gaussian quadrature weight function x

Question

How do I find an orthogonal polynomial for the Gaussian quadrature when the weight function is $x$?

In particular, I need to find $w_0,w_1,x_0, x_1$ such that $$\int_0^1xf(x)dx \approx w_0f(x_0)+w_1f(x_1) $$ is exact when $f \in \mathcal{P}_3$.

Which polynomials can I use to find the abscissae $x_0$ and $x_1$?

Answer 1

You can construct monic orthogonal polynomials using Gram-Schmidt. We can take the first one $p_0(x)$ equal to $1$. We then subtract from the function $g_1(x) = x$ the component in the direction of $p_0(x)$ leaving us with a function that is orthogonal to $p_0(x)$. If we call this function $q_1(x)$, then we have:

$$p_1(x) = g_1(x) - \frac{\left<g_1,p_0\right>}{\left<p_0,p_0\right>} p_0(x) = x - 2\int_0^{1}x^2 dx = x - \frac{2}{3}$$

Then the next polynomial is of second degree, we can find it by starting with $g_2(x) = x^2$ and subtracting the projection on the subspace spanned by $p_0(x)$ and $p_1(x)$:

$$ \begin{split} p_2(x) &= g_2(x) - \frac{\left<g_2,p_0\right>}{\left<p_0,p_0\right>} p_0(x)-\frac{\left<g_2,p_1\right>}{\left<p_1,p_1\right>} p_1(x)\\ & = x^2 - 2\int_0^{1}x^3 dx - 36 \left(x-\frac{2}{3}\right)\int_0^1 x^3 \left(x-\frac{2}{3}\right)dx\\ & = x^2 - \frac{6}{5} x +\frac{3}{10} \end{split} $$

The abscissae are the zeros of $p_2(x)$, the weights are given by this formula.

Answer 2

A shorter way to construct monic orthogonal polynomial $p_2$ is orthogonal to 1 and x, so $p_2(x)=x^2+ax+b$; The orthogonality implies $$ \int_0^1 x p_2(x)dx=\frac{1}{4}-\frac{1}{3}a+\frac{1}{2}b=0; $$ $$ \int_0 x^2 p_2(x)dx=\frac{1}{5}+\frac{1}{4}a+\frac{1}{3}b=0. $$ Solving the system with respect to $a$ and $b$ we obtain $a=-\frac{6}{5}$ and $b=\frac{3}{10}$. The nodes are the roots of $p_2$, i.e. $x_0=\frac{3}{5}-\frac{\sqrt{6}}{10}$ and $x_1=\frac{3}{5}+\frac{\sqrt{6}}{10}$. The weights are the solution of the system $$w_0+w_1=\int_0^1x dx=\frac{1}{2}$$ $$w_0x_0+w_1x_1=\int_0^1x^2 dx=\frac{1}{3},$$ that is, $w_0=\frac{1}{4}-\frac{\sqrt{6}}{36}$ and $w_1=\frac{1}{4}+\frac{\sqrt{6}}{36}$. The error (remainder term) $$R(f)=\frac{f^{(4)}(\xi)}{4!}\int_0^1 x\left[p_2(x) \right]^2 dx= \frac{f^{\left(4\right)}\left(\xi \right)}{14400},$$ for some $\xi\in(0,1)$.