Warning: Matrix is singular to working precision.
Diego
i creat a program that find a intersection in two nonlinear equations...
but, i just find one of them, the other have a problem:
"Warning: Matrix is singular to working precision."
the code :
main.m
function main (a,b)
% newton method for systems
% main part
echo off all;
format long;
x = [[a]; [b]];
err = 10^(-6);
nmax = 1000;
xold = x;
for k=1:nmax
x = x - inv(df(x))*f(x);
if (norm(x-xold,inf)< err )
fprintf('precis?o alcan?ada em %3d itera??es\n', k);
fprintf('x1=%10.6f, x2=%10.6f\n', x(1), x(2));
break
end
xold = x;
end
if (k==nmax)
fprintf('precis?o n?o alcan?ada em %3d itera??es\n', nmax);
end
%gr?fico
figure(1);
clf;
hold on;
whitebg('w');
xlabel ('x');
ylabel('y');
grid;
axis([ -5 5 -5 5 ]);
xx1 = double(x(1));
xx2 = double (x(2));
t= 0:0.025:5;
u= 0:0.025:2;
y1 = eval('log(exp(t)-1)');
y2 = (1024-u.^10).^(1/10);
y3= (1024-u.^10).^(1/10)*-1;
plot (t,y1,'r',xx1,xx2,'ob',u,y2,'b',u,y3,'b');
plot([-5 5],[0 0],'k',[0 0],[-5 5],'k' );
hold off;
% end of main part
f.m
%onde x1 = x e x2 = y
function y = f(x)
y(1,1) = exp(x(1)) - exp(x(2)) - 1;
y(2,1) = x(1)^10 + x(2)^10 - 1024;
% end of f(x)
df.m
function df = df(x)
df(1,1)= exp(x(1)); df (2,1) = exp(x(2));
df(2,1)= 10*x(1)^9; df(2,2) = 10*x(2)^9;
Rune Allnor
> hello folks, first , my english is very bad, so, i'm sory about that.
>
> i creat a program that find a intersection in two nonlinear equations...
> but, i just find one of them, the other have a problem:
> "Warning: Matrix is singular to working precision."
This means that there is a problem with the condition number
of some matrix in your code. The matrix is effectively non-
invertible, so whatever expression it appears in can not be solved.
This is a severe error - it *might* in fact mean that what you
intend to do is impossible, although in some cases work-arounds
are possible.
The way to handle this is to find out exactly what variable and
expression causes the rror, and analyze the situation to find out
if a work-around is possible.
Rune
Diego
main (0,-2); <- i find the correct value
main (0,2); <-problem
i'm using the newtons metlhod to solve this problem
"Diego " <dieg...@hotmail.com> wrote in message <gjoh9i$l9u$1...@fred.mathworks.com>...
Diego
main(0,-2)
x1= 0.126928, x2= -2.00000 (correct)
main (0,2), i have a problem(wrong)
i have a doubt, if the inv funcition have cause this problem....
Diego
http://www.mathworks.com/support/solutions/data/1-15NRJ.html?product=OP&solution=1-15NRJ
Optimization Toolbox
the code:
test2.m
function test2 (a,b)
F = inline ('[exp(V(1))-exp(V(2))-1 ; V(1)^10 + V(2)^10 - 1024 ]','V');
InitialGuess = [a;b];
Options = optimset ('Display', 'iter');
XY = fsolve (F, InitialGuess, Options);
ShouldBeZero = F(XY);
figure(1);
clf;
hold on;
whitebg('w');
xlabel ('x');
ylabel('y');
grid;
ezplot('exp(x)-exp(y) = 1');
ezplot('x^10 + y^10 = 1024');
plot(XY(1), XY(2), 'ro');
hold off;
i have read in manual, that the Optimization Toolbox implements newton method, but when i use the raw algorithm, i had problems to find the solution in interval (0,2) , the famous problem called in inv function:
"Warning: Matrix is singular to working precision.", it's very strange, does anybody have this code implemented ?, i'd like to compare with my code.
Tom mcmurray
hi diego:
you likely have figured out that your jacobian, df, is nearly singular around the origin. any steps that result in an root iteration in the rectangular region, -0.02<x,y<0.02, approximately, produce a jacobian that makes matlab unhappy upon attempted inversion. i tried (x,y)=(0.02,0.02) and got:
jacobian = [1.02 -1.02;5e-15 5e-15]
with an inverse:
inv(jacobian) = [0.49 9.77e13;-0.49 9.77e13]
so, gradient optimizations are typically sensitive to initial root selection and fail for singular jacobians, as you have demonstrated. thanx for the interesting objective function.
tom
