How to sum up the functional sequence correctly?
Afficher commentaires plus anciens
I have a complex function that depends on 't' and 'n' F(n,t) I need to sum a series over 'n' and plot F(t).
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
double T;
%T = 0:366;
m = 9.1093837*10^(-31);
double Tc;
Tc = 1.2;
%T = 1.2;
double t;
t = T/Tc;
int8 n;
%n = 0:1:366; %0:49;
double D;
%D = 10^(-8):10^(-5);
double ksi; % длина когерентности
%D = input('input D');
%ksi = input('input ksi');
D = 10^(-6);
ksi = 10^(-9);
d = D/ksi;
E = Ef/(pi*Kb*Tc);
hp = (626176*10^(-34))/(2*pi);
%complex p1;
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
p2 = sqrt(E+1i.*t.*(2.*n+1));
k0 = (ksi/hp)*sqrt(2.*m.*pi.*Kb.*Tc);
arg_of_lg = (1+exp(-1i*d*k0.*p1))/(1+exp(-1i*d*k0.*p2));
l_g = log(abs(arg_of_lg));
arg_1 = angle(1+exp(-1i*d*k0.*p1));
arg_2 = angle(1+exp(-1i*d*k0.*p2));
F = 0;
%F = 1/(2*n+1).*imag(l_g+1i*d*k0.*((p1-p2)./2)+1i*arg_1-1i*arg_2);
for n = 0:49
%p1 = @Calculate_p1;
%p2 = @Calculate_p2;
F = F + 1/(2*n+1).*imag(l_g+1i*d*k0.*((p1-p2)./2)+1i*arg_1-1i*arg_2);
end
F = -(1/d).*F;
F = F - (hp^2*pi^2*ksi)/(2*m*10^(-7));
1 commentaire
Jan
le 21 Déc 2022
What is the problem with the posted code?
Omit these useless lines:
double Tc;
double t;
int8 n;
double D;
double ksi;
You can simplify 9.1093837*10^(-31); to 9.1093837e-31. Note that the first is an expensive power operation while the second is a cheap constant.
Réponses (1)
t = (100:1000).';
n = 0:49;
Ef = 2.77e3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
m = 9.1093837e-31;
Tc = 1.2;
D = 1e-6;
ksi = 1e-9;
d = D/ksi;
E = Ef/(pi*Kb*Tc);
hp = (626176*10^(-34))/(2*pi);
k0 = (ksi/hp)*sqrt(2.*m.*pi.*Kb.*Tc);
p1 = ((1+1i)/sqrt(2))*sqrt(t.*(2.*n+1));
p2 = sqrt(E+1i*t.*(2.*n+1));
arg_of_lg = (1+exp(-1i*d*k0.*p1))./(1+exp(-1i*d*k0.*p2));
l_g = log(abs(arg_of_lg));
arg_1 = angle(1+exp(-1i*d*k0.*p1));
arg_2 = angle(1+exp(-1i*d*k0.*p2));
F = 1./(2*n+1).*imag(l_g+1i*d*k0.*((p1-p2)./2)+1i*arg_1-1i*arg_2);
format long
F = sum(F,2)
F = -(1/d).*F;
F = F - 1e7*(hp^2*pi^2*ksi)/(2*m);
plot(t,F)
Catégories
En savoir plus sur Programming dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
