User:EGM6341.s11.team1.Chiu/PEA1 F09 Mtg40

Mtg 40: Thu, 19 Nov 09

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Note


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$$ \sum_{n=1}^{ \infty} P'_{n} \rho^{n+1}= \sum_{m=2}^{ \infty} P'_{m-1} \rho^{m} \ $$


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Let $$ m:=n+1 \Rightarrow n=m-1 \ $$ when $$ n=1 \Rightarrow m=2 \ $$


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$$ \sum_{n=1}^{ \infty} P'_{n} \rho^{n+1}= \sum_{n=2}^{ \infty} P'_{n-1} \rho^{n} \ $$


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$$ \sum_{m=2}^{ \infty} P'_{m-1} \rho^{m}= \sum_{n=2}^{ \infty} P'_{n-1} \rho^{n} \ $$ (by changing $$ \color{blue} m \to n \ $$)


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End Note


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2) $$ \frac{1}{ \sqrt{A}}= \sum_{n=0} P_{n} \rho^{n} \ $$ [[media: Egm6321.f09.mtg38.djvu| (5) page38-3]]


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$$ \frac{\mu- \rho}{A^{ \frac{3}{2}}}= \sum_{n=1} P_{n}n \rho^{n-1} \ $$ [[media: Egm6321.f09.mtg39.djvu| (5) page39-3]]


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$$ \Rightarrow \underset{ \color{blue} \underset{(\mu- \rho) \sum_{n=0}P_{n} \rho^{n}}{\downarrow}} \frac{\mu- \rho}{ \sqrt{A}}= \underset{ \color{blue} \underset{(1-2 \mu \rho+ \rho^{2})}{\uparrow}} A \sum_{n=1} P_{n} n \rho^{n-1}=A \sum_{n=0} P_{n+1}(n+1) \rho^{n} \ $$


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[[media: Egm6321.f09.mtg40.djvu| page40-2]]


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$$ \mu \sum_{n=0}P_{n} \rho^{n}- \mathbf{ \sum_{n=0}P_{n} \rho^{n+1}}= \mathbf{ \sum_{n=1}P_{n}n \rho^{n-1}}-2 \mu \sum_{n=1}P_{n}n \rho^{n}+ \mathbf{\sum_{n=1}P_{n}n \rho^{n+1}} \ $$


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$$ \Rightarrow \mu \sum_{n=0}P_{n} \rho^{n}- \sum_{n=1}P_{n-1} \rho^{n}= \sum_{n=0}P_{n+1}(n+1) \rho^{n}-2 \mu \sum_{n=1}P_{n}n \rho^{n}+ \sum_{n=2}P_{n-1}(n-1) \rho^{n} \ $$


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Ignore the initial terms $$ \Rightarrow \ $$ recurrence relation is factors of $$ \rho^{n}: \ $$


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$$ \mu \left [ P_{0} \rho^{0}+P_{1} \rho^{1}+ \sum_{n=2} \mathbf{P_{n} \rho^{n}} \right ]- \left [ P_{0} \rho^{1}+ \sum_{n=2} \mathbf{P_{n-1} \rho^{n}} \right ] \ $$


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[[media: Egm6321.f09.mtg40.djvu| page40-3]]


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$$ =\left [ P_{1} \rho^{0}+P_{2} \cdot 2 \cdot \rho^{1}+ \sum_{n=2} \mathbf{P_{n+1}(n+1)} \rho^{n} \right ]-2 \mu \left [ P_{1} \cdot 1 \cdot \rho+ \sum_{n=2} \mathbf{P_{n}n} \rho^{n} \right ]+ \left [ \sum_{n=2} \mathbf{P_{n-1}(n-1)} \rho^{n} \right ]\ $$


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$$ \mu P_{n}-P_{n-1}= P_{n+1}(n+1)-2 \mu P_{n}n+P_{n-1}(n-1) \ $$


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$$ \Rightarrow (n+1)P_{n+1}-(2n+1) \mu P_{n}+n P_{n-1}=0 \ $$

(1)
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Legendre equation from 2 recurrence relations

[[media: Egm6321.f09.mtg39.djvu| (1) page39-4:]] 1st $$ RR=RR1 \ $$

[[media: Egm6321.f09.mtg40.djvu| (1) page40-3:]] 2nd $$ RR=RR2 \ $$


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$$ \frac{\mathrm{d} }{\mathrm{d} \mu}(RR2)=(n+1)P'_{n+1}-(2n+1) \left [ P_{n}+ \mu P'_{n} \right ]+n P'_{n-1}=0 \ $$


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$$ RR1 \Rightarrow (n+1)P'_{n+1}-(2n+1)(P_{n}+ \mu P'_{n})+n( \mu P'_{n}-n P_{n})=0 \ $$


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[[media: Egm6321.f09.mtg40.djvu| page40-4]]


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$$ \Rightarrow (n+1)P'_{n+1}-(n+1) \mu P'_{n}+ \underbrace{(-2n-1-n^{2})}_{ \color{blue} -(n+1)^{2}} P_{n}=0 \ $$


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$$ \Rightarrow P'_{n+1}- \mu P'_{n}-(n+1)P_{n}=0 \ $$


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Change $$ (n+1) \to n: \ $$


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$$ P'_{n}- \mu \underbrace{P'_{n-1}}_{{ \color{blue} \mu P'_{n}-n P_{n}} \ (RR1)}-n P_{n-1}=0 \ $$


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$$ (1- \mu^{2})P'_{n}+n \mu P_{n}-n P_{n-1}=0 \ $$


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Differentiate the above:


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$$ \underbrace{ \left [ (1- \mu^{2})P'_{n} \right ]'}_{ \color{blue} (1- \mu^{2})P''_{n}-2 \mu P'_{n}}+ \underbrace{ n \left [ \mu P'_{n}+P_{n} \right ]-n \underbrace{P'_{n-1}}_{ \color{blue} \mu P'_{n}-n P_{n}}}_{ \color{red} n(n+1)P_{n}}=0 \ $$


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