Find interpolation polynomial for $f(x) = x^4+3x^2$ of degree $\le3$ , such that $max_{x\in[-1,1]}|f(x)-p(x)|$ is minimal

Question

find interpolation polynomial for $f(x) = x^4+3x^2$ of degree $\le3$ , such that $max_{x\in[-1,1]}|f(x)-p(x)|$ is minimal.

Well I think I don't understand Chebyshev's interpolation points correctly, I tried to take Chebyshev's polynomial of order 4 ($T_4(x)$), to get 4 roots (interpolation points): $\{x_k:=cos(\frac{\pi}{8} +\frac{\pi}{4}k) :k=0,1,2,3 \}$, Yet when calculating $f[x_0],f[x_0,x_1],f[x_0,x_1,x_2], f[x_0,x_1,x_2,x_3]$ in order to get the interpolation polynomial, due to the fact that the roots are $x_0 = -x_3 , x_1 =-x_2$ and the function $x^4+3x^2$, calculated in $x_0,x_1,x_2,x_3$ leads to $f[x_0]=f[x_3] , f[x_1]=f[x_2]$ , thus $f[x_0,x_1,x_2,x_3] =0 $ and I get a interpolation polynomial of order 2, which doesn't best approximate $f(x)$. I was sure that Chebyshev's roots are the only interpolation points which solves the min-max problem of $|f(x)-p(x)|$, but I feel that, there is something which I don't fully understand here.

Any help or reference would be appreciated.

Answer 1

The interpolation method finds the unique polynomial of degree 3 or less that has the same value as $f(x)$ at $x_0,x_1,x_2$ and $x_3$, the roots of $T_4(x)=8x^4-8x^2+1$.

Since $f(x)$ is a polynomial of degree 4, it can be written as $\sum_{i=0}^4 a_i T_i(x)$, where comparison of the coefficients shows that $a_4=\frac18$.

Therefore the required approximation is $f(x)-\frac 18 T_4(x)$.