all of Keats's poems, "To Autumn" most closely describes an actual paradise while focusing on the archetypal images that are connected with autumn. Within the poem, the season of autumn represents the growth, the maturation, and finally an approaching death.[19]"

Source: Bate 1963 pp. 581–583 Derived from: "The result... is also a successful union of the ideal--of the heart's desire--and reality; of the 'greeting of the Spirit' and its objet. What the heart really wants is being found... Here at least is something of a genuine paradise, therefore. It even has its deity--a benevolent deity that wants not only to 'load and bless'... but also to 'spare,' to 'set budding more.' And yet all this is put with concrete exactness and fidelity." The rest come from use of "Eah of the three stanzas oncentrates on a dominant, even arhetypal, aspet of autumn, but, while doing so, admits and absorbs its opposite. The theme of the first is ripeness, of growth now reahing its climax". Use of "mature" and "end is approaching" are found also scattered in the passage.

3."The poem also defends art's role in helping society in a manner similar to Keats's "Ode on Indolence" and "Ode to Psyche". "To Autumn" describes a system in which nature and culture are two separate parts of the universe, and nature is turned into culture by an artist. Civilization is furthered by man's ability to use nature for agricultural cultivation. The artist, like the farmer, has to process nature into a consumable object, which in turn allows people sustenance. The end of the poem is joined in song as nature gives way to civilization, which represents the self-sacrificing of both nature and the artist for society.[20]"

Source: Vendler 1988 pp. 124–125 "Finally, in the ode 'To Autumn,' Keats finds his most comprehensive and adequate symbol for the social value of art. He does this by playing, in this ode, two roles at once. Once again, as in the 'Ode on Indolence' and the 'Ode to Psyche,' he will be playing the role of the artist, the dreamer indolent in reverie on the bedded grass or the gardener Fancy...." and "In 'To Autumn,' in his final understanding of the social function of art, Keats chooses nature and culture as the two poles of his symbolic system. He sees the work of the artist as the transformation of nature into culture, the transmutation of the teeming fields into the garnered grain... Since civilization itself arose from man's domininion over nature, the process of nature by agriculture became the symbol in Greece of the most sacred mysteries."

More on page. 125: "Keats's autumn ode takes as its allegory for art the making of nature into nurtuer. The artist, with reaping hook, gleaning basket, and cider press, denudes nature, we may say, but creates food. We cannot, so to speak, drink apples or eat wheatl we an only consume processed nature... Since the artist in his own teeming field, art, in this allegory, is a process of self-immolation." and "Keats is the audience for the artist-goddess's sarifice of herself into food, as she passes from areless girl through ample maternity and into her own death vigil.. nature has become culture."

4. "The three stanzas of "To Autumn" are able to suggest both a movement from summer to early winter and also day turning into dusk. This progression is joined with a shift from the sensation of touch to sight and then to sound, creating a three part symmetry which is missing in Keats's other odes.[21]"

Source: Sperry 1973 p. 337 Derived from: "As critics have often pointed out, the three stanzas successively proceed from the last growth of late summer through the fullness of high autumn to the spareness of an early winter landscape, just as they suggest the progress of a single day through to its close in sunset. As Bush, among others, has noted, the imagery of the first stanza is mainly tatile, that of the second mainly visual, that of the last hiefly auditory. in these and other respets the ode displays a deliberate symmetry and balance the earlier odes do not possess."

5. "Although there is process and the suggestion of motion within the poem, there is a lack of action. Within the second stanza, autumn is described through metaphor as an exhausted labourer in lines 14–15. Near the end of the stanza, the steadiness of the gleaner in lines 19–20 emphasizes a motionlessness within the poem. The individuals are burdened or merely watch the events surrounding them. The poem as a whole creates within the imagination an image of death and a finality that is welcomed. There are no contrary ideas that are common within the other odes of 1819. Instead, "To Autumn" puts forth the idea that progression is no longer necessary as maturation has taken over, and growth and death are in harmony.[22]"

Source: Bloom 1968 pp. 95–97 Derived from: "As the second stanza of To Autumn opens, we see Autumn already 'amid' her store. The promised overabundance of the first stanza has been fulfilled... Autumn is no longer active process, but a female overcome by the fragrance and soft exhaustion of her own labor. She is passive, an embodiment of the earthly paradise" and "The final four lines of the stanza takes us to the very end of harvest, the gleaner bearing her laden head so steadily as to suggest motionlessness even as she moves, which further suggests he running-down to stasis of a process. Finally, we are shown the girl patiently watching, hours by hours, the meaningful sameness of the 'cyder-press' with its final oozings, the last wealth of complete process itself."

More on page. 96: "Winter descends here as a man might hope to die, with a natural sweetness, a natural movement akin to the extended wings of Stevens' pigeons or the organizing songs of Keats's swallows as they gather together for flight beyond winter. The day dies soft in this great stanza" and continues on page 97 with: "close the poem, whih has climaxed in an acceptance of process beyond the possibility of grief. The last seven lines are all sound; natural music so varied and intense as to preclude even natural lament. We feel that we might be at the end of tragedy or epi, having read only a short ode."

6. "Along with this harmony, the placing of the couplet before the end of each stanza creates a suspension of closing within the poem. This suspension within the poem reinforces the theme of continuation.[23]"

Source: Wagner 1996 pp. 110–111 "The creation of a sort of penultimate couplet has, as a strategy, a familiar feel to it. The new ode form grapples with the old problem of closure in the sonnet by bringing here a haunting reminder of the couplete-closure--and suspending it. 'Suspension' in previous Keas sonnets has meant something like a 'gesture of incompleteness'; here, it means something else: a 'principle of continuation.'"

7. "In a 1979 essay, Jerome McGann argued that while the poem was indirectly influenced by historical events, Keats had deliberately ignored the political landscape of 1819.[24"

Source: McGann 1979 pp. 988–1032 - summary of a long essay.

8. "Countering this view, Andrew Bennett, Nicholas Roe and others focused more on the political aspects of the poem, Roe arguing for a direct connection to the Peterloo Massacre of 1819.[25]"

Source: Strachan 2003 p. 175 - this contains a summary of works and short excerpts. Strachan says: "says McGAnn, is 'great' but 'politically reactionary', and 'To Autumn' 'attempt[s] to 'escape' the period whih provides the poem with its context.' This argument initiated a series of historical interpretations of 'To Autumn' (by such critics as Paul Fry, Andrew Bennett, and Nicholas Roe), many of which repudiate MGann's position and attempt to read the poem in explicitly political terms... Bennet's Keats, Narrative, and Audience argues that the .... while Roe's 'Keats's ommonwealth' offers a riposte to McGann which reads... in the light of the 'discourses of political and social justice after the outrage at Peterloo'."

9. "Later, Paul Fry further argued against McGann's stance when he pointed out, "It scarcely seems pertinent to say that 'To Autumn' is therefore an evasion of social violence when it is so clearly an encounter with death itself [...] it is not a politically encoded escape from history reflecting the coerced betrayal [...] of its author's radicalism. McGann thinks to rescue Keats from the imputation of political naïveté by saying that he was a radical browbeaten into quietism".[26]"

Source: Fry 1995 pp. 123–124 - this is a direct quote and attributed.

10. "In regards to other political aspects, post-colonial critic Alan Bewell interpreted the themes of Keats's ode in the context of British imperialism. He claimed "To Autumn" promoted the moderate climate of Britain over tropical climates.[27]"

Source: Bewell 2008 pp. 635–638 - Summary of statements. Example from 635: "Yet at a time when colonialism had made apparent the connection between health and climate, the seasonal cycle of spring, summer, fall, and winter was not taken for granted." Another on page 636: "More successfully than Hyperion, 'To Autumn' enacts a curing of space by tempering pathogenic extremes."

Thanks
Thanks for this helpful contribution. I have copied it over - I hope that was what you intended. Elen of the Roads 22:01, 17 November 2010 (UTC)


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Wikiversity:Recusal

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title: Wikiversity:Recusal


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