User talk:Egm6321.f10.team5.oh/hw6

=Problem 1 - Showing Using Variation of Parameters = From (meeting 32 page 3)

Find
Show that (1.2) using the variation of parameters method.

Author and proof-reader
[Author]

[Proof-reader]

=Problem 2 - Solve Nonhomogeneous L2_ODE_VC= From (meeting 32 page 3)

Author and proof-reader
[Author]

[Proof-reader]

=Problem 3 - Solve Nonhomogeneous Legendre Equation= From (meeting 33 page 1)

Author and proof-reader
[Author]

[Proof-reader]

=Problem 4 - Solve Axisymmetric Problem= From (meeting 34 page 2)

Author and proof-reader
[Author]

[Proof-reader]

=Problem 5 - Circular Cylindrical Coordinate Problem= From (meeting 35 page 3)

Given

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

\begin{array}{*{20}l} x_1 & = & r \cos \theta & = & \xi_1 \cos \xi_2 \\ x_2 & = & r \sin \theta & = & \xi_1 \sin \xi_2 \\ x_3 & = & z & = & \xi_3 & \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5.1)
 * }
 * }

Find
1. Find $$\displaystyle \{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ in terms of $$\displaystyle \{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ and $$\displaystyle \{{\rm d} \xi_k\} $$

2. Find $$\displaystyle {\rm d} s^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$. Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$

3. Find $$\displaystyle \Delta \psi $$ in cylin. coord.

Part 1. Finding { $$\displaystyle dx_i$$ }
Background knowledge :  Cylindrical Coordinates


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$$\displaystyle \begin{array}{*{20}l} {\rm d}x_1 & = & \cos \xi_2 \cdot {\rm d}\xi_1 - \xi_1 \cdot \sin \xi_2 \cdot {\rm d} \xi_2 \\
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 * style="width:2%; padding:10px; border:2px solid #8888aa" |

{\rm d}x_2 & = & \sin \xi_2 \cdot {\rm d}\xi_1 + \xi_1 \cdot \cos \xi_2 \cdot {\rm d} \xi_2 \\

{\rm d}x_3 & = & {\rm d} \xi_3 \end{array} $$
 * }
 * }
 * }


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 * }
 * }
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===Part 2. Finding $$\displaystyle ds^2$$, Identify { $$\displaystyle h_i$$ }=== 1. Finding $$\displaystyle ds^2$$


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ds^2 = (dx_1)^2 + (dx_2)^2 + (dx_3)^2 $$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle = \underbrace{\cos^2 \cdot \xi_2 \cdot (d \xi_1)^2 - 2 \cdot \xi_1 \cdot \sin \xi_2 \cdot \cos \xi_2 \cdot d \xi_1 \cdot d \xi_2 + \xi_1^2 \cdot \sin^2 \xi_2 \cdot (d \xi_2)^2}_{(dx_1)^2} $$

$$\displaystyle + \underbrace{\sin^2 \xi_2 \cdot (d \xi_1)^2 + 2 \cdot \xi_1 \cdot \sin \xi_2 \cdot \cos \xi_2 \cdot d \xi_1 \cdot d \xi_2 + \xi_1^2 \cdot \cos^2 \xi_2 \cdot (d \xi_2)^2}_{(dx_2)^2} $$

$$\displaystyle + \underbrace{(d \xi_3)^2}_{(dx_3)^2} $$
 * }
 * }
 * }

Therefore,


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$$\displaystyle ds^2 = (d \xi_1)^2 + \xi_1^2 \cdot (d \xi_2)^2 + (d \xi_3)^2 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
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2. Identify { $$\displaystyle h_i$$ }


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ds^2 = (h_1)^2(d \xi_1)^2 + (h_2)^2(d \xi_2)^2 + (h_3)^2(d \xi_3)^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }

According to the lecture [http://upload.wikimedia.org/wikiversity/en/a/ab/2010_11_04_14_51_31.djvu Mtg. 34]

{$$\displaystyle h_i $$} are magnitude of tangent vetors to curv. coord.

So, we just select positive part of them.

Therefore,


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$$\displaystyle h_1 = 1, \quad h_2 = \xi_1, \quad h_3 = 1 $$
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 * }
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Author and proof-reader
[Author]

[Proof-reader]

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

Author and proof-reader
[Author]

[Proof-reader]

=Problem 7 - elliptic Coordinate Problem= From (meeting 36 page 1)

Author and proof-reader
[Author]

[Proof-reader]

=Problem 8 - Verification genernal form of Legendre poly.= From (meeting 36 page 2)

Given

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P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8-1)
 * }
 * }


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

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8-2)
 * }
 * }


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P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8-3)
 * }
 * }


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

P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8-4)
 * }
 * }


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

P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8-5)
 * }
 * }

Find
Verify that (8-1) - (8-5) can be written as


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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 (8-6)
 * }
 * }

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 (8-7)
 * }
 * }

$$\displaystyle [\frac{n}{2}]$$ = interger part of  $$ \displaystyle \ \frac{n}{2} $$

Solution
Background knowledge :  Factorial , Floor and ceiling functions

We tried to discuss about background knowledges and how to verify clearly and efficiently

In this time, we verified using both equations such that $$\displaystyle (8-6) $$ and $$\displaystyle (8-7) $$

Part 1. Verification using $$\displaystyle (8-6)$$
1. In case of $$\displaystyle n = 0 $$


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P_0(x) = \sum_{i=0}^{[0/2]} \frac{(-1)^i (2 (0) - 2 i)! x^{0 - 2i}}{2^0 i! (0-i)! (0-2i)!} \ = \sum_{i=0}^{0} \frac{(-1)^i ( - 2 i)! x^{- 2i}}{i! (-i)! (-2i)!} \ = \frac{(-1)^0 \cdot 0! \cdot x^0}{0! \cdot 0! \cdot 0!}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_0(x) = 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

2. In case of $$\displaystyle n = 1 $$


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P_1(x) = \sum_{i=0}^{[1/2]} \frac{(-1)^i (2 (1) - 2 i)! x^{1 - 2i}}{2^1 i! (1-i)! (1-2i)!}= \sum_{i=0}^{0} \frac{(-1)^i \ (2 - 2 i)! \ x^{1 - 2i}}{2 \ i! \ (1-i)! \ (1-2i)!}= \frac{(-1)^0 \cdot 2! \cdot x^1}{2 \cdot 0! \cdot 1! \cdot 1!} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_1(x) = x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

3. In case of $$\displaystyle n = 2 $$


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P_2(x) = \sum_{i=0}^{[2/2]} \frac{(-1)^i (2 (2) - 2 i)! x^{2 - 2i}}{2^2 i! (2-i)! (2-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i \ (4 - 2 i)! \ x^{2 - 2i}}{4 \ i! \ (2-i)! \ (2-2i)!}= \underbrace{\frac{(-1)^0 \cdot 4! \cdot x^2}{4 \cdot 0! \cdot 2! \cdot 2!}}_{\frac{3}{2}x^2} \ + \ \underbrace{\frac{(-1)^1 \cdot 2! \cdot x^0}{4 \cdot 1! \cdot 1! \cdot 1!}}_{-\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_2(x) = \frac{1}{2}(3x^2-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

4. In case of $$\displaystyle n = 3 $$
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P_3(x) = \sum_{i=0}^{[3/2]} \frac{(-1)^i (2 (3) - 2 i)! x^{3 - 2i}}{2^3 i! (3-i)! (3-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i (6 - 2 i)! x^{3 - 2i}}{8 \ i! \ (3-i)! \ (3-2i)!} = \underbrace{\frac{(-1)^0 \cdot 6! \cdot x^{3}}{8 \cdot 0! \cdot 3! \cdot 3!}}_{\frac{5}{2}x^3} + \underbrace{\frac{(-1)^1 \cdot 4! \cdot x^1}{8 \cdot 1! \cdot 2! \cdot 1!}}_{-\frac{3}{2}x}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_3(x) = \frac{1}{2}(5x^3-3x) $$
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 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
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 * }
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5. In case of $$\displaystyle n = 5 $$
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P_4(x) = \sum_{i=0}^{[4/2]} \frac{(-1)^i (2 (4) - 2 i)! x^{4 - 2i}}{2^4 i! (4-i)! (4-2i)!}= \sum_{i=0}^{2} \frac{(-1)^i \ (8 - 2 i)! \ x^{4 - 2i}}{16 \ i! \ (4-i)! \ (4-2i)!}$$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle = \underbrace{\frac{(-1)^0 \cdot 8! \cdot x^4}{16 \cdot 0! \cdot 4! \cdot 4!}}_{\frac{35}{8}x^4} \ + \ \underbrace{\frac{(-1)^1 \cdot 6! \cdot x^{2}}{16 \cdot 1! \cdot 3! \cdot 2!}}_{- \frac{15}{4}x^2} \ + \ \underbrace{\frac{(-1)^2 \cdot 4! \cdot x^0}{16 \cdot 2! \cdot 2! \cdot 0!}}_{\frac{3}{8}}$$
 * }
 * }
 * }


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

$$\displaystyle P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
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 * }
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Part 2. Verification using $$\displaystyle (8-7)$$
1. In case of $$\displaystyle n = 0 $$
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P_0(x) = \sum_{i=0}^{[0/2]} \frac{1}{2^i \ i! \ (-2i)!} (-1)^i \ x^{-2i} = \sum_{i=0}^{0} \frac{1}{2^0 \cdot 0! \cdot 0!} (-1)^0 \cdot x^0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_0(x) = 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

2. In case of $$\displaystyle n = 1 $$
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P_1(x) = \sum_{i=0}^{[1/2]} \frac{1}{2^i \ i! \ (1-2i)!} (-1)^i \ x^{1-2i} = \sum_{i=0}^{0} \frac{1}{2^0 \cdot 0! \cdot 1!} (-1)^0 \cdot x^1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_1(x) = x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

3. In case of $$\displaystyle n = 2 $$
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P_2(x) = \sum_{i=0}^{[2/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 4-2i-1}{2^i \ i! \ (2-2i)!} (-1)^i \ x^{2-2i} = \underbrace{\frac{1 \cdot 3}{2^0 \cdot 0! \cdot 2!} (-1)^0 \cdot x^2}_{\frac{3}{2}x^2} + \underbrace{\frac{1}{2^1 \cdot 1! \cdot 0!} (-1)^1 \cdot x^0}_{- \frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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$$\displaystyle P_2(x) = \frac{1}{2}(3x^2-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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 * }
 * }
 * }

4. In case of $$\displaystyle n = 3 $$
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P_3(x) = \sum_{i=0}^{[3/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 6-2i-1}{2^i \ i! \ (3-2i)!} (-1)^i \ x^{3-2i} = \underbrace{\frac{1 \cdot 3 \cdot 5}{2^0 \cdot 0! \cdot 3!} (-1)^0 \cdot x^3}_{\frac{5}{2}x^3} + \underbrace{\frac{1 \cdot 3}{2^1 \cdot 1! \cdot 1!} (-1)^1 \cdot x^1}_{- \frac{3}{2}x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_3(x) = \frac{1}{2}(5x^3-3x) $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

5. In case of $$\displaystyle n = 5 $$
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P_4(x) = \sum_{i=0}^{[4/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 8-2i-1}{2^i \ i! \ (4-2i)!} (-1)^i \ x^{4-2i}$$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle \ = \underbrace{\frac{1 \cdot 3 \cdot 5 \cdot 7}{2^0 \cdot 0! \cdot 4!} (-1)^0 \cdot x^4}_{\frac{35}{8}x^4} + \underbrace{\frac{1 \cdot 3 \cdot 5}{2^1 \cdot 1! \cdot 2!} (-1)^1 \cdot x^2}_{- \frac{15}{4}x^2} + \underbrace{\frac{1 \cdot 3}{2^2 \cdot 2! \cdot 0!} (-1)^2 \cdot x^0}_{\frac{3}{8}} $$
 * }
 * }
 * }


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

$$\displaystyle P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
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 * }
 * }
 * }

Author and proof-reader
[Author] [http://en.wikiversity.org/wiki/User_talk:Egm6321.f10.team5.oh/hw6#Problem_8_-_Verification_genernal_form_of_Legendre_poly. Oh, Sang Min]

[Proof-reader]

=Problem 9 - Verification solution of Legendre eq.= From (meeting 36 page 3)

Given

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P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-1)
 * }
 * }


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

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-2)
 * }
 * }


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

P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-3)
 * }
 * }


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

P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-4)
 * }
 * }


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

P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-5)
 * }
 * }

Find
Verify that $$\displaystyle (9-1) \ - \ (9-5)$$ are solution of Legendre equation
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9-6)
 * }
 * }

Solution
1. In case of $$\displaystyle n=0, \ (9-1)$$


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$$\displaystyle y = 1, \ y' = 0, \ y'' = 0$$
 * }
 * }
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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \cdot 0 - 2x \cdot 0 + 0 \cdot (0+1) \cdot 1 = 0 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
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 * }
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2. In case of $$\displaystyle n=1, \ (9-2)$$


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$$\displaystyle y = x, \ y' = 1, \ y'' = 0$$
 * }
 * }
 * }


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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \cdot 0 - 2x \cdot 1 + 1 \cdot (1+1) \cdot x = 0 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
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 * }
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3. In case of $$\displaystyle n=2, \ (9-3)$$


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$$\displaystyle y = \frac{1}{2} (3 x^2 - 1), \ y' = 3x, \ y'' = 3$$
 * }
 * }
 * }


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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = \underbrace{(1-x^2) \cdot 3 - 2x \cdot 3x + 2 \cdot (2+1) \cdot \frac{1}{2} (3 x^2 - 1)}_{(-3 - 6 + 9)x^2 + (3 -3)} = 0 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
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 * }
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 * }
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4. In case of $$\displaystyle n=3, \ (9-4)$$


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$$\displaystyle y = \frac{1}{2} (5 x^3 - 3x), \ y' = \frac{3}{2} (5 x^2 - 1), \ y'' = 15x$$
 * }
 * }
 * }


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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = \underbrace{(1-x^2) \cdot 15 x - 2 x \cdot \frac{3}{2} (5 x^2 - 1)+ 3 \cdot (3+1) \cdot \frac{1}{2} (5 x^3 - 3x)}_{15 x - 15 x^3 - 15x^3 + 3x + 30 x^3 - 18x} = 0 $$
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 * }
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5. In case of $$\displaystyle n=4, \ (9-5)$$


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$$\displaystyle y = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}, \ y' = \frac{35}{2} x^3 - \frac{15}{2} x, \ y'' = \frac{105}{2} x^2 - \frac{15}{2}$$
 * }
 * }
 * }


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

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \left(\frac{105}{2} x^2 - \frac{15}{2} \right) - 2 x \left(\frac{35}{2} x^3 - \frac{15}{2} x\right) + 4(4+1) \left(\frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}\right)$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle \ = \frac{105}{2} x^2 - \frac{15}{2} - \frac{105}{2} x^4 + \frac{15}{2} x^2 - 35 x^4 + 15 x^2 +  \left(\frac{175}{2} x^4 - 75 x^2 + \frac{15}{2}\right)$$

$$\displaystyle \ = (- \frac{105}{2} x^4 + \frac{175}{2} x^4 - \frac{70}{2} x^4) + (\frac{105}{2} x^2 + \frac{15}{2} x^2  + \frac{30}{2}x^2 - \frac{150}{2} x^2) + (\frac{15}{2}- \frac{15}{2}) = 0$$
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Therefore, $$\displaystyle (9-1) - (9-5) $$ are clearly solution of Legendre equation such that $$\displaystyle (9-6) $$
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Author and proof-reader
[Author] Oh, Sang Min

[Proof-reader]