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 $f(x)-\frac 18 T_4(x)$.

Answer 2

The solution to the uniform approximation problem will need to satisfy the Haar conditions $$ f(x_k)-p(x_k)=(-1)^kr $$ for 5 distinct points in the interval which also realize the maximum $|r|$ of $|f(x)-p(x)|$. The Chebyshev approximation usually provides a good starting point for the Remez exchange algorithm. It may be useful to compute the further stages still in terms of the Chebyshev polynomials as basis.

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As result of solving this as general non-linear problem I get the results

xr = -0.707106947599892  -1.1668963128834e-16     0.707106947599893
r =   0.125000072497002
p =  -4.05725754051219e-17  4   2.33379262604217e-16  -0.125000072497002

essentially replacing $x^4$ with $x^2-\frac18$. The plots confirm the solution and Haar conditions

using octave/matlab's fsolve

function eqn = haar(u, f)
    x = sort(u(1:3)); r = u(4); p = u(5:8);
    df = polyder(f); dp=polyder(p);

    fp = @(x) polyval(f,x)-polyval(p,x);
    dfp = @(x) polyval(df,x)-polyval(dp,x);
    eqn = [ fp(-1)-r,
            fp(x(1))+r,
            fp(x(2))-r,
            fp(x(3))+r,
            fp(1)-r,
            dfp(x(1)),
            dfp(x(2)),
            dfp(x(3))
          ];
end

f = [1,0,3,0,0];
u = fsolve(@(u) haar(u,f), [-0.8,0.0, 0.8, 1, 0, 0, 0, 0]);     
xr = u(1:3)
r = u(4)
p = u(5:8)