need help with this summation
1 vue (au cours des 30 derniers jours)
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i cant get the summation correct. is ther any one that can help me or point me in the right dirrection.
clc;
clear;
zeta = 0:.1:2;
fo = [.01;.1;1];
n = 1:2:101;
theta1 = zeros(length(zeta),1);
theta3 = zeros(length(zeta),1);
theta5 = zeros(length(zeta),1);
for i = 1:length(zeta)
for k = 1:length(n)
theta1(i) = theta1(i)+(1/n(k))*exp((-(n(k)^2*pi^2)/4)*fo(1,1))*sin(((n(k)*pi)/2)*zeta(i));
theta3(i) = theta3(i)+(1/n(k))*exp((-(n(k)^2*pi^2)/4)*fo(2,1))*sin(((n(k)*pi)/2)*zeta(i));
theta5(i) = theta5(i)+(1/n(k))*exp((-(n(k)^2*pi^2)/4)*fo(3,1))*sin(((n(k)*pi)/2)*zeta(i));
end
end
theta1 =(4*pi)* theta1;
theta3 =(4*pi)* theta3;
theta5 =(4*pi)* theta5;
plot(zeta,theta1)
hold on
plot(zeta,theta1+theta3)
plot(zeta,theta1+theta3+theta5)
0 commentaires
Réponses (1)
Walter Roberson
le 23 Fév 2021
Modifié(e) : Walter Roberson
le 23 Fév 2021
zeta = reshape(linspace(0,2), 1, [], 1); %row
fo = reshape(logspace(-2,0,50), [], 1, 1); %column
n = reshape(1:2:101, 1, 1, []); %pane
t1 = (1./n);
t2 = exp((-(n.^2.*pi.^2)/4) .* fo);
t3 = sin(n .* (pi/2) .* zeta);
temp = t1 .* t2 .* t3;
theta = sum(temp, 3);
surf(zeta, fo, theta, 'edgecolor', 'none')
3 commentaires
Walter Roberson
le 23 Fév 2021
zeta = reshape(linspace(0,2), 1, [], 1); %row
fo = reshape([0.01, 0.1, 1], [], 1, 1); %column
n = reshape(1:2:5, 1, 1, []); %pane
t1 = (1./n);
t2 = exp((-(n.^2.*pi.^2)/4) .* fo);
t3 = sin(n .* (pi/2) .* zeta);
temp = t1 .* t2 .* t3;
theta = cumsum(temp, 3);
subplot(3,1,1)
surf(zeta, fo, theta(:,:,1), 'edgecolor', 'none')
title(strjoin(string(n(1:1)), '+'))
subplot(3,1,2)
surf(zeta, fo, theta(:,:,2), 'edgecolor', 'none')
title(strjoin(string(n(1:2)), '+'))
subplot(3,1,3)
surf(zeta, fo, theta(:,:,3), 'edgecolor', 'none')
title(strjoin(string(n(1:3)), '+'))
Walter Roberson
le 23 Fév 2021
The problem with your graph is that "the first three terms of the series" is not referring to different fo values: it is referring to n = 1, n = 1 + n = 3, n = 1 + n = 3 + n = 5 . And you are supposed to compare at three different fo values. One way of interpreting that would be to use 3 terms * 3 fo = 9 different plots.
Do not add the results for different fo values: only add the results for the different n values.
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