ppval and polyval giving different results

I created a piecewise polynomial using mkpp. But when I evaluate a set of points I get incorrect results using ppval, although polyval() gives correct result. Looks like I'm missing something. Any idea what that is?

Thanks!

>> ppm
ppm = 
struct with fields:
form: 'pp'
breaks: [749.5 16747 32768 48771 64784]
coefs: [4×2 double]
pieces: 4
order: 2
dim: 1

K>> ppm.coefs
ans =
0.00031269      -10.246
0.00031268      -10.246
0.00031274      -10.247
0.00031265      -10.245

>> ppval(ppm,ppm.breaks(1))
ans =
      -10.246

>> polyval(ppm.coefs(1,:),ppm.breaks(1))
ans =
      -10.012
      
>> polyval(ppm.coefs(1,:),8750)
ans =
        -7.51

>> ppval(ppm,8750)
ans =
      -7.7444      

Answer 1

[Bruno Luong]

Here is a correct way to know how ppval works

pp=spline(cumsum(rand(1,10)),rand(1,10));

x=3;
ppval(pp,x)

ans = 2.1433

% 
i=discretize(x,pp.breaks);
polyval(pp.coefs(i,:),x-pp.breaks(i))

ans = 2.1433

Your evaluation

polyval(ppm.coefs(1,:),8750)

is wrong unless pp.breaks(1) is 0.

Answer 2

[Torsten]

I think you fed "polyval" with the coefficients cut to a certain number of digits.

You must use the coefficients in full precision to get equal results from ppval and polyval.

[Bruno Luong]

How it cuts? He type directly the variable ppm.coefs(1,:)

polyval(ppm.coefs(1,:),8750)

[Torsten]

But it's the correct syntax to use polyval in my opinion.

Result should be

0.00031269*8750-10.246

And I was surprised it really came out to be

-7.5099625

Since the values 0.00031269 and 10.246 seem to be cut, I think the OP fed "polyval" with cut coefficients and worked with ppval with the coefficients in full length.

[Bruno Luong]

"But it's the correct syntax to use polyval in my opinion."

It's not a problem of syntax, it's problem of convention of pp.coefs which is the coefficients of polynomial of shifted variable yi := x-pp.breaks(i), and not directly x.

[Torsten]

breaks = [0 4 10 15];
coefs = [0 1 -1 1 1; 0 0 1 -2 53; -1 6 1 4 77];
pp = mkpp(breaks,coefs)

But you prescribe the coefficients usually. And I don't think you must recalculate them for the shifted polynomial.

[Bruno Luong]

"And I don't think you must recalculate them for the shifted polynomial."

You are wrong here is the doc of mkpp

"Polynomial coefficients, specified as an L-by-k matrix with the ith row coefs(i,:) containing the local coefficients of an order k polynomial on the ith interval, [breaks(i), breaks(i+1)]. In other words, the polynomial is coefs(i,1)*(X-breaks(i))^(k-1) + coefs(i,2)*(X-breaks(i))^(k-2) + ... + coefs(i,k-1)*(X-breaks(i)) + coefs(i,k)."

[Torsten]

You are right. This also explains the deviations in the breakpoints themselves.