# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg39

## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 39: Thurs, 19Nov09

### Page 39-1

Note: Other Application: Quantum Mechanics

discrete variables <-- coding theory

Ref: Nikiforov, et.al (1991)

Probability (Queing theory, birth and death)

Generating functions:
-Legendre Polynomial $P_{n}\$ : Eq.(5) P.38-3
- ${r \choose k}\ =$  "r choose k" :

$(1+x)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{k}\$  for $\left|x\right\vert \leq 1\$

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

From Eq.(6) P.38-3 and Eq.(7) P.38-3 : ${\frac {1}{\sqrt {A(\mu \ ,\rho \ )}}}=\alpha \ _{0}+\alpha \ _{1}(2\mu \ \rho \ -\rho \ ^{2})+\alpha \ _{2}(2\mu \ \rho \ -\rho \ ^{2})^{2}+...=\alpha \ _{0}+(2\mu \ \alpha \ _{1})\rho \ +(-\alpha \ _{1}+4\mu \ ^{2}\alpha \ _{2})\rho \ ^{2}+...\$

Where $(2\mu \ \rho \ -\rho \ ^{2})=4\mu \ ^{2}\rho \ ^{2}-4\mu \ \rho \ ^{3}+\rho \ ^{4}\$

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

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

and $(-\alpha \ _{1}+4\mu \ ^{2}\alpha \ _{2})=P+2(\mu \ )\$

### Page 39-2

 \displaystyle {\begin{aligned}P_{0}(\mu \ )=\alpha \ _{0}=1\end{aligned}} (1)

 \displaystyle {\begin{aligned}P_{1}(\mu \ )=2\mu \ \alpha \ _{1}=\mu \ \end{aligned}} (2)

 \displaystyle {\begin{aligned}P_{2}(\mu \ )={\frac {1}{2}}(3\mu \ ^{2}-1)\end{aligned}} (3)

HW: Continue the power series development to find $P_{3},P_{4},P_{5}\$  and complete result to that obatined by Eq.(6) P.31-3 or Eq.(7) P.31-3

2 recurrence formulas

Plan: Find ${\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}\$  and ${\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}\$

1) ${\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}=-{\frac {1}{2}}A^{\frac {-3}{2}}{\frac {dA}{d\mu \ }}={\frac {\rho \ }{A^{\frac {-3}{2}}}}\$

Recall Eq.(5) P.38-3, now ${\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}={\frac {\rho \ }{A^{\frac {3}{2}}}}=\$

 \displaystyle {\begin{aligned}\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho \ ^{n}\end{aligned}} (4)

Where $P_{0}(\mu \ )=1\Rightarrow \ P_{0}'=0\$  and $P_{n}'(\mu \ )={\frac {d}{d\mu \ }}P_{n}(\mu \ )\$

### Page 39-3

2) \displaystyle {\begin{aligned}{\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}\end{aligned}} (5)

Compare Eq(4) and Eq(5)

 \displaystyle {\begin{aligned}\mu \ -\rho \ =\rho \ \end{aligned}} ${\frac {\rho \ (\mu \ -\rho \ )}{A^{\frac {3}{2}}}}=(\mu \ -\rho \ )\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho \ ^{n}=\rho \ \sum _{n=0}^{\infty }P_{n}(\mu \ )n\rho \ ^{n-1}=\rho \ \sum _{n=1}^{\infty }P_{n}{\mu \ }n\rho \ ^{n-1}$

Where n=1 in summation due to factor u being introduced

$\Rightarrow \ \mu \ \sum _{n=1}P_{n}'\rho \ ''-\sum _{n=1}P_{n}'\rho \ ^{n-1}=\sum _{n=1}P_{n}\mu \ \rho \ ''\$

$\Rightarrow \ \mu \ P_{1}'\rho \ -P_{1}1\rho \ +\sum _{n=2}\left[-P_{n-1}'+\mu \ P_{n}'-nP_{n}\right]\rho \ ^{n}=0\$

### Page 39-4

$\Rightarrow \ -P_{n-1}'+\mu \ P_{n}'-nP_{n}=0\$

 \displaystyle {\begin{aligned}\Rightarrow \ \mu \ P_{n}'-P_{n-1}'=nP_{n}\end{aligned}} (1)