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What Is a Chebyshev Spline?

The Chebyshev spline C=C_t=C_k,t of order k for the knot sequence t=(t_i: i=1:n+k) is the unique element of S_k,t of max-norm 1 that maximally oscillates on the interval [t_k..t_n+1] and is positive near t_n+1. This means that there is a unique strictly increasing n-sequence τ so that the function C=C_t∊S_k,t given by C(τ_i)=(–1)^{n – 1}, all i, has max-norm 1 on [t_k..t_n+1]. This implies that τ₁=t_k,τ_n=t_n+1, and that t_i < τ_i < t_k+i, for all i. In fact, t_i+1 ≤ τ_i ≤ t_i+k–1, all i. This brings up the point that the knot sequence is assumed to make such an inequality possible, i.e., the elements of S_k,t are assumed to be continuous.

In short, the Chebyshev spline C looks just like the Chebyshev polynomial. It performs similar functions. For example, its extreme sites τ are particularly good sites to interpolate at from S_k,t because the norm of the resulting projector is about as small as can be; see the toolbox command chbpnt.

You can run the example Construction of a Chebyshev Spline to construct C for a particular knot sequence t.

Choice of Spline Space

You deal with cubic splines, i.e., with order

k = 4;

and use the break sequence

breaks = [0 1 1.1 3 5 5.5 7 7.1 7.2 8];
lp1 = length(breaks); 

and use simple interior knots, i.e., use the knot sequence

t = breaks([ones(1,k) 2:(lp1-1) lp1(:,ones(1,k))]);
n = length(t)-k;

Note the quadruple knot at each end. Because k = 4, this makes [0..8] = [breaks(1)..breaks(lp1)] the interval [t_k..t_n+1] of interest, with n = length(t)-k the dimension of the resulting spline space S_k,t. The same knot sequence would have been supplied by

t=augknt(breaks,k);

Initial Guess

As the initial guess for the τ, use the knot averages

recommended as good interpolation site choices. These are supplied by

tau=aveknt(t,k); 

Plot the resulting first approximation to C, i.e., the spline c that satisfies c(τ_i)=(–1)^n-–i, all i:

b = cumprod(repmat(-1,1,n)); b = b*b(end);
c = spapi(t,tau,b); 
fnplt(c,'-.') 
grid

Here is the resulting plot.

Remez Iteration

Starting from this approximation, you use the Remez algorithm to produce a sequence of splines converging to C. This means that you construct new τ as the extrema of your current approximation c to C and try again. Here is the entire loop.

You find the new interior τ_i as the zeros of Dc, i.e., the first derivative of c, in several steps. First, differentiate:

Dc = fnder(c);

Next, take the zeros of the control polygon of Dc as your first guess for the zeros of Dc. For this, you must take apart the spline Dc.

[knots,coefs,np,kp] = fnbrk(Dc,'knots','coefs','n','order');

The control polygon has the vertices (tstar(i),coefs(i)), with tstar the knot averages for the spline, provided by aveknt:

tstar = aveknt(knots,kp);

Here are the zeros of the resulting control polygon of Dc:

npp = (1:np-1); 
guess = tstar(npp) -coefs(npp).*(diff(tstar)./diff(coefs));

This provides already a very good first guess for the actual zeros.

Refine this estimate for the zeros of Dc by two steps of the secant method, taking tau and the resulting guess as your first approximations. First, evaluate Dc at both sets:

sites = tau(ones(4,1),2:n-1); 
sites(1,:) = guess; 
values = zeros(4,n-2); 
values(1:2,:) = reshape(fnval(Dc,sites(1:2,:)),2,n-2);

Now come two steps of the secant method. You guard against division by zero by setting the function value difference to 1 in case it is zero. Because Dc is strictly monotone near the sites sought, this is harmless:

for j=2:3 
  rows = [j,j-1];Dcd=diff(values(rows,:));
  Dcd(find(Dcd==0)) = 1; 
  sites(j+1,:) = sites(j,:) ... 
           -values(j,:).*(diff(sites(rows,:))./Dcd); 
  values(j+1,:) = fnval(Dc,sites(j+1,:)); 
end

The check

max(abs(values.')) 
  ans = 4.1176 5.7789 0.4644 0.1178

shows the improvement.

Now take these sites as your new tau,

tau = [tau(1) sites(4,:) tau(n)];

and check the extrema values of your current approximation there:

extremes = abs(fnval(c, tau));

The difference

max(extremes)-min(extremes)
  ans = 0.6905

is an estimate of how far you are from total leveling.

Construct a new spline corresponding to the new choice of tau and plot it on top of the old:

c = spapi(t,tau,b); 
sites = sort([tau (0:100)*(t(n+1)-t(k))/100]); 
values = fnval(c,sites); 
hold on, plot(sites,values)

The following code turns on the grid and plots the locations of the extrema.

grid on
plot( tau(2:end-1), zeros( 1, np-1 ), 'o' )
hold off
legend( 'Initial Guess', 'Current Guess', 'Extreme Locations',...
 'location', 'NorthEastOutside' );

Following is the resulting figure (legend not shown).

If this is not close enough, one simply reiterates the loop. For this example, the next iteration already produces C to graphic accuracy.