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

3 ビュー (過去 30 日間)
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

採用された回答

Walter Roberson
Walter Roberson 2020 年 10 月 27 日
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.

その他の回答 (1 件)

Mukter
Mukter 2023 年 8 月 12 日
編集済み: Walter Roberson 2023 年 8 月 12 日
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

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