MHB Solving Coefficant Matrices with Legendre Polynomials

[ognik]

Whenever a problem seems too easy, I assume I'm missing something :-)

This is in a section on Legendre polynomials ...

Given the series $ \alpha_0 + \alpha_2Cos^2\theta +\alpha_4C^4 +\alpha_6C^6 = a_0P_0 + a_2P_2 + a_4P_4 +a_6P_6 $ (abbreviating $Cos^n\theta$ to $C^n$)

Express both coefficients as col. matrices and find A, B such that $A\vec{\alpha}=\vec{a} $ and $ B\vec{a}=\vec{\alpha} $

I found, almost by inspection, that A was the diagonal matrix with elements $ P_0, \frac{P_0}{C^2}, \frac{P_4}{C^4}, \frac{P_6}{C^6} $, similarly B is diagonal with elements $ \frac{1}{P_0}, \frac{C^2}{P_2}, \frac{C^4}{P_4}, \frac{C^6}{P_6} $ (and $AB=I$) Please confirm/correct?

 

[ognik]

Hi, people are normally so helpful here, so I figure when I don't get a reply I've done something wrong - if I have just let me know what please :-)

 

[ognik]

So my A is $ \begin{bmatrix}P_{0}&0&0&0 \\0&\frac{P_{2}}{C^2}&0&0 \\0&0&\frac{P_4}{C^4}&0\\0&0&0&\frac{P_6}{C^6} \end{bmatrix}$ , if someone can confirm/correct that, I'll be happy with the rest.

I could get an A with just entries in col 1, but I chose the diagonal matrices because it's more useful/important - also a basis for example?

 

[ognik]

Probably best if show the col. matrices I found.

So $ \vec{\alpha} =\left[1, C^2, C^4, C^6\right] $ and $ \vec{a} = \left[P_0, P_2, P_4, P_6\right] $

Any comments on these, or my matrix A?