I'm working on Fortran 90. I need to calculate a recursion like xn = 6*(xn-1) + 7*(xn-2) where (xn-1) is the n-1 step and (xn-2) is the n-2 step. So if we set x1 = 2 and x0 = 1 we get x2 = 6*x1 + 7*x0 and so on for each n.
So I wrote
x0 = 1.
x1 = 2.
do i = 1,20
xn = 6.*x1 + 7.*x0
x1 = xn
x0 = x1
end do
but this code is replacing x0 and x1 for xn and I need to replace x1 for xn and x0 for x1 in each step. I'd tryed many things but I failed. Any idea how to do that?
Though the answer has already been added to this question, let me answer a more general question which is encountered more frequently. Consider the problem where the very next value in the iteration depends on n previous values. In the present case n = 2. The general strategy to solve this problem is to construct another 1-d array of size n and save all the initial values x(1),x(2),..,x(n) in this array. Then in each iteration we use these values to calculate the next value x(n+1) and update the array with x(1) by x(2), x(2) by x(3),...,x(n) by x(n+1) and again use these values to calculate the next value of x and so on. A particular example where such strategy must necessarily be used is the integration of time-delayed systems.
cshift and eoshift are quite useful.@parthian-shot has given the correct answer in the comment. But that leaves the question marked as unanswered, so I am repeating it here:
You are assigning the value of xn to x1, and then the value of x1 (which is now the same as xn) to x0. You just need to flip it around:
do i = 1,20
xn = 6.*x1 + 7.*x0
x0 = x1
x1 = xn
end do
