why does the following code give error?

u=[3 4 30 50];% Desired Vector
b=u;
[R,C]=size(b);
P=C/2;
M=2*C;

% calculate observed vector
xo=zeros(1,M);
for k=1:M 
    for i=1:P 
        xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos^2(u(P+i))*(k-1)^2)));

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

    end 
end
xe=zeros(1,M);
for k=1:M
     for i=1:P
            xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos^2(b(P+i))*(k-1)^2));
     end
end
abc=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;  
end
abc=abc/M;
e=abc

Answer 1

[Torsten]

Use

xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos(u(P+i))^2*(k-1)^2));

xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos(b(P+i))^2*(k-1)^2));

instead of

xo(1,k)=xo(1,k)+1*exp(1i*((-pi/2)*sin(u(P+i))*(k-1)+(pi/(16*u(i)))*cos^2(u(P+i))*(k-1)^2)));

xe(1,k)=xe(1,k)+1*exp(1i*((k-1)*(-pi/2)*sin(b(P+i))+(pi/(16*b(i)))*cos^2(b(P+i))*(k-1)^2));

[Sadiq]

Thanks a lot dear Torsten fro your help. Yes now it works. But now it must give zero error "e" but it's not zero. Instead it gives "e=2.6077e-31". Why it is so?

[Torsten]

The accuracy of floating point arithmetic is limited:

https://uk.mathworks.com/help/matlab/matlab_prog/floating-point-numbers.html

[Sadiq]

Thanks a lot dear Torsten for your prompt response. If it is so, then why does the following code gives zero er

ror?

clear all;clc

u=[3 4 30 50];% Desired vector
b=u;
[R,C]=size(b);
P=C/2;
M=2*C;
% calculate observed vector
xo=zeros(1,M);
for k=1:M 
    for i=1:P 
        xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cos(u(P+i)));
    end 
end
xe=zeros(1,M);
for k=1:M
        for i=1:P
            xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cos(b(P+i)));
        end
end
abc=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2; 
end
abc=abc/M;
e=abc

[Torsten]

Because even the order in which terms are multiplied can matter for the result.

Here, the terms that are summed are absolutly identical while in your previous code, the multiplicative order in both terms differs.

[Sadiq]

Thanks a lot dear Torsten for your prompt response. Yes now I got it. Thank you once again.