Gauss–Kronrod quadrature formula

Remarks

The rule has $3n + 2$ free parameters, which matches the $3n + 2$ conditions of exactness for $1, x, \dots, x^{3n+1}$. The new nodes are the zeros of the Stieltjes polynomial $E_{n+1}$, the polynomial of degree $n + 1$ with $\int_{-1}^{1} E_{n+1}(x) P_n(x) x^k\,dx = 0$ for $k = 0, \dots, n$, where $P_n$ is the Legendre polynomial. For the Legendre weight these zeros are real, lie in $(-1, 1)$, and interlace with the Gauss nodes. In practice $\abs{G_n - K_{2n+1}}$, the difference of the Gauss value and the Kronrod value, serves as an error estimate.

Example: $n = 1$. The Gauss node is $x_1 = 0$, and the rule $\int_{-1}^{1} f \approx \frac{8}{9} f(0) + \frac{5}{9} f(-\sqrt{3/5}) + \frac{5}{9} f(\sqrt{3/5})$ keeps it and adds two nodes. It is exact for $1$ ($\frac89 + \frac{10}{9} = 2$), for $x^2$ ($\frac{10}{9} \cdot \frac35 = \frac23$), for $x^4$ ($\frac{10}{9} \cdot \frac{9}{25} = \frac25$), and for odd powers by symmetry, so up to degree $3n + 1 = 4$. In this smallest case the Kronrod rule coincides with the $3$-point Gauss–Legendre rule.

Non-example: for $n = 2$ the $5$-point Gauss–Legendre rule is not a Gauss–Kronrod formula. Its nodes are $0$ and the zeros of $63x^4 - 70x^2 + 15$, approximately $\pm 0.538$ and $\pm 0.906$, so they do not contain the $2$-point Gauss nodes $\pm 1/\sqrt{3} \approx \pm 0.577$.