About Newton-Cotes numerical integration for degree >8.

Question

I'm studying Newton-Cotes integration for nine points of interpolation, let's say

$$(a_i)_{0 \leq i \leq 8}=\{0,\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}, 1 \}$$

The approximation of $\int_{0}^1 f(t) dt$ is

$$\sum_{i=0}^8 f(a_i) \int_0^1 L_i(t) dt= \sum_{i= 0}^8 f(a_i) w_i$$

where $L_i$ is the lagrange elementary polynôme :

$$L_i(t) = \prod_{j=0, j \neq i}^8 \frac{t-a_j}{a_i-a_j}.$$

Now I am asked to prove that some $w_i=\int_0^1 L_i(t) dt$ are negatives. If we do the whole calculation it is true indeed and can be checked online. However I'm looking for a proof which don't calculate explicitly the weights $w_i$. This proof should be accessible to bachelor student but I can't see it right now...

Any help is welcomed !

Answer 1

There is a beautiful and elegant proof due to Bernstein. Start with $$P_5^{\prime}(x)=\frac1{2^55!}\frac{d^6}{dx^6}(x^2-1)^5=\frac18(21x^4-14x^2+1)$$ Then for any polynomial $Q_3(x)$ of degree at most $3$, $$\int_{-1}^1(1-x^2)P_5^{\prime}(x)Q_3(x)dx=0$$ Proof: just integrate by parts $6$ times. It follows that $$\int_{-1}^1(1-x^2)P_5^{\prime}(x)\frac{P_5^{\prime}(x)}{x-x_4}dx=0$$ Where $$x_4=\sqrt{\frac{7+2\sqrt7}{21}}$$ is the largest zero of $P_5^{\prime}(x)$. Since the $9$-point Newton-Cotes formula is exact for polynomials of degree at most $9$ it must also give this zero result. However, the integrand is only positive for $x_4\lt x\lt1$ so unless it uses a sample point in this interval or all sample points lie on zeros of the integrand, one of the weights must be negative or the integral would be negative.