User:EGM6321.F10.TEAM1.WILKS/Mtg37

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EGM6321 - Principles of Engineering Analysis 1, Fall 2009[edit]

Mtg 37: Thurs, 17Nov09


Page 37-1

[edit]

P.36-4 continued

 P_n(x) \

 P_1(x)=x \

 Q_n(x) \

From P.18-1 :

\displaystyle
\begin{align}
Q_1(x)=\frac{1}{2}xlog  \left ( \frac{1+x}{1-x} \right ) -1 = x \tanh ^{-1}x-1
\end{align}

(1)



HW show  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  Q_2, Q_3...\

\displaystyle
\begin{align}
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{align}

(2)



 Q_0(x)= \tanh ^{-1}(x) \ \ is odd

HW Use Eq(2) to show when  Q_n \ is even or odd, depending on "n" END HW

HW Plot  \left \{ P_0, P_1,...,P_4 \right \} \ and  \left \{ Q_0, Q_1,...,Q_4 \right \} \ END HW

Legendre function  L_n(x)=P_n(x) \ or  Q_n(x) \ solution of Legendre solution

Page 37-2

[edit]

\displaystyle
\begin{align}
\left \langle L_n,L_m \right \rangle = 0 
\end{align}

(1)

for  n \ne m \

for  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  \left \langle L_n,L_n \right \rangle = \left \langle P_n,Q_n \right \rangle \ =  END HW

 \left \langle L_n,L_m \right \rangle =  \left \langle P_n,P_m \right \rangle \ =  Eq.(3) P.33-1

 \left \langle L_n,L_m \right \rangle =  \left \langle P_n,Q_m \right \rangle \ = 0 for  n \ne m \

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

Where

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

(2)



 \Rightarrow \ \left [ (1-x^2)L_n' \right ]'+n(n+1)L_n=0 \

Multiply by  L_m \ and integrate from -1 to +1:

 \int_{-1}^{1} L_m \left [ (1-x^2)L_n' \right ]' \, dx + n(n+1) \int_{-1}^{1} L_mL_n \, dx=0 \

Where  L_m \left [ (1-x^2)L_n' \right ]' = \alpha\ \


Page 37-3

[edit]

Integrate  \alpha\ \ by parts:

\displaystyle
\begin{align}
- \int_{-1}^{1} (1-x^2)L_n'L_m'\, dx +(n)(n+1) \left \langle L_n,L_m \right \rangle = 0 
\end{align}

(1)



Interchange n and m:

\displaystyle
\begin{align}
- \int_{-1}^{1} (1-x^2)L_m'L_n'\, dx +(m)(m+1) \left \langle L_m,L_n \right \rangle = 0 
\end{align}

(2)



Eq(1)-Eq(2):

 \left [ n(n+1)-m(m+1) \right ]  \left \langle L_n,L_m \right \rangle =0

Where  \left [ n(n+1)-m(m+1) \right ] \ne 0 since  n \ne m \

 \Rightarrow \ \left \langle L_n,L_m \right \rangle =0 \ when  n \ne m \

cf.K.p41

References[edit]