>> % cleaned-up diary of matlab computations done by Carl de Boor during
>> % the 02dec99 class of CS 412

>> % first part:  end conditions in cubic spline interpolation.
>> % The gui, called splinetool, is accessible to you provided you add
>> % the following directory to your matlabpath:
>> %    /u/d/e/deboor/public/matlab/splines
>> % For this to work properly, you would have to be (a) on afs, i.e., on one
>> % of the CS department's work stations, and (b) would have to put this 
>> % directory ahead in the path, i.e., use a command like
>> %  path = path('/u/d/e/deboor/public/matlab/splines',path);

>> splinetool

>> % second part: least-squares approximation by cubic splines
>> % If you want to read more about this, try the last two pages of
>> % http://www.cs.wisc.edu/~deboor/412/story6a.ps
>> % (albeit with slightly different notation than used in class).

>> %  1. the noisy data:
>> x = linspace(0,2*pi,101);
>> y = sin(x) + (rand(size(x))-.5)/5;
>> plot(x,y,'o')
>> %  2. choosing the breakpoints for the cubic spline:
>> xi = linspace(0,2*pi,5);
>> %  3. constructing the `Vandermonde-like' matrix:
>> %     since the spline command can deal with vector-valued data, we can
>> %     evaluate all the Lagrange spline in one call:
>> Atrans = spline(xi,eye(5),x);
>> size(Atrans)

ans =  5   101

>> % the rows of this matrix contain the values at the x's of the various
>> % `Lagrange splines', as the following plot shows:
>> plot(x,Atrans(1,:))
>> hold on
>> % plot the xi's
>> plot(xi,zeros(size(xi)),'x')
>> plot(x,Atrans(2,:),'r')
>> % plot also the points (xi,1), to be certain these Lagrange splines do
>> % what they are supposed to: 
>> plot(xi,ones(size(xi)),'x')
>> % plot one more of these Lagrange splines
>> plot(x,Atrans(4,:),'k')
>> hold off

>> %  4. we are ready to compute the least-squares fit:
>> a = Atrans'\y
>> ??? Error using ==> \
>> Matrix dimensions must agree.
>> 
>> % ok, since matlab only deals with matrices, there is always the problem
>> % of knowing whether some vector is a row or a column; a column was 
>> % expected here:

>> a = Atrans'\y(:)

a =
   -0.1454
    0.9732
    0.0043
   -0.9634
    0.1670

>> % These are just the coefficients, but they are the coefficients in the
>> `Lagrange form' for the cubic spline, i.e., the values of that cubic
>> spline at the breaks xi. So, to get the spline itself, we simply interpolate
>> to those data:

>> f = spline(xi,a);
>> % replot the data and the value of the least-squares spline fit on top of it:
>> plot(x,y,'o')
>> hold on
>> fx = ppval(f,x);
>> plot(x,fx,'r','linewi',2)
>> % that's not bad, considering that we used the cubic spline with the 
>> % not-a-knot condition, i.e., only the middle one of our 3 interior breaks is
>> % actually active.
>> % We would have done much better with a break sequence 
>> %  xi = [0,1/4,1/2,1,3/2,7/4,1]
>> % which does have 3 interior active breaks. Try it out :-)
>> diary off