User:Egm6321.f10.team4.Yoon/Mtg38

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_11_11_08_24_37.djvu | Mtg 38:]] Thu, 11 Nov 10

= Legendre functions: Orthogonality, odd-ness, even-ness =

[[media: 2010_11_11_08_24_37.djvu | Page 38-1]]
Now, back to the heat problem.

= Heat conduction on a sphere (cont'd) =

Spotting zero coefficients with no computation
[[media: 2010_11_10_14_40_12.djvu | Eqn(4) p.37-5:]] $$\displaystyle f(\mu) =T_0 \left( 1-\mu^2 \right)^2 $$ is an even function.

$$\displaystyle \Rightarrow A_n=0$$ for $$\displaystyle n=2k+1$$, since $$\displaystyle P_{2k+1}(\mu)$$ is an odd function.

$$\displaystyle \Rightarrow A_1=A_3=A_5=. . . =0$$

Actually, it turns out that $$\displaystyle A_n=0 \ \forall n \geqslant5$$ due to the linear independence of the family $$\displaystyle \mathcal F = \{ P_k \} $$ and orthogonality of $$\displaystyle \mathcal F$$.

$$\displaystyle P_n(x) $$ is a polynomials of order n.

$$\displaystyle P_n(x) \in \mathcal P_n$$ set of all polynomials with $$\displaystyle \underbrace{\text{order}}_{\text{degree}} \leqslant n $$.

[[media: 2010_11_11_08_24_37.djvu | Page 38-3]]
Orthogonality of $$\displaystyle \mathcal F=\{ P_k \}$$ [[media: 2010_11_10_14_40_12.djvu | Eqn(2a) p.37-4]]

[[media: 2010_11_11_08_24_37.djvu | Page 38-4]]
Consider $$\displaystyle f(\mu) = T_0(1-\mu^2)^2 = T_0 \cos^4 \theta$$.



Hence only need to evaluate $$\displaystyle A_0, A_2, A_4$$

[[media: 2010_11_11_08_24_37.djvu | Page 38-5]]
[[media: 2010_11_10_14_40_12.djvu | Eqn(1)p.37-2]]: