how to reduce its computing time?

u=[30 40 50 70];
b=u;
[~,C]=size(b);
P=C/2;
M=10;
xo=zeros(1,M);
yo=zeros(1,M);
zo=zeros(1,M);
for k=1:M
    for i=1:P
        xo(1,k)=xo(1,k)+exp(-1i*(k-1)*pi*sind(u(i))*cosd(u(P+i))); 
        yo(1,k)=yo(1,k)+exp(-1i*(k-1)*pi*sind(u(i))*(cosd(u(P+i))+sind(u(P+i))));
        zo(1,k)=zo(1,k)+exp(-1i*(k-1)*pi*sind(u(i))*cosd(u(P+i))+cosd(u(i)));
    end %end of i
end
xe=zeros(1,M);
ye=zeros(1,M);
ze=zeros(1,M);
for k=1:M
        for i=1:P
            xe(1,k)=xe(1,k)+exp(-1i*(k-1)*pi*sind(b(i))*cosd(b(P+i)));
            ye(1,k)=ye(1,k)+exp(-1i*(k-1)*pi*sind(b(i))*(cosd(b(P+i))+sind(b(P+i))));
            ze(1,k)=ze(1,k)+exp(-1i*(k-1)*pi*sind(b(i))*cosd(b(P+i))+cosd(b(i)));
        end %end of i
end %end of k
abc=0.0;
abcd=0.0;
abcde=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;
  abcd=abcd+(abs(yo(1,m1)-ye(1,m1))).^2;
  abcde=abcde+(abs(zo(1,m1)-ze(1,m1))).^2;
end%end of m1
  err=abc+abcd+abcde;
err=err/(3*M);

Answer 1

[Torsten]

Here is the code for the first loop. You can do it for the other loops in the same manner.

u=[30 40 50 70];
b=u;
[~,C]=size(b);
P=C/2;
M=10;
MM =(0:(M-1)).';
PP=1:P
xo = sum(exp(-1i*MM*pi.*sind(u(PP)).*cosd(u(P+PP))),2).'
yo = sum(exp(-1i*MM*pi.*sind(u(PP)).*(cosd(u(P+PP))+sind(u(P+PP)))),2).'
zo = sum(exp(-1i*MM*pi.*sind(u(PP)).*cosd(u(P+PP))+cosd(u(PP))),2).';

[Sadiq Akbar]

Thank you very much dear Torsten for your kind response. I did the 2nd part like you. But what about the last part? i.e.,

abc=0.0;
abcd=0.0;
abcde=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;
  abcd=abcd+(abs(yo(1,m1)-ye(1,m1))).^2;
  abcde=abcde+(abs(zo(1,m1)-ze(1,m1))).^2;
end%end of m1
  err=abc+abcd+abcde;
err=err/(3*M);

I don't understand how to do it as its different from that.

[Torsten]

abc = sum(abs(xo-xe).^2);
abcd = sum(abs(yo-ye).^2);
abcde = sum(abs(zo-ze).^2);
err = abc+abcd+abcde;
err = err/(3*M);

[Sadiq Akbar]

Thanks a lot dear Torsten for your help.