Which one is the correct one to calculate
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Yamina chbak
am 1 Jul. 2021
Kommentiert: Yamina chbak
am 2 Jul. 2021
I tried both of those instruction to calculate as :
where and with are vectors ( zeros(15,1))
Both leads to different results, i don't know which one is the correct one , Can you hepl me ? .
R=0;
for j=1:L
s(j)=(1-q(j))/(1+q(j));
A(j)=4*sin(pi*alpha)*lamnda(j)/(pi*(1+q(j))^2);
R = R+A(j)*exp(-s(j)*dt);
end
% or
R=sum((4*sin(pi*alpha).*lambda/(pi.*(1+q)^2))*exp(-dt*(1-q)./(1+q)));
1 Kommentar
Jan
am 1 Jul. 2021
Bearbeitet: Jan
am 1 Jul. 2021
If you do not provide L, q, alpha, lamnda and dt, we cannot check your code.
Is "lamnda" a typo? In the 2nd formula it is "lambda".
"Both leads to different results" ist strange, because the 2nd fails with an error. Do you mean:
R=sum((4*sin(pi*alpha).*lambda ./ (pi.*(1+q).^2)).*exp(-dt*(1-q)./(1+q)));
% elementwise: ^ ^ ^
Akzeptierte Antwort
Jan
am 1 Jul. 2021
Bearbeitet: Jan
am 1 Jul. 2021
R = 0;
L = 30; % Some educated guesses
q = rand(1, L);
lambda = rand(1, L);
alpha = rand;
dt = rand;
for j = 1:L
s(j) = (1 - q(j)) / (1 + q(j));
A(j) = 4 * sin(pi * alpha) * lambda(j) / (pi * (1 + q(j))^2);
R = R + A(j) * exp(-s(j) * dt);
end
R
% R2 = sum((4*sin(pi*alpha) .* lambda / ...
% (pi.*(1+q)^2)) * exp(-dt*(1-q) ./ (1+q)))
% Failing: Several operations must be made elementwise:
R2 = sum((4 * sin(pi * alpha) .* lambda ./ ...
(pi* (1 + q) .^ 2)) .* exp(-dt * (1 - q) ./ (1 + q)))
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Logical 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!