My intention with the following code:
step = 0.1;
time = Table[i, {i, 0, 100, step}];
T = Table[0, {i, Length[time]}];
tt = Prepend[T, 90];
temp = 0;
Do[
tr = tt[[i]]+temp;
temp = ((tr + 273.15) - time[[i+1]]) - 273.15
// Print, {i, 1, 100, 1}
]
is to have tr to be updated each time as if you were reading it line by line. In other words, to have tr to be in the first iteration 90, and then in the following iterations whatever the value of temp is in each iteration. As an example, for the first iteration temp should be temp = ((90 + 273.15) - 0.1) - 273.15 =89.9. In the second iteration temp should be temp = ((89.9 + 273.15) - 0.2) - 273.15 =89.7. In the third iteration temp should be temp = ((89.7 + 273.15) - 0.3) - 273.15 =89.4 and so on. I am do not think the do loop is the appropiate loop for this but I am not what would be the best strategy for this. I appreciate your inputs in advanced
{89.9, 89.7, 89.4, 89., 88.5, 87.9, 87.2, 86.4, 85.5, 84.5,etc....}or something else? Answer below gives the above. May be you can give a small explicit example (sayi=1..3only and show what list should the loop generate. This is easier than describing with words. $\endgroup$temp=((tr + 273.15) - time[[i+1]]) - 273.15by using updatingtreach time. So that in the first iterationtr=90, in the secondtr=temp, in the thirdtr=tempand so on. So I want 'tr' to be 90 at first by then to get the value oftempafter the second iteration. I hope that helps clarify it $\endgroup${i, 1, 3 1}instead of using words to describe things. This will make things much more clear. $\endgroup$tempis compute it (which the other code happens to get the answer to but it does not computetemp) $\endgroup$i=itempwill generate 89.9, fori=2tempwill generate 89.7, fori=3tempwill generate 89.4. It would do so by having 'tr' to be 90 fori=itr=tempfor any iteration after that. $\endgroup$