Numerical method with rational nodes and weights to compute exact value of definite integral?

Question

Description

Let $p(x)$ be a polynomial of degree $n$ and rational coefficients.

I'm interested in computing numerically the exact value of the integral $I$, which is also rational

$$I = \int_{a}^{b} p(x) \ dx$$

Usually the numerical integral is computed by the sum of weights $w_i$ and function values at nodes $x_i$

$$\tilde{I} = \sum_{i=1}^{m} w_i \cdot p(x_i)$$

The exact integral can be obtained by

• Newton-cotes formula with $m = 1+2\left\lfloor \dfrac{n}{2}\right\rfloor$ evaluation nodes
• Gauss-legendre quadrature with $m = \left\lceil \dfrac{n+1}{2}\right\rceil$ nodes.

Clearly gauss-legendre uses less points than newton-cotes due to the fact Newton-cotes keeps the nodes fixed at equal distance and find the best weights, while gauss-legendre find the best nodes and weights.

But the problem with gauss-legendre is the fact its nodes are irrational, impossible to represent them in computer, while rational values can.

Then I asked myself if there's a numerical method such we allow the nodes $x_{i}$ to move (different from newton-cotes) while keeping $x_i$ and $w_{i}$ in rationals (different from gauss-legendre). But unfortunatelly I have no clue how to start it

Question

• Is there some numerical method that uses $m \le 2\left\lfloor \dfrac{n}{2}\right\rfloor$ rational nodes and returns the exact value of $I$ ?

An equivalent question is:

• Given constant $m$, what is the maximum degree $n$ such $I$ is equal to $\tilde{I}$?

Notes

Note 1: I don't know the values of the coefficients $c_i$ of $p(x)$ to compute

$$I = \sum_{i=0}^{n} c_{i} \cdot \dfrac{b^{i+1}-a^{i-1}}{i+1}$$

Finding these coefficients means evaluating $p(x)$ at $(n+1)$ nodes, which would be the same as newton-cotes

Note 2: Assume my computer has infinite integer precision and can represent any rational.

Note 3: For many the approximation of the gauss-legendre irrationals by float are enough, but trying to transform these irrationals into rationals (in computer) may lead to error.

Answer 1

It can be done with $n$ abscissae for odd $n$, though I'm not sure it can always be done with positive weights. Take $a=-1$, $b=1$ without loss of generality.

I'll do $n=5$. Consider $$I=w_0(f(-x_0)+f(x_0)) + w_1(f(-x_1)+f(x_1)) + w_2f(0).$$ The symmetry ensures that it integrates odd powers correctly. Make the three equations that say it works for $1$, $x^2$ and $x^4$, and solve for $w_0,w_1,w_2$ as rational functions of $x_0,x_1$. Put in any rational values of $x_0,x_1$ and you get rational weights. What I'm not sure about is whether you can always get positive rational weights.

For $n=5$: $x_0=1,x_1=\frac12$ gives $w_0=\frac{7}{45}$, $w_1=\frac{32}{45}$, $w_2=\frac{4}{45}$.

For $n=7$: $x_0=1$, $x_1=\frac23$, $x_2=\frac13$ gives $w_0=\frac{41}{420}$, $w_1=\frac{18}{35}$, $w_2=\frac{9}{140}$, $w_3=\frac{68}{105}$.

ADDED:

The problem for degree 7 and 5 abscissae symmetric about 0 comes done to whether $y$ and $\sqrt{(21y^2-15)/(35y^2-21)}$ can be rational at the same time. I didn't prove it but I suspect there are no solutions and therefore no rational quadrature rule of that form.