Taking a descend interval

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Jane Smith
Jane Smith il 5 Mar 2021
Commentato: Jane Smith il 5 Mar 2021
for t=1:n
[L(t)]=(1/(50)^(t))
[C(t)]=(1/(800))^(t);
end
For the 1st line (L(t)) i need to use the interval 1:n however for the 2nd line i need from n:1. How do i change t without changing the interval in the for loop?

Risposta accettata

Alan Stevens
Alan Stevens il 5 Mar 2021
Insert
tau = n+1-t;
then raise 1/800 to the power tau.

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