We want to approximate a function $f$ with a second-degree interpolating polynomial in the interval $[-1,1]$. I need to pick three interpolation points such that the interpolation polynomial $p$ has the best possible approximation. I need to argue that it's best to pick the interpolation points symmetrically. Then I am given a hint which is; the solution to the equation $\frac{2}{3\sqrt{3}}\tau^3 = 1 – \tau^2 $ is $\tau = \frac{1}{2}\sqrt{3}$.
The error using second degree interpolation at the interpolation points $x_0, x_1, x_2$ at the point x is given by $$(x – x_0)(x – x_1)(x – x_2)\frac{f^{3}(\xi)}{6}$$ for some $\xi \in [-1,1]$.
We could probably argue that if we did not pick the interpolation points symmetrically, then the term $(x – x_0)(x – x_1)(x – x_2)$ would start behaving very wildly. I do not really see how the given hint applies here.
Best Answer
If you are expected to choose interpolation points only, but not the weights, then for this problem the hint would be: Chebyshev interpolation (or Tschebyschev, depends who you talk to)
The Chebyschev polynomial $T_n$ of degree $n$ has the property, that it has minimal $L_\infty$-Norm among all polynomials of degree $n$ on $[-1,1]$ with maximal coefficient $a_n=2^{n-1}$.
The roots $x_0,\dots,x_n$ of $T_{n+1}$ minimize $\omega_{n+1}(x)=\prod_{i=0}^n (x-x_i)$ (on $[-1,1]$)