How to solve this problem of integeral?

1 次查看（过去 30 天）
Hamzah Faraj
Hamzah Faraj 2020-4-24
评论： Hamzah Faraj 2020-4-27
Can anypody help with this piece of code?
The problem is with the use of the integeral function
%------------------------------------------------------------------------%
clear variables;
close all;
clc;
%------------------------------------------------------------------------%
x=-1:0.1:1;
n=length(x);
y=zeros(1,n);
for i=1:n
y = m_cost(x);
end
plot(y)
and here is my function:
function y =m_cost(x)
fun1=inline('0-x','x');
fun2=inline('x-4','x');
syms fun1;
syms fun2;
syms x;
if (x>=0)
y=x.^2+4*x+8*int(fun1,x,0,10);
else
y=x.^2+4*x+8*int(fun2,x,0,10);
end
  5 个评论
显示 3更早的评论
Hamzah Faraj
Hamzah Faraj 2020-4-27
Thank you. This is really informative.
If I have this cost function, for example, is using the integeral function the right option? I want to integerat int(p) over time dt (axis x).
% Assume this is the results of the temperature function v(t)=v(i-1)+A*(time);
v = -1:0.1:5;
% A is the heaating parameter
A = 4/3;
% Cost at each v(t) is given by the expression:
% y(v(t))=(50*int(p)+10*x+30)/(x/A);
% P = x-4 if x > 4 and x <= 5
% p = 0 if x >= 0 and x <= 4
% p = -x if x >= -1 and x < 0
y = cost(v)
function y= cost(v)
func1 = @(v) v-4;
func2 = @(v) -v;
% Cost of each v(t) at time t is given by the expression:
% y(v(t))=(50*int(p)+10*x+30)/(x/A);
% P = x-4 if x > 4 and x <= 5
% p = 0 if x >= 0 and x <= 4
% p = -x if x >= -1 and x < 0
if (v > 4 & v <= 5)
y = ((50*integeral(func1,0,inf)+((v./A)*10)+30)/(v./A);
elseif ( v>=0 & v<=4)
y = (((v./A)*10)+30)/(v.A);
else
y = ((integeral(func2,0,inf))+((v./A)*10)+30)/(v./A);
end
end

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Ameer Hamza
Ameer Hamza 2020-4-24
There is no need to use inline. You can use an anonymous function. Also, int() is used for symbolic integration. Here you can use numeric integration using integral(). The function can be vectorized so there is no need to use a for-loop
%------------------------------------------------------------------------%
clear variables;
close all;
clc;
%------------------------------------------------------------------------%
x = -1:0.1:1;
n = length(x);
y = m_cost(x);
function y =m_cost(x)
fun= @(x) -x;
y = x.^2+4*x+8*integral(fun,0,10);
end

更多回答（1 个）

Walter Roberson
Walter Roberson 2020-4-24
int is only for use with Symbolic functions or symbolic expressions. inline objects cannot be used with int.
You should only use inline if you have a very specific reason to use it. You should be using symbolic or anonymous functions instead.
However your
syms fun
on the next line overwrites fun the same as if you had done
fun = sym('fun') ;
So when you int that it is just a symbolic constant and the integral is the constant times the difference in time. Your function is going to return a symbolic expression involving the two unresolved values x and fun. x is unresolved because the
syms x
that you did overwrote the x parameter passed to the function.
so you return a symbolic expression that contains unresolved variables and outside the function you assign that into part of y. But y is double precision and you cannot store a symbolic expression with unresolved variables in double precision.

类别

在 Help Center 和 File Exchange 中查找有关 Conversion Between Symbolic and Numeric 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by