Approximating Definite Integrals as Sums
34 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have a function that I want to integrate over and . I know the answer of this integration is . When I use function, I get the correct result but when I try to evaluate this integral using loops, my result is always . I increased points to but still get the same result. By using loops for integration. The sum approximation is . I am following a research paper which claims that they used loops with gaussian meshes and get as answer.
My question is why exactly I am not getting the correct result and what exactly are gaussian meshes? The function and the code runner.m that I am using are given below:
runner.m
%runner.m
clear; clc;
NBZ = 500; %number of x and y points. total points = NBZ^2
JNN = 0; %a parameter
a = 1;
% x and y limits... and x and y points.
xmin = -2*pi/(3*a);
xmax = 4*pi/(3*a);
ymin = -2*pi/(sqrt(3)*a);
ymax = 2*pi/(sqrt(3)*a);
dx = (xmax-xmin)/(NBZ-1); %Delta x
dy = dx; %Delta y
xs = xmin:dx:xmax; %array of x points
ys = ymin:dy:ymax; %array of y points
dsum= 0;
for ny = 1:NBZ
y = ys(ny);
for nx = 1:NBZ
x = xs(nx);
out = F(x,y,JNN);
dsum = dsum + out*dx*dy;
end
end
answer = dsum
%it gives: answer = -0.6778
F(x,y)
function out = F(x,y,JNN)
JN = 4;
D = 1;
s = 1;
h11 = 0;
h12 = -(JN+1i*D)*s*(1+exp(1i*(-x-sqrt(3)*y)/2))...
-JNN*s*(exp(1i*x)+exp(1i*(+x-sqrt(3)*y)/2));
h13 = -(JN-1i*D)*s*(1+exp(1i*(+x-sqrt(3)*y)/2))...
-JNN*s*(exp(1i*x)+exp(1i*(-x-sqrt(3)*y)/2));
h23 = -(JN+1i*D)*s*(1+exp(1i*x))-2*JNN*s*exp(1i*x/2)*cos(sqrt(3)/2*y);
h = [h11 h12 h13;
conj(h12) h11 h23;
conj(h13) conj(h23) h11];
[evecs, evals] = eig(h);
v1 = evecs(:,1);
v2 = evecs(:,2);
v3 = evecs(:,3);
e1 = evals(1,1);
e2 = evals(2,2);
e3 = evals(3,3);
X = [ 0, - exp(- (x*1i)/2 - (3^(1/2)*y*1i)/2)*(1/2 - 2i) - JNN*(exp(x*1i)*1i + (exp((x*1i)/2 - (3^(1/2)*y*1i)/2)*1i)/2), - exp((x*1i)/2 - (3^(1/2)*y*1i)/2)*(1/2 + 2i) - JNN*(exp(x*1i)*1i - (exp(- (x*1i)/2 - (3^(1/2)*y*1i)/2)*1i)/2)
- exp((conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*(1/2 + 2i) + JNN*((exp(- (conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*1i)/2 + exp(-conj(x)*1i)*1i), 0, exp(x*1i)*(1 - 4i) - JNN*exp((x*1i)/2)*cos((3^(1/2)*y)/2)*1i
- exp(- (conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*(1/2 - 2i) - JNN*((exp((conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*1i)/2 - exp(-conj(x)*1i)*1i), exp(-conj(x)*1i)*(1 + 4i) + JNN*exp(-(conj(x)*1i)/2)*cos((3^(1/2)*conj(y))/2)*1i, 0];
Y = [ 0, - 3^(1/2)*exp(- (x*1i)/2 - (3^(1/2)*y*1i)/2)*(1/2 - 2i) + (3^(1/2)*JNN*exp((x*1i)/2 - (3^(1/2)*y*1i)/2)*1i)/2, 3^(1/2)*exp((x*1i)/2 - (3^(1/2)*y*1i)/2)*(1/2 + 2i) + (3^(1/2)*JNN*exp(- (x*1i)/2 - (3^(1/2)*y*1i)/2)*1i)/2
- 3^(1/2)*exp((conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*(1/2 + 2i) - (3^(1/2)*JNN*exp(- (conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*1i)/2, 0, 3^(1/2)*JNN*exp((x*1i)/2)*sin((3^(1/2)*y)/2)
3^(1/2)*exp(- (conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*(1/2 - 2i) - (3^(1/2)*JNN*exp((conj(x)*1i)/2 + (3^(1/2)*conj(y)*1i)/2)*1i)/2, 3^(1/2)*JNN*sin((3^(1/2)*conj(y))/2)*exp(-(conj(x)*1i)/2), 0];
o1a = ( ((v2'*X*v1)*(v1'*Y*v2)) - ((v2'*Y*v1)*(v1'*X*v2)))/((e2-e1)^2);
o1b = ( ((v3'*X*v1)*(v1'*Y*v3)) - ((v3'*Y*v1)*(v1'*X*v3)))/((e3-e1)^2);
o1 = 1i*(o1a+o1b);
out = real(o1/(2*pi));
end
0 commentaires
Réponse acceptée
Matt J
le 2 Nov 2020
Modifié(e) : Matt J
le 2 Nov 2020
I don't know what gaussian meshes are, but the Riemann sum works with a just few fixes to the way the sampling is set up:
NBZ = 600; %number of x and y points. total points = NBZ^2
JNN = 0; %a parameter
a = 1;
% x and y limits... and x and y points.
xmin = -2*pi/(3*a);
xmax = 4*pi/(3*a);
ymin = -2*pi/(sqrt(3)*a);
ymax = 2*pi/(sqrt(3)*a);
xs = linspace(xmin,xmax,NBZ+1); %array of x points
ys = linspace(ymin,ymax,NBZ+1); %array of y points
xs(end)=[]; ys(end)=[];
dx=xs(2)-xs(1);
dy=ys(2)-ys(1);
dsum= 0;
for ny = 1:NBZ
y = ys(ny);
for nx = 1:NBZ
x = xs(nx);
out = F(x,y,JNN);
dsum = dsum + out;
end
end
answer = dsum*dx*dy
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Numerical Integration and Differentiation dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!