User talk:Egm6321.f10.team5.oh/hw7

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

Author and proof-reader
[Author]

[Proof-reader]

=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
[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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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 $$ q $$

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

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

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

$$\displaystyle a_{3} = (11 + \frac{8}{3}) \cdot \frac{2}{5} \ = \ \frac{106}{15}$$

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

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

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

$$\displaystyle a_{1} = -5 + \frac{106}{5} - \frac{10}{7} \ = \ \frac{517}{35}$$

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

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

Therefore


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$$\displaystyle a_{0} = \frac{12}{5}, \ a_{1} = \frac{517}{35}, \ a_{2} = 2, \ a_{3} = \frac{106}{15}, \ a_4 = \frac{8}{5}, \ a_5 = \frac{16}{21} $$
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 2. Plot $$\displaystyle q$$

Author and proof-reader
[Author Oh, Sang Min

[Proof-reader]

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

Author and proof-reader
[Author]

[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 $$\displaystyle (5-4)$$'''

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} $$
 * $$\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
 * $$\displaystyle

\ = \ \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), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5-7)
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In this case, $$ 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
 * $$\displaystyle
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\ = \ x \tanh^{-1}(x) - 1 $$
 * $$\displaystyle
 * $$\displaystyle
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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)

Author and proof-reader
[Author]

[Proof-reader]

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

Author and proof-reader
[Author]

[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
 * <p style="text-align:right;">$$\displaystyle (8-1)
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\begin{pmatrix} r \\ k \end{pmatrix} = \frac{r(r-1) \cdot\cdot\cdot (r-k+1)}{k!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-2)
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(1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-3)
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where
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\alpha_i = \frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-4)
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Find
Use $$\displaystyle (8-1), \ (8-2)$$ to obtain $$\displaystyle (8-3), \ (8-4)$$

Solution
[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
 * <p style="text-align:right;">$$\displaystyle (8-5)
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We can expand using $$\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
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (8-7)
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Change $$\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
 * $$\displaystyle
 * <p style="text-align:right;">$$\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]

=References=

=Contributing members= Oh, Sang min