PPVAL , a short question
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
If I use 2 piecewise polynoms with both y=x and breaks at 1 2 3 and input is 1 2 3
I would expect y = 1 2 3
But I get y = 0 0 1.
Why ?
example :
x = linspace(1,3,3)
breaks=[1 2 3];
coefs=[1 0; 1 0];
pp = mkpp(breaks,coefs);
y=ppval(pp,x);
0 Kommentare
Antworten (3)
Wayne King
am 19 Sep. 2011
Hi Stephan, the polynomials are not y=x.
Your polynomials are
f(x) = x-1 % x-break(1)
g(x) = x-2 % x-break(2)
So that is why you get y = [0 0 1].
From the command line help:
The matrix COEFS must be L-by-K, with the i-th row, COEFS(i,:), representing the local coefficients of the order K polynomial on the interval [BREAKS(i) ... BREAKS(i+1)], i.e., the polynomial 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)
Hope that helps,
Wayne
0 Kommentare
Andrei Bobrov
am 19 Sep. 2011
x = linspace(1,3,3)
breaks=[1 2 3];
coefs=[1 1; 1 2];
pp = mkpp(breaks,coefs)
y=ppval(pp,[1 2 3])
0 Kommentare
Siehe auch
Kategorien
Mehr zu Polynomials finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!