The task is to give for $\cos(x)$ the nodes of the interpolation polynom of degree 1 that approximates the function on $[-\pi,\pi]$ the best as well as the related error.
I want to solve this task explicitly with Chebyshev nodes/Chebyshev polynoms.
We know that $\cos(x) - p_1(x) =$$ -\sin(\xi) (x-x_1)$. Minimizing $||(x-x_1)||_{\infty}$ is achieved by using the Chebyshev node $x_1 = $$(-\pi + \pi)\over2$$ + $$(\pi + \pi)\over2$$ \cos(\pi$$ 2-1 \over 2$) = 0. The error then is $||\cos(x) - p_1(x)||_{\infty} = pi$ .
But by a actualy closed formula I ought to receive for the error $ (2\pi)^2 \over 8$.
Help is much appreciated!