plot an obtained function after several calculus
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Good morning!
I want to plot the folloing function which is obtained after several calculus on Matlab,
syms s t a
Y=(sin(int(cos((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/9))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/18)))^2/100)^2*(exp(-2*s^(1/3)) + cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + 0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s^(2/3)))^2/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/6)))^2/100)^2)^2, s, 0, 1))/100 + sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/18))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/36)))^2/100)*(exp(-2*t^(1/6)) + cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t^(1/3)))^2/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100))^2/100)*(exp(-2*t) + cos(int(sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/12)))^2/100)*(exp(-2*s^(1/2)) + sin(0.00008414709848078965066525023216303*cos(s^(1/2))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/4)))^2/100 + cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + 0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s))^2/100)) + 1, s, 0, t^2))^2/100 + sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100)*(exp(-2*t^(1/2)) + cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t))^2/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/2))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/4)))^2/100))^2/100)
in this function another variable appeared (denoted by ' a' ) so i add it in symbolique variable
To plot this function i used the folowing code:
t_vals=linspace(0, 1, 10);
figure(1)
hold on
y_eval = double(eval(subs(Y(i+1),t, t_vals)))
plot(t_vals, y_eval;
hold off
6 个评论
You did not yet tell us what the problem is with your code. The evaluation for one value for t already lasts too long to be executed here. Is the problem the long time it takes on your PC ?
syms s t a
Y=(sin(int(cos((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
sin(0.00008414709848078965066525023216303*cos(s^(1/9))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/18)))^2/100)^2*(exp(-2*s^(1/3)) + ...
cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + ...
0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s^(2/3)))^2/100 + ...
sin(0.00008414709848078965066525023216303*cos(s^(1/3))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/6)))^2/100)^2)^2, s, 0, 1))/100 + ...
sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
sin(0.00008414709848078965066525023216303*cos(t^(1/18))^2 + ...
0.008414709848078965066525023216303*exp(-2*t^(1/36)))^2/100)*(exp(-2*t^(1/6)) + ...
cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t^(1/3)))^2/100 + ...
sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + ...
0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100))^2/100)*(exp(-2*t) + ...
cos(int(sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
sin(0.00008414709848078965066525023216303*cos(s^(1/6))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/12)))^2/100)*(exp(-2*s^(1/2)) + ...
sin(0.00008414709848078965066525023216303*cos(s^(1/2))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/4)))^2/100 + ...
cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + ...
0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s))^2/100)) + 1, s, 0, t^2))^2/100 + ...
sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + ...
0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100)*(exp(-2*t^(1/2)) + ...
cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + ...
0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t))^2/100 + ...
sin(0.00008414709848078965066525023216303*cos(t^(1/2))^2 + ...
0.008414709848078965066525023216303*exp(-2*t^(1/4)))^2/100))^2/100);
double(subs(Y,t,1))
L.GREEN
2024-8-4
L.GREEN
2024-8-4
Walter Roberson
2024-8-4
y_eval = double(eval(subs(Y(i+1),t, t_vals)))
You should never eval() a symbolic result. eval() of a symbolic result has no documented meaning.
In practice, eval() of a symbolic result is the same as eval() of char() of the symbolic result. The problem with that is that char() of symbolic results produces something that is not in MATLAB expression format and is not in MuPAD internal code: char() of symbolic results is intended to be human readable, and is not intended for machine processing.
What you need to do instead is perform all necessary subs() on the expression, and then double() the result without using eval()
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