Trapezoid rule over trigonometric polynomials

Question

The question is regarding trapezoid rule applied on trigonometric polynomials

Here is the question

Show that the composite trapezoid rule over an equidistant partitioning with interval size $h = 2 * \pi / (n+1)$ is exact for all trigonometric polynomials of period $ 2 \pi$, i.e., for functions of the form $\displaystyle\sum\limits_{k=-n}^n c_{k}e^{ikt}$., where i is the imaginary unit.`

Now, I have no idea what he means by that question. What do they mean by exact?

Thank you for your time.

Cheers

Answer 1

Hint, what if we wrote a formula for trigonometric polynomials as:

$$\displaystyle f(x) = a_0 + \sum_{k = 1}^{n-1} a_k \cos kx + b_k \sin kx$$

Then:

$$\displaystyle \int_0^{2 \pi} f(x) dx = \frac{2 \pi}{n} \sum_{i = 0}^{n-1} f(\frac{2 \pi i}{n})$$

Now, try an example and see if you can show this.

Answer 2

The composite trapezoid rule applied to the interval $[0,2\pi]$ with $n+1$ intervals $(n+2)$ points including the ends is that $\int_0^{2\pi} f(x) dx \approx \frac 1{2(n+1)}\left(f(0)+2f(\frac {2\pi}{n+1}) + 2f(\frac {2*2\pi}{n+1}+\ldots+2f(\frac {n*2\pi}{n+1})+f(2\pi)\right)$

You are to evaluate the integral and the approximation for your functions and show they are equal.