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Hello, everybody,

In my program I need to use fsolve, but don't know how to adress my variables to the x(1), x(2), etc. that needs fsolver.

I give you simplified example:

My program in a certain pointt gives me (x, y are variables):

y = 2*x

y = x^4

and I want to receive the solution: [2,2]

I can get if I do this:

----------------

function F=myfun(x)

F = [x(2)-x(1)^2 ; x(2)-2*x(1)];

end

fsolve(@myfun,[3,3])

--------------------------

But I can't do manually because my program has many and very complex equations. I need a way to adress my equations to the fsolve form.

Matt J
on 24 Aug 2019

Edited: Matt J
on 24 Aug 2019

I don't understand your question very well, but maybe it will help to mention that fsolve does not require you to express the function in terms of individual x(i). fsolve does not know or care how the function is expressed in code. It only cares that the function is differentiable.

For example, I could conceivably solve a linear system of equations as below, which uses no x(i) references at all:

A=rand(5); b=A*[1;2;3;4;5]

x=fsolve(@(x) A*x-b, ones(5,1))

x =

1.000000000000000

1.999999998886354

2.999999996241878

4.000000002442892

5.000000000534329

Diego R
on 25 Aug 2019

Edited: Diego R
on 25 Aug 2019

Matt J
on 25 Aug 2019

But I know in which stretch I have to search....This is the reason of using fsolve.

That's not the right reason to choose fsolve over the symbolic solver. The reason why a non-symbolic solver might be preferable for you is that you said you had many complicated equations. That makes it less likely that a symbolic solution exists.

If you want to try to solve symbolically, you could use the assume() command to specify bounds.

To solve non-symbolically, you should not use fsolve, but rather lsqnonlin, which has input arguments for specifying bounds on the variables. If your question is about how to convert symbolic equations to a form usable by non-symbolic solvers, you should use matlabFunction().

Walter Roberson
on 25 Aug 2019

solve() does not always respect assumptions.

fsolve() does not permit bounds to be specified. fzero() does permit bounds to be specified, but only supports a single function of a single variable.

If I recall correctly, fsolve() and fzero() assume continuity, which could be a problem with piecewise functions.

vpasolve() permits bounds to be specified for each variable.

>> solve(y==2*x, y==x^2,x>0)

ans =

struct with fields:

x: [1×1 sym]

y: [1×1 sym]

>> [ans.x,ans.y]

ans =

[ 2, 4]

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