# Accuracy of the Newton-Cotes formulas for polynomials of degree $n+1$ and even $n$

Let $f$ be a polynomial of degree $n+1$. The Newton-Cotes formula is given by $$\int_{-t}^tf(x)\text{ d}x\approx\sum_{k=0}^nf(x_k)\int_{-t}^t\omega_{n+1}(x)\text{ d}x \tag{*}$$ where $$\omega_{n+1}(x):=\prod_{k=0}^n(x-x_k)$$ denote the Newton basis polynomials and $$x_k:=kh-t\;,\;\;\;h:=2\frac{t}{n}$$ $(*)$ is accurate for polynomials of degree at most $n$. Why is it even accurate for polynomials of degree $n+1$, when $n$ is even?

If $p(f\mid x_0,\cdots,x_n)$ denotes the interpolation polynomial of $f$. The interpolation error is given by $$f(x)-p(f\mid x_0,\cdots,x_n)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\omega_{n+1}(x)$$ Because $f$ has degree $n+1$ $$\frac{f^{(n+1)}(\xi)}{(n+1)!}\equiv c$$ is a constant. Thus, $$\int_{-t}^tp(f\mid x_0,\cdots,x_n)\text{ d}x=\int_{-t}^t f(x)\text{ d}x-c\int_{-t}^t\omega_{n+1}(x)\text{ d}x$$ Because $n$ is even, it holds $x_k=-x_{n-k}$. So, $$\omega_{n+1}(t-x)=\omega_{n+1}(x-t)$$ However, I'm unable to show the desired statement.

*EDIT:*$\;$ I've provided an answer which is pretty close to the solution, but still need some help.

I think I'm pretty close to the solution. Since $n$ is even, it holds $x_k=-x_{n-k}$. That implies $$\omega_{n+1}(x)=(x-x_{n/2})\prod_{k=0}^{\frac{n}{2}-1}(x-x_k)\prod_{k=0}^{\frac{n}{2}-1}(x+x_k)=x\prod_{k=0}^{\frac{n}{2}-1}(x^2-x_k^2)$$ So, it should hold $$\int_{-t}^t\omega_{n+1}(x)=0$$ I've tested this with $n$ even and some $t>0$. It seems to be true. But I'm still unable to argue why.
• It won't work with $f(x_k)$ there. You need those to be the coefficients of the Newton basis polynomials in the interpolant, which are not just $f(x_k)$ (there is a triangular linear system in the way of reading them off from the values of $f$). In the Lagrange basis you get a formula along these lines. (Sorry I know this is an old question and might be trivial to you now 6 years later, but it did come up in the feed...) – Ian Apr 19 '20 at 4:12