Gaussian quadrature with weight function x^2

Question

I would like to get the points and weights of Gaussian quadrature formulas for $$ \int_{-1}^{+1} x^2 f(x)\;\text{d}x. $$ Is this tabulated anywhere yet?

Answer 1

It is not too difficult to derive the coefficients from scratch. The general idea is outlined here.

Up to order 5, the result is:

n=3: $$\begin{array}{cc} x_i & w_i\\\hline 0 & \frac{8}{75}\\ \pm \sqrt{\frac{5}7} & \frac{7}{25} \end{array}$$

n=4: $$\begin{array}{cc} x_i & w_i\\\hline \pm\frac{1}{3} \sqrt{5-2 \sqrt{\frac{10}{7}}} & \frac{1}{300} \left(50-\sqrt{70}\right)\\ \pm\frac{1}{3} \sqrt{5+2 \sqrt{\frac{10}{7}}} & \frac{1}{300} \left(50+\sqrt{70}\right) \end{array}$$

n=5: $$\begin{array}{cc} x_i & w_i\\\hline 0 & \frac{128}{3675}\\ \pm\sqrt{\frac{1}{33} \left(21-2 \sqrt{14}\right)} & \frac{3 \left(258+\sqrt{14}\right)}{4900}\\ \pm\sqrt{\frac{1}{33} \left(21+2 \sqrt{14}\right)} & \frac{3 \left(258-\sqrt{14}\right)}{4900} \end{array}$$

Answer 2

It's possible to use the symmetry of the problem to relate analysis to known problems. Because the weight function and interval are symmetric about $x=0$, we know that the even-ordered orthogonal polynomials are even: $P_{2n}(x)=Q_n(x^2)$ and the odd-ordered polynomials are odd: $P_{2n+1}(x)=xR_n(x^2)$. Any even-ordered polynomial of the family is orthogonal to any even polynomial of lesser order, $g_{n-1}(x^2)$ where $g_{n-1}(y)$ is a polynomial of degree at most $n-1$. Then $$0=\int_{-1}^1x^2Q_n(x^2)g_{n-1}(x^2)dx=2\int_0^1x^2Q_n(x^2)g_{n-1}(x^2)dx =\int_0^1y^{1/2}Q_n(y)g_{n-1}(y)dy=2^{-3/2}\int_{-1}^1(u+1)^{1/2}Q_n\left(\frac{u+1}2\right)g_{n-1}\left(\frac{u+1}2\right)du$$ So $Q_n\left(\frac{u+1}2\right)$ is orthogonal to any polynomial of degree at most $n-1$ over the interval $(-1,1)$ with weight function $w(u)=(u+1)^{1/2}$ so it must be a multiple of the Jacobi polynomial $J_n^{0,1/2}(u)$. Backtracking, we get $P_{2n}(x)=P_n^{0,1/2}(2x^2-1)$.

Similarly, we see that $P_{2n+1}(x)=xP_n^{0,3/2}(2x^2-1).$ Looking back at the Wikipedia page on Gaussian quadrature, the only thing we need to get the weights are the normalization integrals. We can work out that $$\int_{-1}^1x^2\left[P_{2n}(x)\right]^2dx=2^{-3/2}\int_{-1}^1(u+1)^{1/2}\left[P_n^{0,1/2}(u)\right]^2du$$ and $$\int_{-1}^1x^2\left[P_{2n+1}(x)\right]^2dx=2^{-5/2}\int_{-1}^1(u+1)^{3/2}\left[P_n^{0,3/2}(u)\right]^2du$$ So we can leverage what we know about the Jacobi polynomials to derive the Gaussian quadrature formulas without having to start from scratch.

Answer 3

For anyone interested, I've written orthopy for the derivation of Gaussian rules with new weight functions. The main example in the readme gives the points and weights

[-0.978228658146056992803938001123,
 -0.887062599768095299075157769304,
 -0.730152005574049324093416252031,
 -0.519096129206811815925725669458,
 -0.269543155952344972331531985401,
  0.269543155952344972331531985401,
  0.519096129206811815925725669458,
  0.730152005574049324093416252031,
  0.887062599768095299075157769304,
  0.978228658146056992803938001123]

[0.0532709947237135572432759986252,
 0.0988166881454075626728761840589,
 0.0993154007474139787312043384226,
 0.0628365763465911675266984722740,
 0.0190936733702070671592783399524,
 0.0190936733702070671592783399524,
 0.0628365763465911675266984722744,
 0.0993154007474139787312043384225,
 0.0988166881454075626728761840592,
 0.0532709947237135572432759986251]

for the 10-point Gaussian rule with the integral weight $x^2$.