n+1 in formula

B (view profile)

on 3 Jul 2019
Latest activity Answered by Dheeraj Singh

Dheeraj Singh (view profile)

on 17 Jul 2019
Hi, could someone help me to put the next formula in matlab?
The sum goes to n-1: 1/2*T* sum(abs(AP[n+1]*ML[n] - AP[n]*ML[n+1])).
The main problem is the n+1 how can i do this.

Raj

Raj (view profile)

on 3 Jul 2019
Stephen Cobeldick

Stephen Cobeldick (view profile)

on 3 Jul 2019
B's "Answer" moved here:
Is this better? AP and ML are about the x and y coordinates when someone is standing still. Raj

Raj (view profile)

on 3 Jul 2019
" main problem is the n+1 " I am still not quite sure what is the problem. Is this what you are looking for?
count=0;
for n=1:N-1
count=count+abs((AP(n+1)*ML(n))-(AP(n)*ML(n+1)));
end
count=(0.5*count)/T;

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Answer by Dheeraj Singh

Dheeraj Singh (view profile)

on 17 Jul 2019

Hi,
I understand that by “put” you mean you want to implement it without using loops.
The following code implements the above formula without using loops using random values of “AP” and “ML”
T=2;
n=100
AP=rand(n,1);
ML=rand(n,1);
i=1:n-1;
j=i+1;
val=sum(abs(AP(j).*ML(i)-AP(i).*ML(j)))/(2*T)