Error: Function definitions are not permitted in this context.
6 次查看(过去 30 天)
显示 更早的评论
I am not able to use function as all in my code, I am confused now that is using because I am suing R2015b version? Do I need to buy the toolbox ?
0 个评论
采纳的回答
Walter Roberson
2019-6-19
It is never valid to copy and paste a "function" definition to the command line. It is also not valid to eval() a character vector that contains a "function" definition.
In releases up to R2015a, it was never valid to have a "function" definition inside a script file (a .m file in which the first executable word was not "function" and not "classdef"). In R2015b, it became legal to define functions inside of script files, provided that they were after the script, and provided that every "function" definition ended with a corresponding "end" statement.
Whether in functions or scripts, there are places where it is not valid to have a function definition. For example it is not valid to have a function definition inside a for loop or if statement.
if true
function result = test %invalid function
result = 5;
end
end
0 个评论
更多回答(2 个)
Nishaben Desai
2019-6-19
2 个评论
Walter Roberson
2019-6-19
No, you cannot run that code without having the symbolic toolbox.
You would have to numeric methods, such as using ode45(). That would not, however, give you the equation of the result.
Nishaben Desai
2019-6-19
2 个评论
Walter Roberson
2019-6-19
If U is an unknown variable and dy/dt = u - y and presuming that U and u are the same thing, then you have three values to be concerned with: t, y(t), and u
If u is a constant, then integrating dy/dt = u - y on both sides, we get y = u*t - 1/2*y^2 + C for some boundary condition value C. y(0) = 0 tells us that u*0 + 1/2*0^2 + C = 0 which tells us that C = 0, so y = u*t - 1/2*y^2 which gives us that 1/2*y^2 + y - u*t = 0 which gives us that y = sqrt(2*t*u + 1) - 1 . We know this must stay in the range 0 to 90, and by examination we can see that it is largest when u is largest, so we can solve 90 == sqrt(2*t*u + 1) - 1 which would give us u = 4140/t . But time is potentially unlimited and this is not a constant in time.
From this we conclude that either u is not a constant or else time for the system is not unlimited.
Either way, we do not have enough information to determine u .
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Special Values 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!