Let's say you want to build: u(n+1)=f(u(n)) with u(0)=u0
One solution is to define a simple recursive function:
u0 = ...
def f(x):
...
def u(n):
if n==0: return u0
return f(u(n-1))
Unfortunately, if you want to calculate high values of u, you will run into a stack overflow error.
Another solution is a simple loop:
def u(n):
ux = u0
for i in xrange(n):
ux=f(ux)
return ux
But if you want multiple values of u for different values of n, this is suboptimal. You could cache all values in an array, but you may run into an out of memory error. You may want to use generators instead:
def u(n):
ux = u0
for i in xrange(n):
ux=f(ux)
yield ux
for val in u(1000):
print val
There are many other options, but I guess these are the main ones.