# EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 37: Thurs, 17Nov09

### Page 37-1

P.36-4 continued

${\displaystyle P_{n}(x)\ }$

${\displaystyle P_{1}(x)=x\ }$

${\displaystyle Q_{n}(x)\ }$

From P.18-1 :

 {\displaystyle \displaystyle {\begin{aligned}Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\end{aligned}}} (1)

HW show ${\displaystyle Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\ }$ END HW

ref K p33 for ${\displaystyle Q_{2},Q_{3}...\ }$

 {\displaystyle \displaystyle {\begin{aligned}Q_{n}(x)=P_{n}(x)\tanh ^{-1}x-2\sum _{j=1,3,5}^{J}{\frac {2n-2j+1}{(2n-j+1)j}}P_{n-j}(x)\end{aligned}}} (2)

${\displaystyle Q_{0}(x)=\tanh ^{-1}(x)\ \ }$ is odd

HW Use Eq(2) to show when ${\displaystyle Q_{n}\ }$ is even or odd, depending on "n" END HW

HW Plot ${\displaystyle \left\{P_{0},P_{1},...,P_{4}\right\}\ }$ and ${\displaystyle \left\{Q_{0},Q_{1},...,Q_{4}\right\}\ }$ END HW

Legendre function ${\displaystyle L_{n}(x)=P_{n}(x)\ }$ or ${\displaystyle Q_{n}(x)\ }$ solution of Legendre solution

### Page 37-2

 {\displaystyle \displaystyle {\begin{aligned}\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}}} (1)

for ${\displaystyle n\neq m\ }$

for ${\displaystyle L_{n}=P_{n},\left\langle L_{n},L_{n}\right\rangle =\left\langle P_{n},P_{n}\right\rangle ={\frac {2}{2n+1}}\ }$

HW for ${\displaystyle \left\langle L_{n},L_{n}\right\rangle =\left\langle P_{n},Q_{n}\right\rangle \ =}$ END HW

${\displaystyle \left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},P_{m}\right\rangle \ =}$ Eq.(3) P.33-1

${\displaystyle \left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},Q_{m}\right\rangle \ =0}$ for ${\displaystyle n\neq m\ }$

Proof: Legendre equation, Eq.(1) P.14-2 : 2) ${\displaystyle \left[(1-x^{2})y'\right]'+n(n+1)y=0\ }$

Where

 {\displaystyle \displaystyle {\begin{aligned}\left[(1-x^{2})y'\right]'=(1-x^{2})y''-2xy'\end{aligned}}} (2)

${\displaystyle \Rightarrow \ \left[(1-x^{2})L_{n}'\right]'+n(n+1)L_{n}=0\ }$

Multiply by ${\displaystyle L_{m}\ }$ and integrate from -1 to +1:

${\displaystyle \int _{-1}^{1}L_{m}\left[(1-x^{2})L_{n}'\right]'\,dx+n(n+1)\int _{-1}^{1}L_{m}L_{n}\,dx=0\ }$

Where ${\displaystyle L_{m}\left[(1-x^{2})L_{n}'\right]'=\alpha \ \ }$

### Page 37-3

Integrate ${\displaystyle \alpha \ \ }$ by parts:

 {\displaystyle \displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{n}'L_{m}'\,dx+(n)(n+1)\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}}} (1)

Interchange n and m:

 {\displaystyle \displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{m}'L_{n}'\,dx+(m)(m+1)\left\langle L_{m},L_{n}\right\rangle =0\end{aligned}}} (2)

Eq(1)-Eq(2):

${\displaystyle \left[n(n+1)-m(m+1)\right]\left\langle L_{n},L_{m}\right\rangle =0}$

Where ${\displaystyle \left[n(n+1)-m(m+1)\right]\neq 0}$ since ${\displaystyle n\neq m\ }$

${\displaystyle \Rightarrow \ \left\langle L_{n},L_{m}\right\rangle =0\ }$ when ${\displaystyle n\neq m\ }$

cf.K.p41