Dimension issues when using fsolve

Hello everyone, I face some "dimension issues" when using fsolve. Thus, I tried to use the exact example in the matlab help: http://www.mathworks.fr/help/toolbox/optim/ug/fsolve.html (example 1). Following the example, my initial condition is always given by:

x0 = [-5; -5];

After playing around a bit, I discovered that when I put the function F as follows:

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

I get the error message: ??? Error using ==> vertcat CAT arguments dimensions are not consistent.

When I type:

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

it works! The difference? In the second case, there is a space between the "+" and 2*x(2). First, I do not understand why these two ways of typing F should be any different. Second, it seems that fsolve is sensitive to dimensions. What should I be careful of when using it?

Thanks a lot!

Kyriacos

[Ning Wu]

I had the same problem when using fsolve. Titus is correct on this. To avoid this problem, just set tp1=2*x(1) - x(2) - exp(-x(1)); tp2=-x(1) +2*x(2) - exp(-x(2)); and write F=[tp1;tp2]

Answer 1

[Titus Edelhofer]

Hi Kyriacos,

if you leave out the first half of F you see the difference:

F = [-x(1) +2*x(2) - exp(-x(2))]
F =
  5.0000 -158.4132
F = [-x(1) + 2*x(2) - exp(-x(2))]
F =
-153.4132

In the first case the + is interpreted as a unary plus (like if you write -2).Therefore you get the error because you can't combine one number and a vector of length 2 into a matrix.

Titus

[Titus Edelhofer]

Forgot to mention: although generally speaking spaces make code easier to read, this is a good example of being careful to use spaces in expressions using [], because space works as column separator.

Answer 2

[Kyriacos]

Hi, Thanks for your reply. I am not sure I understand well what you mean by "unary plus", Can you please elaborate a bit more?

Also, regarding dimensions: my system is one of 21 equations and 21 unknowns. The way I do it, I write F by separating each equation by a ";" which is like changing a line in the vector of outputs right? So I am not sure how leaving a space between expressions works here, since as you say it is interpreted as a column operator but for me both my initial conditions and the output are a column-vector x of dimension 21.

I am giving you the whole code to be a bit more clear.

function f = myss(x)

is=0.15 ; theta=0.5 ; rho = 0.44 ; yT_y = 0.5; yN_y = 1-yT_y ; cT_c = 0.43 ; cN_c = 1-cT_c ; beta = 1/1.04 ; delta = 0.1 ; alphaT = (1-0.61) ; alphaN = (1-0.56) ;

%x0 = [1; 1; 1; 1; 1; 1; 1; 1; 1; 0.5; 0.5; 1; 1; 1; 0.2; 0.2; 0.2; 1; 1; 1; 1; ] ;

f = [x(13) - 1 ; x(2) - x(8) - x(9) ; x(3) - x(6) ; x(10)*x(2)/x(1) - yT_y ; x(11)*x(9)/x(1) - is ; x(5) + x(7) - x(10)*(x(8) + x(9)) ; x(5) + x(7) - (x(15)*x(8)^((theta-1)/theta) + (1-x(15))*x(9)^((theta-1)/theta))^(theta/(theta-1)) ; x(11) - x(10) ; x(14) - x(11)/x(10) ; x(5)/(x(14)*x(4)) - cT_c ; x(13)*x(6)/(x(14)*x(4)) - cN_c ; x(13) - ((1 - x(16))/x(16))*(x(5)/x(6))^(1/rho) ; x(4) - (x(16)*x(5)^((rho-1)/rho) + (1 - x(16))*x(6)^((rho-1)/rho))^(rho/(rho-1)) ; x(14)*x(4) - x(5) - x(13)*x(6) ; x(17) - x(15)*(x(8)^(-1/theta))*(x(15)*x(8)^((theta-1)/theta) + (1 - x(15))*x(9)^((theta-1)/theta))^(1/(theta-1)) ; x(2)/x(19)*beta*alphaT*x(17) - (1-beta*(1-delta)) ; x(18) - (x(16)/(1 - x(16)))*(x(5)/x(6))^(-1/rho) ; x(3)/x(20)*alphaN*beta - (1-beta*(1-delta))*x(18) ; x(21) - x(19) - x(20) ; x(7)/x(21) - delta ; x(1) - (x(10)*x(2) + x(13)*x(3)) ];

Then, I run it as follows: [x,fval, exitflag] = fsolve(@myss,x0)

Btw, it does not converge but I am currently working on it.

Thanks again!

Kyriacos

Answer 3

[Titus Edelhofer]

Hi Kyriacos,

unary plus is similar to unary minus: instead of

x = 42.0;

you can just as well write

x = +42.0;

Having said this, MATLAB interprets

F = [2.0 +3.0]

equally to

F = [2.0  3.0]

instead of (what you wanted to have)

F = [2.0+3.0]

My recommendation for making your code somewhat more robust and easier to read: seperate the individual functions

f = zeros(21, 1);
f(1) = x(13) - 1; 
f(2) = x(2) - x(8) - x(9); 
f(3) = x(3) - x(6);
...

 or (doing basically the same):

f1 = x(13) - 1;
f2 = x(2) - x(8) - x(9); 
f3 = x(3) - x(6);
...
f = [f1; f2; f3; ...]

Hope this helps,

Titus