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
- $(\genfrac{}{}{0ex}{}{r}{k})=$ "r choose k" :

$(1+x{)}^r=\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{r}{k})x^k$ for $\left|x\right|\le 1$

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

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 $(-\alpha {}_1+4\mu {}^2\alpha {}_2)=P+2(\mu )$

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$\begin{array}{c}{P}_0(\mu )=\alpha {}_0=1\end{array}$

$\begin{array}{c}{P}_1(\mu )=2\mu \alpha {}_1=\mu \end{array}$

$\begin{array}{c}{P}_2(\mu )=\frac{1}{2}(3\mu {}^2-1)\end{array}$

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

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{P_n}^{\prime }(\mu )\rho {}^n\end{array}$

Where $P_0(\mu )=1\Rightarrow {P_0}^{\prime }=0$ and ${P_n}^{\prime }(\mu )=\frac{d}{d\mu }{P}_n(\mu )$

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2) $\begin{array}{c}\frac{d}{d\rho }\frac{1}{\sqrt{A}}=\frac{-\rho +\mu }{A^{\frac{3}{2}}}=\sum _{n=0}^{\mathrm{\infty }}{P}_n(\mu )n\rho {}^{n-1}\end{array}$

Compare Eq(4) and Eq(5)

$\begin{array}{c}\mu -\rho =\rho \end{array}$

$\frac{\rho (\mu -\rho )}{A^{\frac{3}{2}}}=(\mu -\rho )\sum _{n=1}^{\mathrm{\infty }}{P_n}^{\prime }(\mu )\rho {}^n=\rho \sum _{n=0}^{\mathrm{\infty }}{P}_n(\mu )n\rho {}^{n-1}=\rho \sum _{n=1}^{\mathrm{\infty }}{P}_n\mu n\rho {}^{n-1}$

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

$\Rightarrow \mu \sum _{n=1}{P_n}^{\prime }\rho {}^{″}-\sum _{n=1}{P_n}^{\prime }\rho {}^{n-1}=\sum _{n=1}{P}_n\mu \rho {}^{″}$

$\Rightarrow \mu {P_1}^{\prime }\rho -P_{1}1\rho +\sum _{n=2}[-P_{n^{\prime }-1}+\mu {P_n}^{\prime }-n{P}_n]\rho {}^n=0$

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$\Rightarrow -P_{n^{\prime }-1}+\mu {P_n}^{\prime }-n{P}_n=0$

$\begin{array}{c}\Rightarrow \mu {P_n}^{\prime }-P_{n^{\prime }-1}=n{P}_n\end{array}$

References

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