Serial summation within a for() loop

Summation Equations are a particular Achilles' Heel for me plus I'm a bit rusty on MATLAB these days. How I can algorithmize this equation?

where Δt_sup, q and t are each vector arrays. I was given this expansion as an example for Δt_sup(3):

Now I know that I need to do this within a for loop to iterate values for each corresponding value of Δt_sup(i), q(i) and t(i), but I'm not sure how to go about tackling this beyond the basic setup as such:

t_sup=zeros(length(q),1);
for i=1:length(q)
  t_sup(i)= ???
end

Any help would be much appreciated.

[Stefan Raab]

You could use another for-loop within the for-loop which runs from 1 to i and adds the actual j-value in each step:

t_sup(i) = t_sup(i) + ((q(j) - q(j-1))/q(i))*log(t(i)-t(j-1)); % Caution: log10 for log to the base 10, log is for base e

Answer 1

[Star Strider]

You might not need any loops. See if this works:

q = randi(99, 10, 1);                       % Create Data (Column Vector)
t = [1:10]';                                % Create Data (Column Vector)
dq = [diff(q); q(1)]/q(end);                % ‘q-difference’ Vector
dt = [(t(end)-t(1:end-1)); t(end)];         % ‘t-difference’ Vector
Dt_sup = sum(dq .* log(dt));                % Δt_sup

[Ali Dhukka]

Correct me if my understanding of what your script does is off, but I think I still need a for loop as I need to be able to find a value for Dt_sup at each time t and this script results with a scalar value for Dt_sup.

[Star Strider]

My code calculates ‘Dt_sup’ for equal-length vectors of ‘t’ and ‘q’ defined in the code. My understanding of your code is that it is a scalar for given equal-length vectors of ‘t’ and ‘q’. It does not appear to be a function of ‘t’ otherwise. You can certainly loop through it with different vectors for ‘t’ and ‘q’, the only restriction being that the ‘t’ vector must be strictly monotonically increasing.

I just took the equation in your Question and coded it as presented. I have no idea what you’re doing, or the larger context of this equation.