User:Egm6321.f10.team2.dube/HW7

=Problem 7.1=

From the lecture slide Mtg 37-1

Given
The first four legendre polynomials are given as
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$$  \displaystyle P_{0}(x)=1 $$     (Eq 1.1)
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$$  \displaystyle P_{1}(x)=x $$     (Eq 1.2)
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$$  \displaystyle P_{2}(x)=\frac{1}{2}\left ( 3x^2-1 \right ) $$     (Eq 1.3)
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$$  \displaystyle P_{3}(x)=\frac{1}{2}\left ( 5x^3-3x \right ) $$     (Eq 1.4)
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$$  \displaystyle P_{4}(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} $$     (Eq 1.5)
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Find
Plot$$\displaystyle \left \{ P_{0},P_{1},P_{2},P_{3} \right \} $$,$$\displaystyle\left \{ Q_{0},Q_{1},Q_{2},Q_{3} \right \}$$,$$\displaystyle \left \{ P_{0},Q_{0} \right \}$$....$$\displaystyle\left \{ P_{4},Q_{4} \right \}$$ and observe evenness and oddness.

Solution
Example
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$$  \displaystyle sin(-\varphi )=-sin(\varphi ) $$     (Eq 1.6) The sine function is an odd function. Notice how the graph of the sine function is rotation symmetric about the origin as it goes into itself under a rotation of 180 degrees. 'Odd functions possess this type of antisymmetry.'
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$$  \displaystyle cos(-\varphi )=cos(\varphi ) $$     (Eq 1.7) The cosine function is an even function. Notice how the graph of the cosine function is reflection symmetric about the vertical axis throught 0 degrees. 'All even functions possess this type of symmetry.'
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Legendre Polynomials


As we see from the graph $$\displaystyle P_{0}(x)$$,$$\displaystyle P_{2}(x)$$,$$\displaystyle P_{4}(x)$$ are symmetric about the vertical axis. They all possess the characteristic of even functions. On tht other hand $$\displaystyle P_{1}(x)$$ and $$\displaystyle P_{3}(x)$$ are symmetric about the origin. They posess the characteristic of odd functions.

Legendre Functions


Controversely from polynomials, as we see from the graph $$\displaystyle Q_{0}(x)$$,$$\displaystyle Q_{2}(x)$$,$$\displaystyle Q_{4}(x)$$ are symmetric about the origin. So they are odd functions. And same with same approach $$\displaystyle Q_{1}(x)$$ and $$\displaystyle Q_{3}(x)$$ are symmetric about the vertical axis. They are even functions.

Functions Vs. Polynomials


The graph shows that whenever for the same indices$$\displaystyle P_{i}(x)$$ is even then $$\displaystyle Q_{i}(x)$$ is odd and controversely whenever $$\displaystyle P_{i}(x)$$ is odd then $$\displaystyle Q_{i}(x)$$ is even. So if we look at the multiplication of two functions,


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$$  \displaystyle F(x)=P_{i}(x)*Q_{i}(x) $$     (Eq 1.8)
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is always be odd.For i=1,2,3,4,.....

For i =2k
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$$  \displaystyle F(-x)=P_{i}(-x)*Q_{i}(-x)=P_{i}(x)*[-Q_{i}(x)] $$     (Eq 1.9) For i=2k+1
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$$  \displaystyle F(-x)=P_{i}(-x)*Q_{i}(-x)=[-P_{i}(x)]*Q_{i}(x) $$     (Eq 1.10)
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$$  \displaystyle \int P_{i}(x)Q_{i}(x)\Rightarrow even $$     (Eq 1.11)
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From the integral of multiplication of two functions we have even functions. And for integral boundries +1 and -1 These integrals will be same and their difference will be zero.
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$$  \displaystyle \int_{-1}^{1}P_{i}(x)Q_{i}(x)=R(1)-R(-1)=0 $$     (Eq 1.12) Where R is an even function.
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=Problem 7.2=

From the lecture slide Mtg 38-1

Given
In meeting 38-1 the equation is given as


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$$  \displaystyle f=\sum_{i}g_{i } $$     (Eq 2.1) Where f is a linear combination of g(x).
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Find
Show that Eq(2.1) is odd when gi is odd and Eg(2.1) is even when gi is even.

Solution
For even functions if we change the independent variable x into -x the relationship between two functions can be written by definition


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$$  \displaystyle f(x)=f(-x) $$     (Eq 2.2)
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And for odd functions if we change the independent variable x into -x the relationship for odd functions by definition


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$$  \displaystyle f(-x)=-f(x) $$     (Eq 2.3)
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So now we can expand the Eq(2.1) as


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$$  \displaystyle f(x)=g_{1}(x)+g_{2}(x)+g_{3}(x)+............. $$     (Eq 2.4)
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Then we can change the independent variable with its minus sign
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$$  \displaystyle f(-x)=g_{1}(-x)+g_{2}(-x)+g_{3}(-x)+............. $$     (Eq 2.5)
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Case 1
If $$\displaystyle g(x)$$ is even then by definition $$\displaystyle g(x)=g(-x)$$ so we can write
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$$  \displaystyle f(-x)=g_{1}(x)+g_{2}(x)+g_{3}(x)+............. $$     (Eq 2.6) Where the RHS of the function is the same as Eq(2.4). Hence f is an even function.
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Case 2
And If $$\displaystyle g(x)$$ is an odd function then $$\displaystyle g(x)=-g(x)$$. Substituting into Eq(2.2) yields;
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$$  \displaystyle f(-x)=-g_{1}(x)-g_{2}(x)-g_{3}(x)+............. $$     (Eq 2.7)
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Where the RHS of the functions is minus f. So $$\displaystyle f(x)=-f(x)$$ when g(x) is odd then f(x) is odd.

=Problem 7.3=

From the lecture slide Mtg 38-1

Given

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$$  \displaystyle P_{n}(\mu )=\sum_{i=0}^{[n/2]}(-1)^i\frac{(2n-2i)!\mu^{n-2i}}{2^ni!(n-i)!(n-2i)!} $$     (Eq 3.1) or
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$$  \displaystyle P_{n}(\mu )=\sum_{i=0}^{[n/2]}\frac{1,3,...(2n-2i-1)}{2^ii!(n-2i)!}(-1)^i\mu^{n-2i} $$     (Eq 3.2) where $$ \displaystyle [n/2] $$ = integer part of $$ \displaystyle n/2 $$ and
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$$  \displaystyle q(x)=\sum_{i=0}^{4}c_ix^i $$     (Eq 3.3) where $$ \displaystyle c_0=2, c_1=-5, c_2=-3, c_3=11, c_4=7, c_5=6 $$
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Find
Show that
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$$  \displaystyle P_{2k}(\mu) $$     (Eq 3.4) is even ($$ \displaystyle k=0,1,2... $$) and
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$$  \displaystyle P_{2k+1}(\mu) $$     (Eq 3.5) is odd.
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Find $$ \displaystyle {a_i} $$ such that $$ \displaystyle q(x)=\sum_{i=0}^{5}a_iP_i(x) $$

and plot $$ \displaystyle q=\sum_{i}^{}c_ix^i=\sum_{i}^{}a_iP_i $$

Solution
An equation is even when $$ \displaystyle f(x)=f(-x) $$.

Taking Equation 3.1 and plugging in $$ \displaystyle n=2k $$ we get:
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$$  \displaystyle P_{2k}(\mu )=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!\mu^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.6) Now let us plug in $$ \displaystyle \mu=x $$ and $$ \displaystyle \mu=-x $$ and compare $$ \displaystyle P_{2k}(x) $$ and $$ \displaystyle P_{2k}(-x) $$.
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$$  \displaystyle P_{2k}(x)=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!x^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.7) and
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$$  \displaystyle P_{2k}(-x)=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!(-x)^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.8) As you can see, $$ \displaystyle x $$ and $$ \displaystyle -x $$ are both getting raised to an even power, so they will both be equal and satisify the even clause, $$ \displaystyle f(x)=f(-x) $$.
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An equation is odd when $$ \displaystyle f(-x)=-f(x) $$.

Taking Equation 3.1 and plugging in $$ \displaystyle n=2k+1 $$ we get:
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$$  \displaystyle P_{2k+1}(\mu )=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!\mu^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.9) Now let us plug in $$ \displaystyle \mu=x $$ and $$ \displaystyle \mu=-x $$ and compare $$ \displaystyle P_{2k+1}(x) $$ and $$ \displaystyle P_{2k+1}(-x) $$.
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$$  \displaystyle P_{2k+1}(x)=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!x^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.10) and
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$$  \displaystyle P_{2k+1}(-x)=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!(-x)^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.11) As you can see, $$ \displaystyle P_{2k+1}(x) $$ and $$ \displaystyle P_{2k+1}(-x) $$ satisfy the odd clause, $$ \displaystyle f(-x)=-f(x) $$, therefore they are odd.
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We can see from (36-2) that
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$$ \displaystyle P_0(x)=1 $$ (Eq 3.12)
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$$ \displaystyle P_1(x)=x $$ (Eq 3.13)
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$$ \displaystyle P_2(x)=\frac{3x^2}{2}-\frac{1}{2} $$ (Eq 3.14)
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$$ \displaystyle P_3(x)=\frac{5x^3}{2}-\frac{3}{2}x $$ (Eq 3.15)
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$$ \displaystyle P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} $$    (Eq 3.16)
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$$ \displaystyle P_5(x)=\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x $$    (Eq 3.17)
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We know that $$ \displaystyle P_i(x)=A*x_i $$ so:
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$$ \displaystyle \begin{bmatrix}P_0 \\P_1 \\ P_2 \\ P_3 \\ P_4 \\ P_5 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 & 0\\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} & 0\\ 0 & \frac{15}{8} & 0 & -\frac{35}{4} & 0 & \frac{63}{8} \end{bmatrix} \begin{bmatrix}1 \\ x \\ x^2 \\ x^3 \\ x^4 \\ x^5
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\end{bmatrix} $$    (Eq 3.18)
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We also know that $$ \displaystyle q(x)=a_iP(x)=(c_i)^Tx^i $$.

If $$ \displaystyle P_i(x)=Ax^i $$,

then

$$ \displaystyle (a_i)^TAx_i=(c_i)^Tx_i $$

and

$$ \displaystyle (a_i)^T=A^{-1}(c_i)^T $$.

We know from (Mtg38-3) that:
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$$ \displaystyle c_i=\begin{bmatrix}2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{bmatrix} $$ (Eq 3.19)
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and
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$$ \displaystyle A^{-1}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 & 0\\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} & 0\\ 0 & \frac{27}{63} & 0 & \frac{28}{63} & 0 & \frac{8}{63} \end{bmatrix} $$    (Eq 3.20) Therefore:
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$$ \displaystyle (a_i)^T=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 & 0\\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} & 0\\ 0 & \frac{27}{63} & 0 & \frac{28}{63} & 0 & \frac{8}{63} \end{bmatrix} \begin{bmatrix}2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{bmatrix}^T $$    (Eq 3.21)
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$$ \displaystyle a_i=\begin{bmatrix}\frac{12}{5} \\ \frac{146}{35} \\ 2 \\ \frac{106}{15} \\ \frac{8}{5} \\ \frac{16}{21} \end{bmatrix} $$ (Eq 3.22) <div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #FF00FF; background-color: #C0C0C0; text-align: left;"> This problem was solved by referencing [| Problem #11].
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= Problem 7.4 =

From the lecture slide Mtg 38-3

Given

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$$ \displaystyle

f(\theta )={{T}_{0}}{{\cos }^{6}}\theta $$

(Eq 4.1)
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$$ \displaystyle

f(\theta )={{T}_{0}}\exp \left[ -{{(\frac{2\theta }{\pi })}^{2}} \right]$$

(Eq 4.2)
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Find
i) Without calculation find property of $$\displaystyle {{A}_{n}}$$


 * $$\displaystyle \left. \begin{align}

& {{A}_{2k}}=0 \\ & {{A}_{2k+1}}=0 \\ \end{align} \right\}k=0,1,2,3...$$

ii) Evaluate non-zero coefficients.

Part a

 * $$\displaystyle f(\mu )={{T}_{0}}{{(1-{{\mu }^{2}})}^{3}}$$

where $$\displaystyle \mu =\sin \theta $$.

When $$\displaystyle n=2k+1$$ and when $$n\displaystyle \ge 7$$, we have,


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& k=0\Rightarrow n=1 \\ & k=1\Rightarrow n=3 \\ & k=2\Rightarrow n=5 \\ & k=3\Rightarrow n=7 \\ \end{align} \right\}{{A}_{n}}=0$$

Then for evaluating non-zero coefficients.

For simplicity setting $$\displaystyle \mu =x$$ we get $$\displaystyle \mu =x$$.

Therefore we derive coefficients using WolframAlpha ,


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$$ \displaystyle
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\begin{align} & {{A}_{0}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{1}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*1*dx=}0.457{{T}_{0}} \\ & {{A}_{2}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{2}}(x)dx}=\frac{5}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{3{{x}^{2}}}{2}-\frac{1}{2})*dx}=-0.762{{T}_{0}} \\ & {{A}_{4}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{4}}(x)dx=}\frac{9}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{35{{x}^{4}}}{8}-\frac{15{{x}^{2}}}{4}+\frac{3}{8})*dx}=0.374{{T}_{0}} \\ & {{A}_{6}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{13}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{231{{x}^{6}}}{16}-\frac{315{{x}^{4}}}{16}+\frac{105{{x}^{2}}}{16}-\frac{5}{16})*dx}=0.069{{T}_{0}} \\ \end{align}$$

(Eqs 4.3)
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Part b

 * $$\displaystyle f(\theta )={{T}_{0}}\exp \left[ -{{(\frac{2\theta }{\pi })}^{2}} \right]$$

Then,


 * $$\displaystyle f(\mu )={{T}_{0}}\exp \left[ -{{(\frac{2\mu }{\pi })}^{2}} \right]={{T}_{0}}\exp \left[ -(\frac{4{{\mu }^{2}}}) \right]$$

where $$\displaystyle \mu =\sin \theta $$.

Since f(x) is an even function for k=0,1,2,3… we get $$\displaystyle {{A}_{2k}}$$.

As to evaluating non-zero coefficients we will repeat the same procedure by setting $$\displaystyle \mu =x$$ to get $$\displaystyle f(x)={{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]$$.


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$$ \displaystyle
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\begin{align} & {{A}_{0}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{1}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*1*dx=}1.173{{T}_{0}} \\ & {{A}_{2}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{2}}(x)dx}=\frac{5}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{3{{x}^{2}}}{2}-\frac{1}{2})*dx}=-0.639{{T}_{0}} \\ & {{A}_{4}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{4}}(x)dx=}\frac{9}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{35{{x}^{4}}}{8}-\frac{15{{x}^{2}}}{4}+\frac{3}{8})*dx}=0.714{{T}_{0}} \\ & {{A}_{6}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{13}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{231{{x}^{6}}}{16}-\frac{315{{x}^{4}}}{16}+\frac{105{{x}^{2}}}{16}-\frac{5}{16})*dx}=-0.726{{T}_{0}} \\ \end{align}$$

(Eqs 4.4)
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= Problem 7.5 Finding second homogenous solution of Legendre equations of order 0 and 1 =

From the lecture slide Mtg39-1

Given
The Legendre equation is,


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

with its first homogeneous solution with respect to order 0 and 1,


 * $$\displaystyle \begin{align}

& {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ \end{align}$$

Find
Prove the second homogenous solutions of Legendre equation with respect to order 0 and 1 are,


 * $$\displaystyle \begin{align}

& {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x \\ & {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1 \\ \end{align}$$

Solution
During the derivation of Legendre equation we assumed that $$\displaystyle x=\sin \theta $$ meaning $$\displaystyle x$$ has the domain of $$\displaystyle [-1,+1]$$ in which +1 and -1 are its singularity point.

Verification of $$\displaystyle {{Q}_{0}}(x)$$
When the order of Legendre equation equals 0,


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$$ \displaystyle

(1-{{x}^{2}}){y}''-2x{y}'=0$$

(Eq 5.1)
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Eq 5.1 can be immediately reduced to a linear first ODE with varing coefficients by taking $$\displaystyle {y}'=z$$,


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$$ \displaystyle

{z}'+\frac{2x}{{{x}^{2}}-1}z=0$$

(Eq 5.2)
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The integrating factor is,


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$$ \displaystyle

\begin{align} h(x) & =\exp \left[ \int_ – ^{x}{\frac{2x}{{{x}^{2}}-1}dx} \right] \\ & =\exp \left[ \int_ – ^{x}{\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] \\ & =\exp [\log (1-{{x}^{2}})] \\ & =1-{{x}^{2}} \\ \end{align}$$

(Eq 5.3)
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Hence,


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$$ \displaystyle

\begin{align} z & ={{h}^{-1}}(x)\int_ – ^{x}{h(x)*0dx} \\ & =\frac{1-{{x}^{2}}} \\ \end{align}$$

(Eq 5.4)
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Then we can derive $$\displaystyle y$$,


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

$$ \displaystyle

\begin{align} y & =\int{zdx}+{{k}_{1}} \\ & ={{k}_{1}}+\int_ – ^{}{\frac{1-{{x}^{2}}}dx} \\ & ={{k}_{1}}+\frac{2}\int_ – ^{}{\left( \frac{1}{1+x}+\frac{1}{1-x} \right)dx} \\ & ={{k}_{1}}+{{k}_{2}}\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 5.5)
 * <p style="text-align:right">
 * }

Since $$\displaystyle {{P}_{0}}(x)$$ and $$\displaystyle {{Q}_{0}}(x)$$ are two homogeneous solutions of Eq 5.1, we can say,


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

$$ \displaystyle

{{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$

(Eq 5.6)
 * <p style="text-align:right">
 * }

On the other hand,


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

$$ \displaystyle

\begin{align} & \tanh x=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \\ & \Rightarrow x=\frac{1}{2}\log \left( \frac{1+\tanh x}{1-\tanh x} \right) \\ & \Rightarrow {{\tanh }^{-1}}x=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 5.7)
 * <p style="text-align:right">
 * }

Thus $$\displaystyle {{Q}_{0}}(x)$$ can also be written as


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

{{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x$$

(Eq 5.8)
 * <p style="text-align:right">
 * }

Verification of $$\displaystyle {{Q}_{1}}(x)$$
When the order of Legendre is set to 1,


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

or


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

$$ \displaystyle

{y}''+\frac{2x}{({{x}^{2}}-1)}{y}'+\frac{2}{(1-{{x}^{2}})}y=0$$

(Eq 5.9)
 * <p style="text-align:right">
 * }

By direct method in Mtg29-3 and Mtg30-1, we derive,


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

$$ \displaystyle

\begin{align} & h(x)={{u}_{1}}^{2}(x)\exp \left[ \int_ – ^ – {{{a}_{1}}(x)dx} \right] \\ & ={{[{{P}_{1}}(x)]}^{2}}\exp \left[ \int_ – ^ – {\frac{2x}{{{x}^{2}}-1}dx} \right] \\ & ={{x}^{2}}\exp \left[ \int_ – ^ – {\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] \\ & ={{x}^{2}}(1-{{x}^{2}}) \\ \end{align}$$

(Eq 5.10)
 * <p style="text-align:right">
 * }

and then,


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

$$ \displaystyle

\begin{align} {{Q}_{1}}(x) & ={{u}_{2}}(x) \\ & ={{u}_{1}}(x)\int_ – ^ – {\frac{1}{h(x)}dx} \\ & =x\int{\frac{1}{{{x}^{2}}(1-{{x}^{2}})}dx} \\ & =x\int{\frac{1}+\frac{1}{(1-{{x}^{2}})}dx} \\ & =x\int{\frac{1}{(1-{{x}^{2}})}dx+}x\int{\frac{1}dx} \\ & =\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \\ \end{align}$$

(Eq 5.11)
 * <p style="text-align:right">
 * }

By Eq 5.7,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

{{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1$$

(Eq 5.12)
 * <p style="text-align:right">
 * }

= Problem 7.6 The oddness of $$\displaystyle {{Q}_{n}}(x)$$ with respect to $$\displaystyle n$$ =

From the lecture slide Mtg 39-1

Given
In general the second catogary of Legendre functions have such expression,


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

$$ \displaystyle

{{Q}_{n}}(x)={{P}_{n}}(x){{\tanh }^{-1}}(x)-2\sum\nolimits_{j=1,3,5\cdots }^{J}{\frac{2n-2j+1}{(2n-j+1)j}{{P}_{n-j}}(x)}$$

(Eq 6.1)
 * <p style="text-align:right">
 * }

where $$\displaystyle J=1+2\left[ \frac{n-1}{2} \right]$$.

Find
Prove that $$\displaystyle {{Q}_{n}}(x)$$ has oddness depending on the value $$\displaystyle n$$.

When $$\displaystyle n$$ is even
From problem 7.4 we have known that if $$\displaystyle n$$ is even then $$\displaystyle {{P}_{n}}(x)$$ is even and vice versa.

On the other hand $$\displaystyle {{\tanh }^{-1}}(x)$$ is always odd, then the first tern of Eq 6.1 is odd.

Since $$\displaystyle J=1+2\left[ \frac{n-1}{2} \right]$$, it's clear that $$\displaystyle J$$ is always odd.

Also $$\displaystyle j$$ is always odd making every $$\displaystyle n-j$$ in the summation odd.

Therefore every $$\displaystyle {{P}_{n-j}}(x)$$ in the summation is odd.

Drawn together, when $$\displaystyle n$$ is even, $$\displaystyle {{Q}_{n}}(x)$$ consists of several odd terms. By problem 7.2 we know that $$\displaystyle {{Q}_{n}}(x)$$ is eventually odd.

When $$\displaystyle n$$ is odd
Going the same process like above, we know that:

$$\displaystyle {{P}_{n}}(x)$$ is odd;

$$\displaystyle {{\tanh }^{-1}}(x)$$ is always odd;

the first tern of Eq 6.1 is even;

$$\displaystyle J$$ is always odd;

every $$\displaystyle n-j$$ in the summation even;

every $$\displaystyle {{P}_{n-j}}(x)$$ in the summation is even;

$$\displaystyle {{Q}_{n}}(x)$$ consists of several even terms thus it's even.

Conclusion
The oddness of $$\displaystyle {{Q}_{n}}(x)$$ depend on the value of $$\displaystyle n$$, and $$\displaystyle {{P}_{n}}(x)$$ and $$\displaystyle {{Q}_{n}}(x)$$ have opposote oddness.

= Problem 7.7 Distance Between Two Spheres =

From the lecture slide Mtg 40-3

Given
$$\displaystyle P\left( {{r}_{P}},{{\varphi }_{P}},{{\theta }_{P}} \right)$$ and $$\displaystyle Q\left( {{r}_{Q}},{{\varphi }_{Q}},{{\theta }_{Q}} \right)$$ for spherical coordinate system.

In Astronomical Convention we have,


 * $$\displaystyle \begin{align}

& x_{P}^{1}=r_{P}^ – \cos {{\theta }_{P}}\cos {{\varphi }_{P}} \\ & x_{P}^{2}=r_{P}^ – \cos {{\theta }_{P}}\sin {{\varphi }_{P}} \\ & x_{P}^{3}=r_{P}^ – \sin {{\theta }_{P}} \\ \end{align}$$


 * $$\displaystyle \begin{align}

& x_{Q}^{1}=r_{Q}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}} \\ & x_{Q}^{2}=r_{Q}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}} \\ & x_{Q}^{3}=r_{Q}^ – \sin {{\theta }_{Q}} \\ \end{align}$$


 * $$\displaystyle \cos \gamma =\cos {{\theta }_{Q}}\cos {{\theta }_{P}}\cos ({{\varphi }_{Q}}-{{\varphi }_{P}})+\sin {{\theta }_{Q}}\sin {{\theta }_{P}}$$


 * $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}={{\left\| \overrightarrow{PQ} \right\|}^{2}}=\sum\limits_{i=1}^{3}={{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}+{{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}+{{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}$$

Find
Show that,


 * $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}={{\left( r_{Q}^ –  \right)}^{2}}\left[ {{\left( \frac \right)}^{2}}+1-2\left( \frac \right)\cos \gamma  \right]$$

Solution
Since,


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

$$ \displaystyle

{{\left( r_{PQ}^ – \right)}^{2}}={{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}+{{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}+{{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}$$

(Eq 7.1)
 * <p style="text-align:right">
 * }

so we need to calculate each terms.

First Term:


 * $$\displaystyle {{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}={{\left( r_{Q}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}}-r_{P}^ – \cos {{\theta }_{P}}\cos {{\varphi }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}=r_{Q}^{2}{{\cos }^{2}}{{\theta }_{Q}}{{\cos }^{2}}{{\varphi }_{Q}}+r_{P}^{2}{{\cos }^{2}}{{\theta }_{P}}{{\cos }^{2}}{{\varphi }_{P}}-2r_{Q}^ – r_{P}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}}\cos {{\theta }_{P}}\cos {{\varphi }_{P}}$$

Second Term:


 * $$\displaystyle {{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}={{\left( r_{Q}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}}-r_{P}^ – \cos {{\theta }_{P}}\sin {{\varphi }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}=r_{Q}^{2}{{\cos }^{2}}{{\theta }_{Q}}{{\sin }^{2}}{{\varphi }_{Q}}+r_{P}^{2}{{\cos }^{2}}{{\theta }_{P}}{{\sin }^{2}}{{\varphi }_{P}}-2r_{Q}^ – r_{P}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}}\cos {{\theta }_{P}}\sin {{\varphi }_{P}}$$

Third Term:


 * $$\displaystyle {{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}={{\left( r_{Q}^ – \sin {{\theta }_{Q}}-r_{P}^ – \sin {{\theta }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}=r_{Q}^{2}{{\sin }^{2}}{{\theta }_{Q}}+r_{P}^{2}{{\sin }^{2}}{{\theta }_{P}}-2\sin {{\theta }_{Q}}\sin {{\theta }_{P}}$$

Add all three terms together, then we get


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

$$ \displaystyle

\begin{align} {{\left( r_{PQ}^ – \right)}^{2}}= & r_{Q}^{2}\left[ {{\cos }^{2}}{{\theta }_{Q}}\left( {{\cos }^{2}}{{\varphi }_{Q}}+{{\sin }^{2}}{{\varphi }_{Q}} \right)+{{\sin }^{2}}{{\theta }_{Q}} \right]+r_{P}^{2}\left[ {{\cos }^{2}}{{\theta }_{P}}\left( {{\cos }^{2}}{{\varphi }_{P}}+{{\sin }^{2}}{{\varphi }_{P}} \right)+{{\sin }^{2}}{{\theta }_{P}} \right] \\ & -2r_{Q}^ – r_{P}^ – \left[ \cos {{\theta }_{Q}}\cos {{\theta }_{P}}\left( \cos {{\varphi }_{Q}}\cos {{\varphi }_{P}}+\sin {{\varphi }_{Q}}\sin {{\varphi }_{P}} \right)+\sin {{\theta }_{Q}}\sin {{\theta }_{P}} \right] \\ \end{align}$$

(Eq 7.2)
 * <p style="text-align:right">
 * }

From trigonometric equations, we know that


 * $$\displaystyle \left( {{\cos }^{2}}{{\varphi }_{Q}}+{{\sin }^{2}}{{\varphi }_{Q}} \right)=1$$


 * $$\displaystyle \left( {{\cos }^{2}}{{\varphi }_{P}}+{{\sin }^{2}}{{\varphi }_{P}} \right)=1$$


 * $$\displaystyle \left[ {{\cos }^{2}}{{\theta }_{Q}}+{{\sin }^{2}}{{\theta }_{Q}} \right]=1$$


 * $$\displaystyle \left[ {{\cos }^{2}}{{\theta }_{P}}+{{\sin }^{2}}{{\theta }_{P}} \right]=1$$


 * $$\displaystyle \left( \cos {{\varphi }_{Q}}\cos {{\varphi }_{P}}+\sin {{\varphi }_{Q}}\sin {{\varphi }_{P}} \right)=\cos \left( {{\varphi }_{Q}}-{{\varphi }_{P}} \right)$$

So putting equations above in the $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}$$, we obtain


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

$$ \displaystyle

\begin{align} {{\left( r_{PQ}^ – \right)}^{2}} & =r_{Q}^{2}+r_{P}^{2}-2r_{Q}^ – r_{P}^ – \left[ \cos {{\theta }_{Q}}\cos {{\theta }_{P}}\left( \cos ({{\varphi }_{Q}}-{{\varphi }_{P}}) \right)+\sin {{\theta }_{Q}}\sin {{\theta }_{P}} \right] \\ & =r_{Q}^{2}+r_{P}^{2}-2r_{Q}^ – r_{P}^ – \cos \gamma \\ \end{align}$$

(Eq 7.3)
 * <p style="text-align:right">
 * }

If we take common parenthesis of $$\displaystyle {{\left( r_{Q}^ – \right)}^{2}}$$, it will be shown


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

{{\left( r_{PQ}^ – \right)}^{2}}={{\left( r_{Q}^ –  \right)}^{2}}\left[ {{\left( \frac \right)}^{2}}+1-2\left( \frac \right)\cos \gamma  \right]$$

(Eq 7.4)
 * <p style="text-align:right">
 * }

= Problem 7.8: Derivation of a Specialized Case of the Binomial Series =

From the lecture slide Mtg 40-4

Given
Given the following binomial series equation:


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

$$ \displaystyle \left( x +y \right)^r=\sum_{k=0}^{\infty}\dbinom{r}{k}x^{r-k}y^{k} $$ (Eq 8.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where the binomial coefficient, "r choose k", is defined as follows:


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

$$ \displaystyle \dbinom{r}{k}=\frac{r(r-1)\cdots (r-k+1)}{k!} $$ (Eq 8.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

and r is a real number, $$\displaystyle r \in \mathbb{R}$$

Derive the following equation:


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

$$ \displaystyle \left(1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\alpha_{i}x^{i}
 * style="width:95%" |
 * style="width:95%" |

$$ (Eq 8.3)
 * <p style="text-align:right">
 * }

where


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

$$ \displaystyle \alpha_i = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2i - 1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2i)} $$ (Eq 8.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
In comparing the left hand side (LHS) of Eq 8.3 to the LHS of Eq 8.1, the following values for r, x, and y are obtained:


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

$$ \displaystyle r=-\frac{1}{2}, \qquad x=1, \qquad y=(-x) $$ (Eq 8.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plugging these values into the right hand side (RHS) of Eq. 8.1 yields the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\dbinom{-\frac{1}{2}}{k}1^{\left(-\frac{1}{2}-k\right)}(-x)^{k}

$$ (Eq 8.6)
 * <p style="text-align:right">
 * }

Note that for every k

$$ \displaystyle 1^{\left(-\frac{1}{2}-k\right)}=1,\quad \forall k $$

Next, the binomial coefficient in Eq 8.6 is expanded by using Eq 8.2 and the values in Eq 8.5 to yield the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)\cdots (-\frac{1}{2}-k+1)}{k!} (-x)^{k}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})\cdots (-k+\frac{1}{2})}{k!} (-x)^{k}

$$ (Eq 8.7)
 * <p style="text-align:right">
 * }

Next, factor out -1/2 from each term in the numerator and change the summation index k to i as follows:


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

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{\left(-\frac{1}{2}\right)[1]\left(-\frac{1}{2}\right)[3]\left(-\frac{1}{2}\right)[5]\cdots \left(-\frac{1}{2}\right)[2i-1]}{i!} (-x)^{i}=\sum_{i=0}^{\infty}\left(\left(-\frac{1}{2}\right)(-x)\right)^i\frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (Eq 8.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eq 8.8 can be simplified as follows:


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

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{x^i}{2^i}\cdot \frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (Eq 8.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Note that the terms in the denominator can be rewritten as:
 * {| style="width:100%" border="0"

$$ \displaystyle (2^i)(i!)=\underbrace{(2\cdot 2\cdot 2\cdot \ldots)}_{i\ times}(1\cdot 2\cdot 3\cdot \ldots \cdot i)=2\cdot 4\cdot 6\cdot \ldots \cdot 2i $$ (Eq 8.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The RHS of Eq 8.10 can replace the denominator of Eq 8.9 resulting in the equation found in Eq 8.3:


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

$$ \displaystyle \begin{align} \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\cdot \underbrace{\frac{(1)(3)(5)\cdots (2i-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2i}}_{\alpha_i}x^i
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\end{align}$$ (Eq 8.11)
 * <p style="text-align:right">
 * }


 * The following link was consulted after this problem was solved in an attempt to provide the most clear and comprehensive presentation :

http://en.wikiversity.org/w/index.php?title=User:Egm6321.f09.Team1/HW5&oldid=510466#Problem_6:_Binomial_Series

= Problem 7.9 Deriving Legendre polynominals using recurrence relation =

From the lecture slide Mtg 41-2

Given
The second recurrence relation of Legendre polynominals is,


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

$$ \displaystyle

(n+1){{P}_{n+1}}-(2n+1)x{{P}_{n}}+n{{P}_{n-1}}=0$$

(Eq 9.1)
 * <p style="text-align:right">
 * }

And Legendre polynominal of order 0 and 1 are,


 * $$\displaystyle \begin{align}

& {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ \end{align}$$

Find
Legendre polynominals of order 2,3,4,5,6.

Solution
From Eq 9.1 we have,


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

$$ \displaystyle

{{P}_{n+1}}=\frac{2n+1}{n+1}x{{P}_{n}}-\frac{n}{n+1}{{P}_{n-1}}$$

(Eq 9.2)
 * <p style="text-align:right">
 * }

Take $$\displaystyle n$$ as $$\displaystyle 2,3,4,5,6$$ we have,


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

$$ \displaystyle

\begin{align} & {{P}_{2}}=\frac{3}{2}x{{P}_{1}}-\frac{1}{2}{{P}_{0}} \\ & {{P}_{3}}=\frac{5}{3}x{{P}_{2}}-\frac{2}{3}{{P}_{1}} \\ & {{P}_{4}}=\frac{7}{4}x{{P}_{3}}-\frac{3}{4}{{P}_{2}} \\ & {{P}_{5}}=\frac{9}{5}x{{P}_{4}}-\frac{4}{5}{{P}_{3}} \\ & {{P}_{6}}=\frac{11}{6}x{{P}_{5}}-\frac{5}{6}{{P}_{4}} \\ \end{align}$$

(Eqs 9.3)
 * <p style="text-align:right">
 * }

Solve Eqs 9.3 by sequence we have,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{P}_{2}}=\frac{3}{2}x*x-\frac{1}{2}*1=\frac{3}{2}{{x}^{2}}-\frac{1}{2} \\ & {{P}_{3}}=\frac{5}{3}x\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)-\frac{2}{3}x=\frac{5}{2}{{x}^{3}}-\frac{3}{2}x \\ & {{P}_{4}}=\frac{7}{4}x\left( \frac{5}{2}{{x}^{3}}-\frac{3}{2}x \right)-\frac{3}{4}\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ & {{P}_{5}}=\frac{9}{5}x\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)-\frac{4}{5}\left( \frac{5}{2}{{x}^{3}}-\frac{3}{2}x \right)=\frac{63}{8}{{x}^{5}}-\frac{35}{4}{{x}^{3}}+\frac{15}{8}x \\ & {{P}_{6}}=\frac{11}{6}x\left( \frac{63}{8}{{x}^{5}}-\frac{35}{4}{{x}^{3}}+\frac{15}{8}x \right)-\frac{5}{6}\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)=\frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-5 \\ \end{align}$$

(Eqs 9.4)
 * <p style="text-align:right">
 * }

=Problem 7.10=

From the lecture slide Mtg 41-3

Given
To derive $$ \displaystyle P_{3}(x),P_{4}(x),P_{5}(x),P_{6}(x)$$ from the expansion of binomial expansion we have to start with giving generation function of legendre series. Generating function was derived in meeting 40 form Newtonian potential.


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

$$  \displaystyle \left ( 1-2\mu \rho +\rho ^{2} \right )^{-\frac{1}{2}}=\sum_{n=0}^{\infty }P_{n}(\mu )\rho ^{n} $$     (Eq 10.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle -2\mu \rho +\rho ^{2}=-x $$     (Eq 10.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle (1-x)^{-\frac{1}{2}}=\sum_{n=0}^{\infty }\alpha _{i}x^{i} $$     (Eq 10.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we have to use binmial expansion. But our power is not an integer. So we use 'r choose k' for general derivation.


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

$$  \displaystyle (x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k} $$     (Eq 10.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \binom{n}{k}=\frac{n(n-1)....(n-k+1)}{k!} $$     (Eq 10.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Where our n designates real number.

Solution
Now we can use Eq(10.4) and Eq (10.5) to obtain expansion of generating function.


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

$$  \displaystyle \begin{align} &(1-(2\mu \rho +\rho ^{2}))^{-\frac{1}{2}}=1+(-\frac{1}{2})(\rho ^{2}-2\mu \rho )+ \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(\rho ^{2}-2\mu \rho ) ^{2}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{^3!}(\rho ^{2}-2\mu \rho )^{3}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})}{4!}(\rho ^{2}-2\mu \rho )^{4}\\
 * style="width:95%" |
 * style="width:95%" |

&+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})(-\frac{9}{2})}{5!}(\rho ^{2}-2\mu \rho )^{5}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})(-\frac{9}{2})(-\frac{13}{2})}{6!}(\rho ^{2}-2\mu \rho )^{6}\\ \end{align} $$     (Eq 10.6) Since we need sixth power of $$\displaystyle \rho $$ ,6th binomial expansion is enough for observing sixth P.
 * <p style="text-align:right">
 * }


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

$$  \displaystyle 1+\mu \rho -\frac{1}{2}\rho ^{2}+\frac{3}{2}\mu ^{2}\rho ^{2}+\frac{5}{2}\mu ^{3}\rho ^{3}-\frac{3}{2}\mu \rho ^{3}+\frac{3}{8}\rho ^{4}-\frac{15}{4}\mu ^{2}\rho ^{4}+\frac{35}{8}\mu ^{4}\rho ^{4}+\frac{15}{8}\mu \rho ^{5}-\frac{35}{4}\mu ^{3}\rho ^{5}+\frac{63}{8}\mu ^{5}\rho ^{5}-\frac{5}{16}\rho ^{6}+\frac{105}{16}\mu ^{2}\rho ^{6}-\frac{315}{16}\mu ^{4}\rho ^{4}+\frac{273}{16}\mu ^{6}\rho ^{6}.....
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 10.7)
 * <p style="text-align:right">
 * }

Eq(10.7) has also other higher order terms. Since the definition is given in Eq(10.1) we have to collect same orders togerher. Then we can observe six legendre polynomials.


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$$  \displaystyle 1+\mu \rho +\frac{1}{2}(3\mu ^{2}-1)\rho ^{2}+\frac{1}{2}(5\mu ^{3}-3\mu) \rho ^{3}+(\frac{35}{8}\mu ^{4}-\frac{15}{4}\mu ^{2}+\frac{3}{8})\rho ^{4}+(\frac{63}{8}\mu ^{5} -\frac{35}{4}\mu ^{3}+\frac{15}{8}\mu)\rho ^{5}+(\frac{273}{16}\mu ^{6}-\frac{315}{16}\mu ^{4}+\frac{105}{16}\mu ^{2}-\frac{5}{16})\rho ^{6} $$     (Eq 10.8)
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Manipulating Eq(10.1) abit more ,
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$$  \displaystyle (1-2\mu \rho +\rho ^{2})^{-\frac{1}{2}}=P_{0}(\mu )\rho ^{0}+P_{1}(\mu )\rho ^{1}+P_{2}(\mu )\rho ^{2}+P_{3}(\mu )\rho ^{3}+P_{4}(\mu )\rho ^{4}+P_{5}(\mu )\rho ^{5}+P_{6}(\mu )\rho ^{6}.... $$     (Eq 10.9)
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If we compare with Eq (10.8)and Eq(10.9) we can see that coefficients of different powers of $$ \displaystyle\rho$$'s are our legendre polynomials. These polynomials are the same as what we have found from Eq.(7)and(8) from the meeting 36-2. Where
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$$  \displaystyle P_3(x) = \sum_{i=0}^1 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} =\frac{5}{2} x^3 -\frac{3}{2} x $$ (Eq 10.10)
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$$  \displaystyle P_4(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{35}{8} x^4-\frac{15}{4} x^2+\frac{3}{8} $$     (Eq 10.11)
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$$  \displaystyle P_5(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{63}{8} x^5 - \frac{35}{4} x^3 +\frac{15}{8} x $$ (Eq 10.12)
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$$  \displaystyle P_6(x) = \sum_{i=0}^3 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{273}{16} x^6 - \frac{315}{16} x^4 + \frac{105}{16} x^2 - \frac{5}{16} $$     (Eq 10.13) =Contributions Of Team Members=
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 * Problem 1:    Solved by: Oztekin

Author: Oztekin

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 * Problem 2:    Solved by: Oztekin

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 * Problem 3:    Solved by: Dube

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 * Problem 4:    Solved by: Abudaram

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 * Problem 5:    Solved by: Zou

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 * Problem 6:    Solved by: Zou

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 * Problem 7:    Solved by: Ismail

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 * Problem 8:    Solved by: Zou

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 * Problem 9:    Solved by: Patterson

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 * Problem 10:    Solved by: Oztekin

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= References =