Error using * Inner matrix dimensions must agree.

   clc 
clear all
close all
format short
% define function
A = 0.42748; B = 0.08664;
P_r = (0:2:10);
T_r = [1 1.2 1.4 2 0 0];
% Function in f(x)=0
f =@(Z)Z^3-Z^2+(A.*(P_r./T_r.^(5/2))-B.*(P_r./T_r.*(1+B.*(P_r./T_r)))).*Z-A*B*(P_r.^2./T_r.^(7/2));
xu = 6;% Upper initial guess
xl = 0;% Lower initial guess
xm=0; % first midpoint value
fprintf('\n\n\n Bisection using a For Loop\n')
fprintf('The initial guess of xl, xu %10.4f, %10.4f\n', xl,xu)
for i = 1:15
 xold = xm; 
  if f(xm)*f(xu)<0
 xl=xm;
 else
 xu=xm; 
 end
 xm=(xu+xl)/2;
 ea=abs((xm-xold)/xm)*100;
 fprintf(['iteration ',num2str(i),' xl, xu, ea: %10.4f, %10.4f, %10.4f\n'],xl,xu,ea)
 xold = xm; 
end

please help, im getting Error using * Inner matrix dimensions must agree

Answer 1

[KSSV]

Replace line:

f(xm)*f(xu)<0

with

f(xm).*f(xu)<0

Read about element by element operations in MATLAB. Note that you are multiplying two vectors there. Think whether you expect a array as output or a scalar as output.

If you expect a scalar, that line should be :

f(xm)*f(xu)'<0

[Trevor Potter]

Thank you that worked, may i also ask why this code

   clear all ; close all ;
clc ;
format long
% Function in f(x)=0
f=@(x)-26+85*x-91*x.^2+44*x.^3-8*x.^4+x.^5;
xu = 1;% Upper initial guess
xl = 0.5;% Lower initial guess
xm=0; % first midpoint value
fprintf('\n\n\n Bisection using a For Loop\n')
fprintf('The initial guess of xl, xu %10.4f, %10.4f\n', xl,xu)
for i = 1:5
 xold = xm; % save the mid point
 % Check the interval between which the root lies
 if f(xm)*f(xu)<0
 xl=xm; % the midpoint becomes the lower guess
 else
 xu=xm; % the midpoint becomes the upper guess
 end
 xm=(xu+xl)/2; % compute the midpoint
 % Calculate relative approximate error
 ea=abs((xm-xold)/xm)*100;
 fprintf(['iteration ',num2str(i),' xl, xu, ea: %10.4f, %10.4f, %10.4f\n'],xl,xu,ea)
 xold = xm; % save the mid point
end

...works and actually find roots, however my equation returns zeros for all values of xu, xl, and ea

[KSSV]

In the above code when you substitute xm , xu in f it gives a number, so * worked. Where as in the old script on substituting those values into f, the result was a vector. I think you need to check your f, rethink on your code.

Answer 2

[Torsten]

f(x) with a scalar argument x returns a vector of length 6.

For two vectors of length 6, the product f(xm)*f(xu) is not defined.

Formally, f(xm).*f(xu) should work, but I don't know if it makes sense in your context.

Best wishes

Torsten.