How can i call an equation and it's derivative inside a matlab function?

Question

Ömer Utku Örengül 2021-11-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1580784-how-can-i-call-an-equation-and-it-s-derivative-inside-a-matlab-function

编辑： Ömer Utku Örengül 2021-11-8

As a newbie, i like to ask a simple question. I am trying to impliment a newton-rapson method for a simple equation as an example. I create a different matlab function from main function for the equation and call it inside the main function. However when I try to call the functions derivative it gives an error. I am aimin to not to take the derivative inside the main function for optimization concerns. I did try different methods but they give errors all the same.

function nr(x0,TC)

% TC is given in terms of percentage!

if nargin<2, x0=0; TC=10^-4;end

error=TC+1; i=0;

x(1)=x0;

while(error>TC)

x(i+2)=x(i+1)-f(x(i+1))/fd(x(i+1));

error=100*abs((x(i+2)-x(i+1))/x(i+2));

i=i+1;

end

fprintf('After %d iterations an approximate root is %f',i,x(i));

end

function [fx]=f(x)

fx=exp(-x)-x;

end

function fd=fd(x)

% syms x %These parts where i need help.

% fx=exp(-x)-x;

% fd=matlabFunction(diff(fx))

fd=-exp(-x)-1;

end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Ömer Utku Örengül 2021-11-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1580784-how-can-i-call-an-equation-and-it-s-derivative-inside-a-matlab-function#answer_826524

编辑：Ömer Utku Örengül 2021-11-8

在 MATLAB Online 中打开

So I have it worked like this.

function fd=fd(a)
syms a
fx=exp(-a)-a;
fd=diff(fx);
end

and put "syms a" at the main function and finally changed "fd(x(i+1))" with;

...
x(i+2)=x(i+1)-f(x(i+1))/subs(fd,a,x(i+1));          %   subs(fd,a,x(i+1))
...

It worked as intended.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Alan Stevens 2021-11-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1580784-how-can-i-call-an-equation-and-it-s-derivative-inside-a-matlab-function#answer_825704

在 MATLAB Online 中打开

Like this, perhaps:

% TC is given in terms of percentage!
x0=0; TC=10^-4;
error=TC+1; i=0;
x(1)=x0;
while(error>TC)
    [fx, fd] = f(x(i+1));
    x(i+2)=x(i+1)-fx/fd;    
    error=100*abs((x(i+2)-x(i+1))/x(i+2));
    i=i+1;
end
fprintf('After %d iterations an approximate root is %f',i,x(i));
After 4 iterations an approximate root is 0.567143
function [fx, fd]=f(x)
fx=exp(-x)-x;
fd = -exp(-x)-1;
end

3 个评论
显示 1更早的评论隐藏 1更早的评论

Walter Roberson 2021-11-7

编辑：Walter Roberson 2021-11-8

在 MATLAB Online 中打开

There are two notable diff() functions. One of them only applies if the first parameter is symbolic or symbolic function.

syms x
fd = matlabFunction(diff(f(x),x))
fd = function_handle with value:
    @(x)-exp(-x)-1.0
function [fx]=f(x)
fx=exp(-x)-x;
end

Alan Stevens 2021-11-8

编辑：Alan Stevens 2021-11-8

在 MATLAB Online 中打开

You could always try something like this:

% TC is given in terms of percentage!
fx = @(x) exp(-x)-x;
dx = 10^-10;   % Choose a suitably small value
fd = @(x) (fx(x+dx) - fx(x))/dx;
x0=0; TC=10^-4;
error=TC+1; i=0;
x(1)=x0;
while(error>TC)
    x(i+2)=x(i+1)-fx(x(i+1))/fd(x(i+1));    
    error=100*abs((x(i+2)-x(i+1))/x(i+2));
    i=i+1;
end
fprintf('After %d iterations an approximate root is %f',i,x(i));
After 4 iterations an approximate root is 0.567143