User:Egm6322.s12.team1.zheng.zx/R9

Problem 8: Generate{P2,...,P6}using RR2 and from generating function expansion
 Solved without assistance 

Given
1.Recurrence relation RR2:
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$$ \displaystyle (n+1)P_{n+1}-(2n+1) x P_n+nP_{n-1}=0 $$
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with:
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$$ \displaystyle P_0(x)=1 $$
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$$ \displaystyle P_1(x)=x $$
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2.Generating function for Legendre polynomials
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$$ \displaystyle \mathcal G_L(\mu,\rho)=A^{-1/2}(\mu,\rho)=(1-x)^{-1/2} $$
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$$ \displaystyle A(\mu ,\rho):=1-2\mu \rho+\rho^2=:1-x $$
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with:
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$$ \displaystyle -x:=-2\mu\rho+\rho^2 $$
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$$ \displaystyle x:=2\mu\rho-\rho^2 $$
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Find

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$$ \displaystyle {P_0,...,P_6} $$ Using both two methods
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Solution
Part I

Using RR2:
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$$ \displaystyle (n+1)P_{n+1}-(2n+1)xP_n+nP_{n-1}=0 $$
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$$ \displaystyle P_0=1, \ P_1=x $$
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n=1
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$$ \displaystyle 2P_2-3xP_1+P_0=0 $$
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$$ \displaystyle 2P_2-3x^2+1=0 $$
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$$ \displaystyle P_2=\frac{1}{2}(3x^2-1) $$
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n=2
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$$ \displaystyle 3P_3-5xP_2+2P_1=0 $$
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$$ \displaystyle 3P_3-\frac{5x}{2}(3x^2-1)+2x=0 $$
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$$ \displaystyle P_3=\frac{1}{2}(5x^3-3x) $$
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n=3
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$$ \displaystyle 4P_4-7xP_3+3P_2=0 $$
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$$ \displaystyle 4P_4-\frac{7x}{2}(5x^3-3x)+\frac{3}{2}3x^2-1=0 $$
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$$ \displaystyle P_4=\frac{1}{8}(35x^4-30x^2+3) $$
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n=4
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$$ \displaystyle 5P_5-9xP_4+4P_3=0 $$
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$$ \displaystyle 5P_5-\frac{9}{8}x(35x^4-30x^2+3)+\frac{4}{2}(5x^3-3x)=0 $$
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$$ \displaystyle P_5=\frac{1}{8}(63x^5-70x^3+15x) $$
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n=5
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$$ \displaystyle 6P_6-11xP_5+5P_4=0 $$
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$$ \displaystyle 6P_6-\frac{11}{8}x(63x^5-70x^3+15x)+\frac{5}{8}(35x^4-30x^2+3)=0 $$
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$$ \displaystyle P_6=\frac{1}{16}(231x^6-315x^4+105x^2-5) $$
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Part II

Using generating function expansion:
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$$ \displaystyle \begin{align} \mathcal G_L(\mu,\rho) &=A^{-1/2}(\mu,\rho)=(1-x)^{-1/2} \\ &=1+(-\frac{1}{2})(-x)+\frac{3}{8}(-x)^2+(-\frac{5}{16})(-x)^3+\frac{35}{128}(-x)^4... \\ &=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3+\frac{35}{128}x^4... \\ &=1+\frac{1}{2}(2\mu \rho-\rho^2)+\frac{3}{8}(4\mu^2\rho^2-4\mu \rho^3+\rho^4)+\frac{5}{16}(8\mu^3\rho^3-12\mu^2\rho^4+6\mu \rho^5-\rho^6) \\ &\ \ \ \ +\frac{35}{128}(16\mu^4\rho^4-32\mu^3\rho^5+24\mu^2\rho^6-6\mu \rho) \\ &=1+\mu \rho \\ &\ \ \ \ +\frac{1}{2}(3\mu^2-1)\rho^2 \\ &\ \ \ \ +\frac{1}{2}(5\mu^3-3\mu)\rho^3 \\ &\ \ \ \ +\frac{1}{8}(35\mu^4-30\mu^2+3)\rho^4 \\ &\ \ \ \ +\frac{1}{8}(63\mu^5-70\mu^3+15\mu)\rho^5 \\ &\ \ \ \ +\frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5)\rho^6 \end{align} $$
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So:
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$$ \displaystyle \begin{align} P_0 &= 1\\ P_1 &= \mu \\ P_2 &= \frac{1}{2}(3\mu^2-1) \\ P_3 &= \frac{1}{2}(5\mu^3-3\mu) \\ P_4 &= \frac{1}{8}(35\mu^4-30\mu^2+3) \\ P_5 &= \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \\ P_6 &= \frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5) \end{align} $$
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Comparing the two series of results, we can find whatever which method we use, the Legendre polynomials we get are exactly same.(Replace variable $$\mu$$ with x )