# Choosing interpolation points

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.

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 $[-1,1]$)
• My understanding is that Chebyshev interpolation is a representation of a function in terms of Chebyshev polynomials of degree up to n, i.e. $f(x) = \sum a_k \multT_k$. Since this representation involves all Chebyshev polynomials of degree less than n, whose zeros are different from the zeros of the polynomial of order n, it seems that the above argument does not quite apply. Is this right or am I missing something? – Confounded Oct 10 '16 at 14:41
• I think you are thinking about Chebyshev approximation, i.e., expansion in a Chebyschev basis. What I have said above pertains, if memory serves, to classical Lagrangian interpolation. $\omega_{n+1}$ is (part of) the interpolation error you seek to minimize. You might want to have a look at "Approximation Theory and Approximation Practice" by N. Trefethen. – Nox Oct 11 '16 at 10:02