Thread: Quadrature Using Legendre Polynomials

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...

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

Use $\displaystyle p_{n-1}$ to derive a quadrature method for the integral $\displaystyle I$

The solution I have seen is as follows

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

Define
$\displaystyle w_i = \int_{-1}^1\! \prod_{\stackrel{j=1}{j\ne i}}^n \frac{x-x_j}{x_i-x_j}\, dx \Rightarrow Q(f) = \sum_{i=1}^n w_i f(x_i)$

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.

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

I figured that was the case. Thank you.