How to solve recurrence relations in Python

Question

I am trying to write code to give numerical answers to a recurrence relation. The relation itself is simple and is defined as follows. The variable x is an integer

• p(i) = p(i+2)/2 + p(i-1)/2 if i > 0 and i < x
• p(0) = p(2)/2
• p(i) = 1 if i >= x

This is also in this code.

from __future__ import division
def p(i):
    if (i == 0):
        return p(2)/2
    if (i >= x):
        return 1
    return p(i-1)/2+p(i+2)/2


x = 4
#We would like to print p(0)  for example.

This of course doesn't actually let you compute p(0). How can you do this in python?

---

Is it possible to set up a system of simultaneous equations which numpy.linalg.solve can then solve?

Answer 1

You're right this can be solved using linear algebra. What I've done below is a simple hard-coded translation. Your equations for p(0) to p(3) are coded up by rearranging them so that the right hand side is =0. For p(4) and p(5) which appear in the recurrence relations as base cases, there is an =1 on the right hand side.

• -p(0) + p(2)/2 = 0

• p(i-1)/2 - p(i) + p(i+2)/2 = 0 for i > 0 and i < x

• p(i) = 1 if i >= x

Here is the program hardcoded for n=4

import numpy
a=numpy.array([[-1,   0, 0.5,  0,   0,   0], # 0
               [0.5, -1,   0,0.5,   0,   0], # 1
               [0,  0.5,  -1,  0, 0.5,   0], # 2
               [0,    0, 0.5, -1,   0, 0.5], # 3
               [0,    0,   0,  0,   1,   0], # 4
               [0,    0,   0,  0,   0,   1], # 5
              ])
b=numpy.array([0,0,0,0,1,1])
# solve ax=b
x = numpy.linalg.solve(a, b)
print x

Edit, here is the code which constructs the matrix programmatically, only tested for n=4!

n = 4

# construct a
diag = [-1]*n + [1]*2
lowdiag = [0.5]*(n-1) + [0]*2
updiag = [0.5]*n
a=numpy.diag(diag) + numpy.diag(lowdiag, -1) + numpy.diag(updiag, 2)

# solve ax=b
b=numpy.array([0]*n + [1]*2)
x = numpy.linalg.solve(a, b)

print a
print x[:n]

This outputs

[[-1.   0.   0.5  0.   0.   0. ]
 [ 0.5 -1.   0.   0.5  0.   0. ]
 [ 0.   0.5 -1.   0.   0.5  0. ]
 [ 0.   0.   0.5 -1.   0.   0.5]
 [ 0.   0.   0.   0.   1.   0. ]
 [ 0.   0.   0.   0.   0.   1. ]]
[ 0.41666667  0.66666667  0.83333333  0.91666667]

which matches the solution in your comment under your question.