2
$\begingroup$

Let $x_0,\cdots,x_k$ be roots of the Legendre polynomial $L_{k+1}(x)$. Show that for any $y\in(-1,1)$, the error of the Gauss-Legendre quadrature rule using $x_0,\cdots,x_k$ for approximating $\int_{-1}^{1}f(x)\,dx$ equals

$\int_{-1}^1 f[x_0,\cdots,x_k,y,x](x-x_0)\cdots(x-x_k)(x-y)\,dx$

Attempt

I presume the error term for $f(x) \approx \sum_{k=0}^{n}c_k(x-x_k)$ is $f[x_0,\dots,x_k,x](x-x_0)\dots(x-x_k)$, but the expected outcome differs by some terms. I am not sure why.

$\endgroup$
1
$\begingroup$

We need to use a trick here and reorder the terms. Notice that

$$\begin{aligned} f[x_0,\dots,x_k,y,x] &= f[y,x_0,x_1,\dots,x_k,x] \\ &= \frac{f[x_0,\dots,x_k,x]-f[x_0,\dots,x_k,y]}{x-y} \end{aligned}$$

We can write the function $f(x)$ as

$$f(x)=q(x)L_{k+1}(x)+r(x)$$

where $r(x)$ has degree at most $k$ and $L_{k+1}(x)$ has degree $k+1$. The degree of precision of the Gauss-Legendre quadrature rule is $2k+1$ that interpolates $k+1$ points, $x_0,\dots,x_k$. If $q(x)$ has degree at most $k$, then by orthonormality of the quadrature rule, $\int_{-1}^1q(x)L_{k+1}(x)\, dx=0$ and hence this term becomes our error term. We see that this is precisely the case when

$$\scriptsize \int_{-1}^1 f[x_0,\dots,x_k,y,x](x-x_0)\dots(x-x_k)(x-y)\, dx = \int_{-1}^1(f[x_0,\dots,x_k,x]-f[x_0,\dots,x_k,y])\cdot(x-x_0)\dots(x-x_k)\, dx $$

where $(f[x_0,\dots,x_k,x]-f[x_0,\dots,x_k,y])$ is a constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.