Bisection Method Error - Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

I'm trying to Run this Below Code with But I am getting this error.Please help me finding the error

tol = 1.e-10; 
a = 1.0;  
b = 2.0; 
nmax = 100; 
% Initialization  
itcount = 0; 
error = 1.0; 
% Graph of the function 
xval = linspace(a,b,100); 
for i=1:100 
    fval(i) = func(xval(i)); 
    rtns=makereturns(20,2);
end 
plot(xval,fval); 
grid on;  
hold on; 
% iteration begins here 
while (itcount <= nmax && error >= tol) 
itcount = itcount + 1; 
% Generate and save iteratres 
x = a + (b-a)/2; 
z(itcount) = x; 
fa = func(a);  
fb = func(b); 
fx = func(x);  
error = abs(fx); 
%  error = abs(x - xold); 
if (error < tol) 
x_final = x; 
else 
if (fa*fx < 0) 
% root is between a and x 
b = x; 
else 
% root is between x and b 
a = x; 
end 
end 
plot(z(1:itcount),zeros(itcount,1),'r+'); 
pause(5)  
end 
if (itcount < nmax); 
val = func(x); 
fprintf(1,'Converged solution after %5d iterations',itcount); 
fprintf(1,' is %15.7e, %e \n',x_final, val); 
else fprintf(1,'Not converged after %5d iterations',nmax); 
end 
function val = func(x) 
%val = x^3 + 4 * x^2 - 10; 
val = x^3 - x - 3; 
%val = sin(x);  
end 

Answer 1

[Walter Roberson]

rtns=makereturns(20,2);

20 x 2

Q = cov(rtns)

2 x 2

a = [0;1]; B = [1;-1];

2 x 1 each

V = a.*Q*a + 2*a.*Q*B*x + B.*Q*B*x^2

a is 2 x 1, Q is 2 x 2, so a.*Q is going to be 2 x 2 because of implicit expansion. Then you * it by a 2 x 1 and the inner dimensions match for that, so you get back 2 x 1.

Likewise 2*a.*Q*B is 2 x 1 and x is a scalar, so the second term is 2 x 1.

SImilarly B.*Q*B is 2 x 1 and x^2 is scalar so the third term is 2 x 1.

You are adding three 2 x 1 vectors, so V is going to be 2 x 1.

You are now throwing around 2 x 1 vectors and 1 x 2 vectors in the same arithmetic expressions, so you end up getting a 2 x 2 output.

V = a.*Q*a + 2*a.*Q*B*x + B.*Q*B*x^2

Perhaps the expression for that is instead

V = a'*Q*a + 2*a'*Q*B*x + B'*Q*B*x^2

Then that would be 2 x 1 transposed giving 1 x 2, matrix multiply by 2 x 2, giving a 1 x 2 result; matrix multiply by a 2 x 1 getting a 1 x 1 result; likewise for all three terms.

Answer 2

[Mukter]

tol = 1.e-10; 
a = 1.0;  
b = 2.0; 
nmax = 100; 
% Initialization  
itcount = 0; 
error = 1.0; 
% Graph of the function 
xval = linspace(a,b,100); 
for i=1:100 
    fval(i) = func(xval(i)); 
    rtns=makereturns(20,2);
end 
plot(xval,fval); 
grid on;  
hold on; 
% iteration begins here 
while (itcount <= nmax && error >= tol) 
    itcount = itcount + 1; 
    % Generate and save iteratres 
    x = a + (b-a)/2; 
    z(itcount) = x; 
    fa = func(a);  
    fb = func(b); 
    fx = func(x);  
    error = abs(fx); 
    %  error = abs(x - xold); 
    if (error < tol) 
        x_final = x; 
    else 
        if (fa*fx < 0) 
            % root is between a and x 
            b = x; 
        else 
            % root is between x and b 
            a = x; 
        end 
    end 
    plot(z(1:itcount),zeros(itcount,1),'r+'); 
    pause(5)  
end 
if (itcount < nmax); 
    val = func(x); 
    fprintf(1,'Converged solution after %5d iterations',itcount); 
    fprintf(1,' is %15.7e, %e \n',x_final, val); 
else fprintf(1,'Not converged after %5d iterations',nmax); 
end 
function val = func(x) 
%val = x^3 + 4 * x^2 - 10; 
val = x^3 - x - 3; 
%val = sin(x);  
end