Integrating an anonymous function in a for loop
显示 更早的评论
Hello All,
I have a relatively simple task. I am trying to integrate an anonymous function in a for loop. The following is the code:
%% Mode Shape Test %%
wn1 = 3.516015284068505; c1=-0.73410;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% What I am Trying to Accomplish with a For Loop %%
% Phi_1 = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^1;
% Phi_2 = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^2;
% Phi_3 = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^3;
% Phi_4 = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^4;
%
% g1=integral(Phi_1,0,1); % g1 = 0.7829
% g2=integral(Phi_2,0,1); % g2 = 0.9999
% g3=integral(Phi_3,0,1); % g3 = 1.4778
% g4=integral(Phi_4,0,1); % g4= 2.3486
% % I have to do this maybe over 100 times and it would be convinient to do
% % this in a loop structure.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% For loop Attempt %%
for i=1:4
Phi_(i) = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^i;
g(i) = integral(Phi_(i),0,1);
end
The commented top section is the way that I have been doing it which works just fine, but I will now need the values for over Phi_100 and so on. When I try to run the For loop I get the following error:
Error using integral (line 82)
First input argument must be a function handle.
Error in Mode_Shape_Test (line 19)
g(i) = integral(Phi_(i),0,1);
Any Ideas?
Thanks
Christopher Reyes
采纳的回答
madhan ravi
2018-12-13
g=zeros(1,4); % preallocate
%% For loop Attempt %%
for i=1:4
Phi = @(z)(-(cos(sqrt(wn1)*z) - cosh(sqrt(wn1)*z) + c1*(sin(sqrt(wn1)*z) - sinh(sqrt(wn1)*z)))).^i;
g(i) = integral(Phi,0,1);
end
g % - > results saved
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
