How to input new value as old value using while loop to compute relative error?

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Ja
Ja 2022-10-8
评论： Ja 2022-10-8
I need help on how can I input new value as an old value
Whenever I run the code, the r_e (relative error) equation goes up and gives me a wrong answer
How do I fix this?
f_x= sin (5x) + cos (2x)
for i=1:n
f_l = f_x(x_l);
f_u = f_x(x_u);
p = (x_l+x_u)/2;
f_m = f_x(p);
if (f_l*f_m)<0
x_u=p;
elseif (f_u*f_m)<0
x_l=p;
end
p_old = p;
r_e = abs(p-p_old/(p));
table(i, :)={i-1 p r_e};
fprintf('%d %d %d \n', table{i, :});
end

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采纳的回答

VBBV
VBBV 2022-10-8
r_e = abs((p-p_old)/(p));
  3 个评论
显示 1更早的评论
VBBV
VBBV 2022-10-8
编辑：VBBV 2022-10-8
Ran in:
f_x= @(x) sin (5*x) + cos (2*x)
f_x = function_handle with value:
@(x)sin(5*x)+cos(2*x)
p_old = 0; % give an initial value to it
x_l =2; x_u = 10;
n = 20;
for i=1:n
f_l = f_x(x_l);
f_u = f_x(x_u);
p = (x_l+x_u)/2;
f_m = f_x(p);
if (f_l*f_m)<0
x_u=p;
elseif (f_u*f_m)<0
x_l=p;
end
r_e = abs((p-p_old)/(p));
p_old = p; % assign the p_old after relative error
table(i, :)={i-1 p r_e};
fprintf('%d %d %d \n', table{i, :});
end
0 6 1 1 8 2.500000e-01 2 9 1.111111e-01 3 8.500000e+00 5.882353e-02 4 8.750000e+00 2.857143e-02 5 8.875000e+00 1.408451e-02 6 8.812500e+00 7.092199e-03 7 8.781250e+00 3.558719e-03 8 8.765625e+00 1.782531e-03 9 8.757812e+00 8.920607e-04 10 8.753906e+00 4.462294e-04 11 8.751953e+00 2.231645e-04 12 8.750977e+00 1.115947e-04 13 8.751465e+00 5.579423e-05 14 8.751709e+00 2.789634e-05 15 8.751587e+00 1.394836e-05 16 8.751526e+00 6.974230e-06 17 8.751556e+00 3.487103e-06 18 8.751572e+00 1.743548e-06 19 8.751579e+00 8.717735e-07
give an initial value to p_old and assign the value to it after relative error inside the for loop

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