Questions about ppval in piecewise polynomial

I have a question about piecewise polynomial.

If I use t=pp.coefs to get the coefficients about the polynomial.

Then I use

[m n]=size(t);
for i=1:n-1
    t(:,i)=i.*t(:,i);
end

to get the new coefficients after differentiate it. How can I use ppval again to evaluate the polynomial with this new coefficients?

Answer 1

[John D'Errico]

You only think that this differentiates the polynomial. But it does not. In fact, look at the result. For example, consider the polynomial represented by coefficients:

t = [1 2 3 4]

so the polynomial is

x^3 + 2^x^2 + 3*x + 4

Remember in MATLAB, the polynomial is represented by the coefficients starting from the highest order first.

Now, try the code you wrote. What vector t results?

[m n]=size(t);
for i=1:n-1
    t(:,i)=i.*t(:,i);
end

t
t =
     1     4     9     4

so:

x^3 + 4*x + 9*x^2 + 4

Clearly that is the wrong derivative. Instead, try this:

M = diag([3 2 1],1);
t = [1 2 3 4];

t*M
ans =
     0     3     4     3

That is the polynomial

0*x^3 + 3*x^2 + 4*x + 3

Which is the proper derivative polynomial. Just stuff it back into the proper field in pp.

pp.coefs = pp.coefs*M;