User:Egm6321.f10.team4.Yoon/Mtg44

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_11_30_19_50_24.djvu | Mtg 44:]] Fri, 3 Dec 10

Step 3: Eliminate $$\displaystyle A^{\frac{3}{2}}$$
Compare [[media: 2010_11_30_14_54_10.djvu | Eqn(1) p.41.4]] and [[media: 2010_11_30_14_54_10.djvu | Eqn(6)p.41.3:]]

$$\displaystyle (\mu - \rho) \times $$ $$\displaystyle [$$ [[media: 2010_11_30_14_54_10.djvu | Eqn(6)p.41.3]] $$\displaystyle ]$$ $$\displaystyle = \rho \times$$ $$\displaystyle [$$ [[media: 2010_11_30_14_54_10.djvu | Eqn(1) p.41.4]] $$\displaystyle ]$$

$$\displaystyle \Rightarrow (\mu - \rho) \times \frac{\rho}{A^{\frac{3}{2}}} = \rho \sum^{\infty}_{n=1} P_n(\mu) n \rho^{n-1} $$

The summation index $$\displaystyle n$$ starts from $$\displaystyle 1$$ instead of $$\displaystyle 0$$ due to the presence of $$\displaystyle n$$ in the summand.

$$\displaystyle \Rightarrow (\mu - \rho) \sum_{n=1}^{\infty} P_n'(\mu) \rho^n = \rho \sum_{n=1}^{\infty} P_n(\mu) n \rho^{n-1}$$

$$\displaystyle \Rightarrow \mu \sum^{\infty}_{n=1} P_n' \rho^n - \sum^{\infty}_{n=1} P_n' \rho^{n+1} = \sum^{\infty}_{n=1} P_n n \rho^n$$

$$\displaystyle \Rightarrow \underbrace{\mu \underbrace{P_1'}_{\color{blue}{1}} \rho - \underbrace{P_1}_{\color{blue}{\mu}} \cdot 1 \cdot \rho}_{\color{red}{0}} + \sum_{n=2} [-P_{n-1}' + \mu P_n' - nP_n]\rho^n = 0 $$

$$\displaystyle \Rightarrow -P_{n-1}' + \mu P_n' - n P = 0$$

[[media: 2010_11_30_19_50_24.djvu | Page 44-2]]
RR1:

2nd recurrence relation: Derivation
[[media: 2010_11_23_15_05_23.djvu | Eqn(5) p.40-3]]:

[[media: 2010_11_30_14_54_10.djvu | Eqn(1) p.41-4]]:

[[media: 2010_11_30_19_50_24.djvu | Page 44-3]]
Want to have $$\displaystyle \rho^n$$ as common factor.

Ignore the initial terms => recurrence rel. is from factors of $$\displaystyle \rho^n $$

[[media: 2010_11_30_19_50_24.djvu | Page 44-4]]
$$\displaystyle -2\mu[P_1 \cdot 1 \cdot \rho + \sum_{n=2} P_n n \color{red}{\rho^n} \color{black}{] + [\sum_{n=2} P_{n-1}(n-1)} \color{red}{\rho^n} \color{black}{]} $$

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

$$\displaystyle \Rightarrow (n+1)P_{n+1} - (2n+1)\mu P_n + n P_{n-1} = 0 $$ (RR2)

Now Legendre differential equation.

[[media: 2010_11_30_19_50_24.djvu | Page 44-5]]
$$\displaystyle \color{red}{(RR1)} \color{black}{ \Rightarrow (n+1)P_{n+1}' - (2n+1)[P_n + \mu P_n' ] + n\underbrace{(\mu P_n' - nP_n)}_{\color{blue}{P_{n-1}'}}} = 0$$

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

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

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

$$\displaystyle \Rightarrow P_n' - \mu P_{n-1}' - n P_{n-1} = 0$$

Transform index to $$\displaystyle n$$ using (RR1) again: $$\displaystyle P_{n-1}' = \mu P_n' - n P_n$$

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

Want to transform the index of $$\displaystyle P_{n-1}$$ in the last term to $$\displaystyle n$$.

[[media: 2010_11_30_19_50_24.djvu | Page 44-6]]
Note:

To write the 2nd term in [[media: 2010_11_30_19_50_24.djvu | Eqn(1) p.44-1]] with the factor $$\displaystyle \rho^n$$, first define the index $$\displaystyle m:=n+1$$ to have

$$\displaystyle \sum^{\infty}_{n=1} P_{n}' \rho^{n+1} = \sum^{\infty}_{m=2} P_{m-1}' \rho^{m}$$

Then change the index $$\displaystyle m$$ back to $$\displaystyle n$$ to obtain

$$\displaystyle \sum^{\infty}_{n=1} P_{n}' \rho^{n+1} = \sum^{\infty}_{n=2} P_{n-1}' \color{red}{\rho^{n}}$$

end Note