error Conversion to double from function_handle is not possible.

Question

Ivan 2023-10-24

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2038111-error-conversion-to-double-from-function_handle-is-not-possible

评论： Star Strider 2023-10-25

t = 0:0.01:10;
nt = length(t);
% Inicialización de u y u'
u = zeros(size(t));
p = zeros(1,nt);
for i = 1:nt
    p(i) = @(t) (10./(t(i)+1));
    u(i) = integral(@(tau) p(t(i)-tau)*h(tau),0,t(i));
end

me aparece el error Conversion to double from function_handle is not possible.

p(i) = @(t) (10./(t(i)+1));

como solucionarlo?

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Answer 1

Star Strider 2023-10-24

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https://ww2.mathworks.cn/matlabcentral/answers/2038111-error-conversion-to-double-from-function_handle-is-not-possible#answer_1339816

在 MATLAB Online 中打开

The ‘p’ function was incorrect, and the ‘h’ funciton is completely missing (so I created it).

Try this (with the correct ‘h’ function) —

t = 0:0.01:10;
nt = length(t);
% Inicialización de u y u'
u = zeros(size(t));
p = zeros(1,nt);
p = @(t) (10./(t+1));
h = @(x) x;                                             % Create Function
for i = 1:nt
    u(i) = integral(@(tau) p(t(i)-tau).*h(tau),0,t(i));
end
u
u = 1×1001
         0    0.0005    0.0020    0.0045    0.0079    0.0123    0.0177    0.0239    0.0312    0.0393    0.0484    0.0584    0.0693    0.0811    0.0937    0.1073    0.1217    0.1369    0.1531    0.1700    0.1879    0.2065    0.2260    0.2463    0.2674    0.2893    0.3120    0.3355    0.3598    0.3849

.

2 个评论
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Walter Roberson 2023-10-24

在 MATLAB Online 中打开

p = zeros(1,nt);
p = @(t) (10./(t+1));

Why bother to initialize p with zeros there?

Star Strider 2023-10-25

It was initially written as:

p(i) = @(t) (10./(t(i)+1));

and I didn’t catch the preallocation when I corrected the ‘p’ function. (I was concentrating on it and the absent ‘h’ function.)

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Answer 2

Sulaymon Eshkabilov 2023-10-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2038111-error-conversion-to-double-from-function_handle-is-not-possible#answer_1339821

在 MATLAB Online 中打开

Here is the corrected solution:

t = 0:0.01:10;
nt = length(t);
% Inicialización de u y u'
u = zeros(size(t));
p = zeros(1,nt);
p = @(t) (10./(t+1));
Pval=p(t);
% Note that your include h() is unknown. Thus it is removed from the
% formulation. Predefine it if it is to be included.
for i = 1:nt  
    u(i) = integral(@(tau) Pval(i)*(t(i)-tau).*(tau),0,t(i));
end