For a function p(0) = 10000, p(n) = p(n-1) + 0.02*p(n-1),
the code should be like this:
def p(n,v=10000):
if n == 0:
return v
else:
return p(n-1,1.02*v)
But if p(0) = 10000, p(n) = p(n-1) + 10**(n-1),
then how to write this tail recursion?
This should solve your problem
def p(n):
if n==0:
return 10000
else: # n!=0
return p(n-1) + 0.02 * p(n-1)
print(p(0)) # The result is 10000
print(p(1)) # The result is 10200.0
the first if will be the base of the recursion which is p(0) and the else will be the recursion function
Here's the tali recursion code for the function that you wanted
def p(n,v=10000):
if n == 0:
return v
else:
return p(n-1, v + 10**(n-1))
Here, we use the v as the value from the previous function recursion call, and then add the 10**(n-1) to it.
Well.. you already have tail recursion with the if n == 0: return v. You just need to rework the non constant return value. Try this:
def p(n, v=10000):
if n == 0:
return v
else:
return p(n - 1, v) + 10**(n - 1)
