User:EGM6341.s11.TEAM1.WILKS/Mtg41

EGM6321 - Principles of Engineering Analysis 1, Fall 2010

Mtg 41: Tue, 30 Nov 10

Generating functions

Page 41-1

Eq.(5)p.40-3 : The generating function for $\displaystyle \{P_{n}\}$ is $\displaystyle (1-2\mu \ \rho \ +\rho ^{2})^{-{\frac {1}{2}}}$.

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 p.40-1.

Newton's law is a claim---that could have been wrong---about the actual relation between the force $\displaystyle \mathbf {F}$ on a particle with mass m and its acceleration $\displaystyle \mathbf {a}$. One tests it by calculating the acceleration with a presumed force and comparing it to the measured value.

The test will fail if either Newton's law or the presumed force is wrong. Could $\displaystyle \mathbf {F} =m\mathbf {a}$ be tested more generally, without recourse to positing forces and looking at actual solutions? It seems not.

— Kane, [String Theory], Physics Today, November 2010.

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

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

Two recurrence relations for Legendre polynomials

Page 41-2

$\displaystyle \mu \,P_{n}'-P_{n-1}'=nP_{n}$

(RR1)

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

$\displaystyle (n+1)P_{n+1}-(2n+1)\mu \,P_{n}+nP_{n-1}=0$

(RR2)

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}\$ 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 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}\$

Page 41-3

$\displaystyle {\frac {1}{\sqrt {A}}}=\alpha _{0}+(2\mu \alpha _{1})\rho +(-\alpha _{1}+4\mu ^{2}\alpha _{2})\rho ^{2}+...$

(1)

$\displaystyle P_{0}(\mu )=\alpha _{0}=1$

(2)

$\displaystyle P_{1}(\mu )=2\mu \alpha _{1}=\mu$

(3)

$\displaystyle P_{2}(\mu )=-\alpha _{1}+4\mu ^{2}\alpha _{2}={\frac {1}{2}}(3\mu ^{2}-1)$

(4)

HW7.10

Continue the power series expansion to find $\left\{P_{3},...,P_{6}\right\}\$ and compare the results to those obtained by (a) Eq.(7) and Eq.(8) p.36-2 and (b) HW7.9.

END HW7.10

1st recurrence relation: Derivation

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 $\displaystyle \mu$

$\displaystyle {\begin{aligned}{\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}=-{\frac {1}{2}}A^{-{\frac {3}{2}}}{\frac {dA}{d\mu \ }}={\frac {\rho \ }{A^{\frac {3}{2}}}}\end{aligned}}$

(5)

with ${\frac {dA}{d\mu \ }}=-2\rho \ \$

From Eq.(5) p.40-3 :

$\displaystyle {\begin{aligned}{\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}={\frac {\rho \ }{A^{\frac {3}{2}}}}=\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho ^{n}\end{aligned}}$

(6)

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 Eq.(6) p.41-3

Step 2. Derivative with respect to $\displaystyle \rho$

$\displaystyle {\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}={\frac {-\rho \ +\mu \ }{A^{\frac {3}{2}}}}=\sum _{n=0}^{\infty }P_{n}(\mu \ )n\rho ^{n-1}$

(1)