• Dec 28th 2010, 09:55 AM
lvleph
I am working on studying for my last Qualifying Exam and ran across a problem that has stumped me. I have seen someone else's work for this, but I am pretty it is wrong. Anyway...

Quote:

Let $f(x)$ be a smooth function and consider the integral
$I=\int_{-1}^1\! f(x) \, dx$
and consider $p_{n-1}$ that interpolates $f$ at the roots of the Legendre polynomial of degree n.

Use $p_{n-1}$ to derive a quadrature method for the integral $I$
The solution I have seen is as follows
Quote:

Let $\left\{x_i\right\}_{i=1}^n$ be the $n$ distinct roots of $\widetilde{p}_n(x)$ (the Legendre Polynomial of degree $n$)

Define
$w_i = \int_{-1}^1\! \prod_{\stackrel{j=1}{j\ne i}}^n \frac{x-x_j}{x_i-x_j}\, dx \Rightarrow Q(f) = \sum_{i=1}^n w_i f(x_i)$
So my issue is that he didn't really use Legendre polynomials. Instead he bipassed the Legendre polynomials in favor of a Lagrange polynomial that interpolates at the Legendre roots. Maybe I am missing something or just confused. Could someone help me either understand the work above or help me with the problem? Thanks in advanced.
• Dec 28th 2010, 09:37 PM
CaptainBlack
Quote:

Originally Posted by lvleph

So my issue is that he didn't really use Legendre polynomials. Instead he bipassed the Legendre polynomials in favor of a Lagrange polynomial that interpolates at the Legendre roots. Maybe I am missing something or just confused. Could someone help me either understand the work above or help me with the problem? Thanks in advanced.

A polynomial with the same roots as a Legendre polynomial differs from the Legendre polynomial only by a multiplicative constant (or it is a Legendre polynomial with a different normalisation).

CB
• Dec 29th 2010, 07:36 AM
lvleph
I figured that was the case. Thank you.