MATLAB代码不起作用（梯度下降算法）

Question

I am trying gradient descent, I wrote following however not getting any answer,

n=0;            %initialize iteration counter 
eps=1;          %initialize error 
a=0.8;         %set iteration parameter 
x=[1;1];        %set starting value
f=6*x(1)^2+8*x(2)^2-3*x(1)*x(2);
%Computation loop 
while eps>1e-12||n<100 
   gradf=[12*x(1)-3*x(2); 16*x(2)-3*x(1)];  %gradf(x) 
   eps=(norm(gradf)/(1+abs(f)));                                %error 
   y=x-a*gradf;                                                 %iterate 
   x=y;                                                         %update x 
   n=n+1;                                                       %counter+1 
end 
n;x;eps;        %display end values

When I add this file to path and type x it shows NaN, NaN. what is wrong?

Answer 1

There are several errors in your code. Consider this (I put comments where corrections were needed)

 n=0;            
 eps=1;         
 a=0.1;                    %You need a way smaller parameter to converge!
 x=[1;1];       

 A = [6 -3/2 ; -3/2 8];     %You have a bilinear positive definite form,
                            %you may use matrix form for convenience


while eps>1e-12 && n<100    %You had wrong termination conditions!!
    gradf=2*A*x;            %(gradf in terms of matrix)
    f=x'*A*x;               %you need to update f every iteration!!
    eps=(norm(gradf)/(1+abs(f)))                              
    disp(eps > 1e-12)
    x=x-a*gradf;                                               

       %Now you can see the orbit  towards minimum                                     
    plot(x(1),x(2),'o'); hold on        
    n=n+1;                                                        
end 
n
x
eps        

for instance with the current value a=.1 I get

n = 100
eps = 1.2308e-011

x = 
  1.0e-012 *

  -0.2509
  0.4688

That is I had to perform 100 iteration because my epsilon is still above the threshold. If I allow 200 iterations I get

n =  110
eps = 
   7.9705e-013

x = 
1.0e-013 *
  -0.1625
   0.3036

Ie 110 iterations are sufficient.

---

Case of a general f (ie not a quadratic form).

You can use, for instance, function handles , ie you define (before the while )

foo = @(x) 6*x(1)^2+8*x(2)^2-3*x(1)*x(2);
foo_x = @(x) 12*x(1)-3*x(2); 
foo_y = @(x) 16*x(2)-3*x(1);

then, in the while you substitute

gradf = [foo_x(x);foo_y(x)];
f = foo(x);

---

PS for what concerns the while cycle, please keep in mind that you keep on iterating while you are not satisfied with your precision ( eps>1e-12 ) AND your total number of iteration is below a given threshold ( n<100 ).

Consider also that you are working in finite precision : a numerical algorithm can never reach the analytic solution (ie what you have with infinite precision and infinite iterations), therefore, you always have to set a threshold ( eps , which should be above the machine precision \\approx 1e-16 ) and that is your 0 .