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 !

  • 1
    $\begingroup$ You basically want to arrive at equation 22 in mathworld.wolfram.com/Newton-CotesFormulas.html As you can see its only at order 9 and odd orders beyond that we get negative coefficients. Unfortunately I can't see any simple argument for exactly why this is the case other than doing this direct calculation (which is not pleasant to do by hand). But you can simplify it by deriving equation 26 and using this to show that $w_2$ is negative. $\endgroup$ – Winther Nov 18 '20 at 18:46

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.


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