If $2 \sum_{j=0}^{n}a_jf(x_j)$ is a quadrature for $\int_{-1}^1f(x)dx$ exact for polynomials of degree $\le n$ and $x_j=-x_{n-j}$, show $a_j=a_{n-j}$.

Question

This was an problem in today's exam in numerical methods and I would like to know if my solution is correct.

For approximating the integral $I(f) = \int_{-1}^{1} f(x) dx$ we take the quadrature formula $$ Q(f) = 2 \sum_{j = 0}^{n} \lambda_j f(x_j) $$ with supporting points $x_0, \ldots, x_n \in [-1, 1]$, which is exact for polynomials with degree up to $n$. Show that if $x_j = -x_{n - j}$ for $j \in \{ 0, \ldots, n \}$, we have $\lambda_j = \lambda_{n - j}$ for $j \in \{ 0, \ldots, n \}$.

What I've tried: For an odd function $f$, we have $$ Q(f) = \sum_{j = 0}^{n} \lambda_j f(x_j) + \sum_{j = 0}^{n} \lambda_j f(x_j) = \sum_{j = 0}^{n} \lambda_j f(x_j) + \sum_{j = 0}^{n} \lambda_{n - j} f(x_{n - j}) = \sum_{j = 0}^{n} (\lambda_j - \lambda_{n - j}) f(x_j). $$ Now let $f$ be an odd polynomial of degree less than or equal to $n$. Then the quadrature formula is equal to the integral and as $f$ is odd, we have $$ 0 = I(f) = Q(f) = \sum_{j = 0}^{n} (\lambda_j - \lambda_{n - j}) f(x_j). $$ Since this holds for all odd polynomials of degree less than or equal to $n$, I want to conclude that the coefficients $\lambda_j - \lambda_{n - j}$ are zero, i.e. $\lambda_j = \lambda_{n - j}$, but I am not sure if this is valid.

Answer 1

There are several ways to proceed:

1. Note that terms $(\lambda_j - \lambda_{n-j}) f(x_j)$ and $(\lambda_{n-j} - \lambda_{j}) f(x_{n-j})$ are in fact the same for odd $f(x)$. So the summation may be reduced to $Q(f) = 2 \sum_{j=0}^{\lfloor \frac{n-1}{2}\rfloor} (\lambda_j - \lambda_{n-j}) f(x_j)$. Since all $x_j$ in this sum are negative, one may choose arbitrary values for $f(x_j)$, mirror them for positive $x$ and take $f(x)$ as interpolating polynomial.
2. We might use Lagrange basis interpolating polynomials $\ell_j(x)$. One of the equivalent definitions of these polynomials is given as $$ \deg \ell_j(x) = n\\ \ell_j(x_k) = \delta_{jk} \equiv \begin{cases} 1, &k = j\\ 0, &k \neq j \end{cases} $$ It is easy to see now that $\ell_j(-x)$ is also an $n$-degree polynomial and at the nodes it satisfies $$ \ell_j(-x_k) = \ell_j(x_{n-k}) = \delta_{j,n-k} = \delta_{n-j,k} = \ell_{n-j}(x_k). $$ Due to uniqueness we conclude that $\ell_j(-x) = \ell_{n-j}(x)$. Consider now $$ f(x) = \ell_j(x) - \ell_{n-j}(x). $$ It is clear now that $f(x)$ is odd. Plugging it to $Q(f)$ yields $$ 0 = Q(f) = (\lambda_j - \lambda_{n-j}) - (\lambda_{n-j} - \lambda_j) = 2(\lambda_j - \lambda_{n-j}). $$ so $\lambda_j = \lambda_{n-j}$.

There is also different approach to the problem. If $Q$ is exact for all polynomials of degree $n$ then we might apply it to $\ell_k(x)$: $$ \int_{-1}^1 \ell_k(x) dx = I(f) = Q(f) = \sum_{j=0}^n \lambda_j \ell_k(x_j) = \lambda_j. $$ On the other hand $$ \int_{-1}^1 \ell_k(x) dx = \int_{-1}^1 \ell_k(-x) dx = \int_{-1}^1 \ell_{n-k}(x) dx, $$ so it is necessary that $\lambda_j = \lambda_{n-j}$.