Vectorization the Legendre Polynomial Without loop
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi All,
I would like to speed up this Legendre polynomial code without using loop. It has a significant delay in my associated code
I would like to alwasy keep L as option which can be changed
t=[1/60:1/60:3];
L=221;
P=zeros(L+1,length(t));
P(1,1:length(t)) = 1;
P(2,1:length(t)) = t;
for i = 2:L
P(i+1,:) = -(i-1)./i.*P(i-1)+(2.*i-1)./i.*t.*P(i);
end
The equation for the Legendre polynomial is
where and
PS: I corected the mathematical equation, sorry for the inconvenience
Many thanks in advance
0 Kommentare
Akzeptierte Antwort
Jan
am 14 Feb. 2019
Bearbeitet: Jan
am 14 Feb. 2019
Are you sure the the code is working? You calculate the vectors P(i, :) but use the first element only: P(i):
P(i+1,:) = -(i-1)./i.*P(i-1)+(2.*i-1)./i.*t.*P(i);
% I assume you want:
P(i+1,:) = -(i-1)./i.*P(i-1, :)+(2.*i-1)./i.*t.*P(i, :);
% ^ ^
I do not think that this can be vectorized, because the values of P depend on former values. cumsum etc. is not a solution also. So a small simplification only:
t = 1/60:1/60:3; % No square brackets needs
L = 221;
P = zeros(L+1, length(t));
P(1, :) = 1;
P(2, :) = t;
for n = 2:L
P(n+1, :) = (1 - n) / n * P(n - 1, :) + (2 * n - 1) / n * t .* P(n, :);
end
A columnwise processing is faster in general - in my measurements maybe 10%:
t = (1/60:1/60:3).';
L = 221;
P = zeros(length(t), L+1);
P(:, 1) = 1;
P(:, 2) = t;
for n = 2:L
P(:, n+1) = (1 - n) / n * P(:, n - 1) + (2 * n - 1) / n * t .* P(:, n);
end
0 Kommentare
Weitere Antworten (1)
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!