User:EGM6341.s11.TEAM1.WILKS/Mtg41

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

[[media: 2010_11_30_14_54_10.djvu | Mtg 41:]] Tue, 30 Nov 10

= Generating functions =

[[media: 2010_11_30_14_54_10.djvu | Page 41-1]]
[[media: 2010_11_23_15_05_23.djvu | Eq.(5)p.40-3]] : The generating function for $$\displaystyle \{ P_n \}$$ is $$\displaystyle (1 - 2 \mu\ \rho\ + \rho^2 )^{-\frac{1}{2}}$$.

[[media: 2010_11_23_15_05_23.djvu | Eq.(6)p.40-4]] : The generating function for "r choose k" $$\displaystyle {r \choose k}$$ is $$\displaystyle (1+x)^r $$.

= Inverse square law =

See lecture notes [[media: 2010_11_23_15_05_23.djvu | p.40-1]].



where $$ F=kr \ $$ and $$ d(t)=r \ $$



where $$ \mathbf{f} = \frac{-GMm \mathbf{ \hat r \ } }{r^2} \ $$.

= Two recurrence relations for Legendre polynomials =

[[media: 2010_11_30_14_54_10.djvu | Page 41-2]]
NOTE: The RR1, even though not useful to generate $$ P_n \ $$ from previously known $$ P_k, k=0,1,...,n-1 \ $$, is useful to obtain the Legendre differential equation together with the RR2. END NOTE

NOTE: RR2 useful to generate $$ \left \{ P_n \right \} \ $$ knowing $$ P_0(x)=1, P_1(x)=x \ $$ END NOTE HW7.9 Generate $$ \left \{ P_2,...,P_6 \right \} \ $$ using RR2 starting from $$ P_0,P_1 \ $$ [[media: 2010_11_09_15_00_14.djvu | cf. Eq.(4) - Eq.(6) p.36-2]] END HW7.9

Generating Legendre polynomials
Define:

$$ A( \mu, \rho\ ) := 1 - 2 \mu \rho + \rho^2 =: 1 - x \ $$

where

$$ -x := -2 \mu \rho + \rho^2 \ $$.

Then from [[media: 2010_11_23_15_05_23.djvu | Eq.(6) and Eq.(7) p.40-3]]:

$$ \frac{1}{\sqrt{A( \mu, \rho ) }}=\alpha_0 + \alpha_1 (2 \mu \rho - \rho^2 ) + \alpha_2 (2 \mu \rho - \rho^2 )^2 + ... \ $$

where $$ ( 2 \mu \rho - \rho^2 )^2 = 4 \mu^2 \rho^2 - 4 \mu \rho^3 + \rho^4 \ $$

[[media: 2010_11_30_14_54_10.djvu | Page 41-3]]
HW7.10

Continue the power series expansion to find $$ \left \{ P_3, ..., P_6 \right \} \ $$ and compare the results to those obtained by (a) [[media: 2010_11_09_15_00_14.djvu | Eq.(7) and Eq.(8) p.36-2]] and (b) HW7.9.

END HW7.10

Two (partial) derivatives with respect to the parameters $$\displaystyle (\mu, \rho)$$
Plan: Find

Step 1. $$ \frac{\partial}{\partial \mu\ }\frac{1}{ \sqrt{A} } = \frac{d}{d \mu\ }\frac{1}{ \sqrt{A} } \ $$ and

Step 2. $$ \frac{\partial}{\partial \rho\ }\frac{1}{ \sqrt{A} } = \frac{d}{d \rho\ }\frac{1}{ \sqrt{A} } \ $$

Step 1. Derivative with respect to $$\displaystyle \mu$$
with $$ \frac{dA}{d \mu\ } = -2 \rho\ \ $$ From [[media:2010_11_23_15_05_23.djvu | Eq.(5) p.40-3]] :

[[media: 2010_11_30_14_54_10.djvu | Page 41-4]]
$$ P_n'( \mu\ ):=\frac{d}{d \mu\ }P_n( \mu\ ) \ $$ Recall $$ P_0( \mu\ )=1 \Rightarrow \ P_0'( \mu\ ) = 0 \ $$ $$ \Rightarrow \ \ $$ n starts from 1 in [[media: 2010_11_30_14_54_10.djvu | Eq.(6) p.41-3]]