Plotting the function by the points that need to be determined

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Dmitry
Dmitry 2023-1-12
编辑： Torsten 2023-1-15
It is necessary to plot F(t) by points. The function F(t) is a sum from 0 to nD(t)(nD is the upper limit of the sum) which depends on 't', i.e. I have an array 't' and an array nD(t) is formed from it, it contains 20 values [496, 248, 165, ...], the first point will be the final sum of F(t) with an upper limit of 496, the first point will be the final sum of F(t) with an upper limit of 248, etc., it is necessary to plot F(t) at these 20 points.
My code:
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
E = Ef/(pi*Kb*Tc);
h_bar = (1.0545726*10^(-34));
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*Kb.*Tc);
C_2 = 0;
for n = 0:49
C_2 = C_2 + (1/(2.*n+1)).*k0.*real(sqrt(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
plot(t,F, '-r');
%% F(t)
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = 0;
for i = 1:20
n = nD(1,i);
F = F + 1/(2*n+1).*(k0.*real(((f_p1(n,t)-f_p2(n,t))./2))+(f_arg_2(n,t)-f_arg_1(n,t))./d);
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end

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回答（2 个）

Torsten
Torsten 2023-1-12
编辑：Torsten 2023-1-15
Ran in:
Maybe something like this ?
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
E = Ef/(pi*Kb*Tc);
h_bar = (1.0545726*10^(-34));
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*Kb.*Tc);
C_2 = 0;
for n = 0:49
C_2 = C_2 + (1/(2.*n+1)).*k0.*real(sqrt(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
hold on
plot(t,F(:,1),"Color","red");
plot(t,F(:,numel(nD)),"Color","blue");
hold off
grid on
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = zeros(numel(t),numel(nD));
for i = 1:numel(t)
for j = 1:numel(nD)
n = nD(j);
for k = 0:n
F(i,j) = F(i,j) + 1/(2*k+1).*(k0.*real(((f_p1(k,t(i))-f_p2(k,t(i)))./2))+(f_arg_2(k,t(i))-f_arg_1(k,t(i)))./d);
end
end
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end
  7 个评论
显示 5更早的评论
Dmitry
Dmitry 2023-1-15
I think we misunderstood each other:
We have:
Each of these sums is finite, there are 20 such sums in total, since there are twenty values in the array t[]. Therefore, our final graph will consist of these 20 points.
Torsten
Torsten 2023-1-15
编辑：Torsten 2023-1-15
And what's the sense that you take a different number of elements for the sum depending on t ?
Do you know in advance how fast the infinite series converges depending on t ?
See the answer below.

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Torsten
Torsten 2023-1-15
Ran in:
%% initial conditions
global d k0 h_bar ksi m E;
Ef = 2.77*10^3;
Kb = physconst('boltzmann'); % 1.38*10^(-23)
T = 0.12:0.24:6.4;
m = 9.1093837*10^(-31);
Tc = 1.2;
%t = T./Tc;
t = 0.1:0.1:2;
nD = floor(375./(2.*pi.*t.*1.2) - 0.5);
D = 10^(-8); % толщина пленки
ksi = 10^(-9);
%d = D/ksi;
d = 1000;
E = Ef/(pi*Kb*Tc);
h_bar = (1.0545726*10^(-34));
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*Kb.*Tc);
C_2 = 0;
for n = 0:49
C_2 = C_2 + (1/(2.*n+1)).*k0.*real(sqrt(3601+1i.*(2.*n+1))-((1+1i)./sqrt(2)).*sqrt(2.*n+1)); % константа
end
%% calculation
F = f_calc(t,nD);
plot(t,F)
grid on
function F = f_calc(t,nD)
global d k0 h_bar ksi m;
F = zeros(1,numel(t));
for i = 1:numel(t)
for k = 0:nD(i)
F(i) = F(i) + 1/(2*k+1).*(k0.*real(((f_p1(k,t(i))-f_p2(k,t(i)))./2))+(f_arg_2(k,t(i))-f_arg_1(k,t(i)))./d);
end
end
F = -F;
%F = -(1/d).*F;
%F = F - C_2;
end
function p1 = f_p1(n,t)
p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));
end
function p2 = f_p2(n,t)
global E;
p2 = sqrt(3601+1i.*t.*(2.*n+1));
end
function n_lg = f_lg(n,t)
global d k0;
arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));
n_lg = log(abs(arg_of_lg));
end
function arg_1 = f_arg_1(n,t)
global d k0;
arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));
end
function arg_2 = f_arg_2(n,t)
global d k0;
arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));
end

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