n+1 in formula

20 Ansichten (letzte 30 Tage)
B
B am 3 Jul. 2019
Beantwortet: Dheeraj Singh am 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.
  3 Kommentare
Stephen23
Stephen23 am 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 am 3 Jul. 2019
Bearbeitet: Raj am 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;

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Dheeraj Singh
Dheeraj Singh am 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)

Kategorien

Mehr zu MATLAB finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by