How can i coda this process?

x=((0*-0.15)+(-7.50*-1.25)+(-9.50*-1.40)+(-5.81*-1.32)+(2.15*-1.28)+(10.20*-0.15)+(7.50*0))
%
h=((0*-7.50)+(-0.15*-9.50)+(-1.25*-5.81)+(-1.40*2.15)+(-1.32*10.20)+(-1.28*7.50)+(-0.15*0))
%
2*F=x-h

How can i coda this process?

[John D'Errico]

What process? Only you know what the notation you have there means. Anyway, it seems like you have written something, although

2*F = x-h

is not valid syntax.

[Muhendisleksi]

If the numbers in the direction of the arrow are multiplied, then x and h are obtained. later: F=(x-h)/2

[John D'Errico]

Ok. So you wrote it. What is the problem?

[Muhendisleksi]

The problem is I write a lot of ridiculous code.

clear
clc
% 1. Alan verileri
ust=[0;-0.15;-1.25; -1.40; -1.32; -1.28; -0.15;0];
alt=[0;-7.50; -9.50; -5.81; 2.15; 10.20;7.50;0];
% 2. Alan verileri
ust2=[0;-0.15;-1.27;-1.22;-1.36;-1.12;-1.23;-0.15;0];
alt2=[0;-7.50;-9.75;-6.05;1.95;6.74;10.12;7.50;0];
s=20;
% Alan heseplamaları    
[u]=size(ust);
m=u-1;
for i=1:m
    for j=1:m      
        F(i)=ust(i+1)*alt(i);
    end
end
for i=1:m
    for j=1:m   
        F1(i)=ust(i)*alt(i+1);
    end
end
for i=1:m
    for j=1:m      
        f(i)=ust2(i+1)*alt2(i);
    end
end
for i=1:m
    for j=1:m   
        f1(i)=ust2(i)*alt2(i+1);
    end
end
Alan_1=(sum(F)-sum(F1))/2
Alan_2=(sum(f)-sum(f1))/2

I think a better code can be written.