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Problem

A second order recurrence relation is defined by u_n+1 = (u_n+u_n1)/2; that is, each new term is the mean of the previous two terms.

For example, when u₁=2 and u₂=5, we generate the following sequence:

2, 5, 3.5, 4.25, 3.875, ... .

Find the limit of the sequence for different starting values.

Solution

Consider the following diagram.

By the definition of the sequence, the next term, u_k+1 will lie between the two previous terms, u_k and u_k1. Thus, as this process continues, each iteration will reduce the upper and lower bounds, demonstrating that the sequence converges towards a value.

Let the distance between the first and second term be defined as one unit, so that, from the first term u₁, the terms follow the following series:

S = 1 1/2 + 1/4 1/8 + ...
2S = 2 1 + 1/2 1/4 + 1/8 ...
3S = 2 S = 2/3

That is, the sequence converges towards a value, L, which is 2/3 above the first term: L = u₁+2(u₂u₁)/3 = (u₁+2u₂)/3.

If u₁ u₂ then the sequence follows the series:

S = -1 + 1/2 1/4 + ...
2S = -2 + 1 1/2 + 1/4 ...
3S = -2 S = -2/3

And we get the same limit, L = u₁2(u₁u₂)/3 = (u₁+u₂)/3.

What about the third order recurrence relation: u_n+1 = (u_n+u_n1+u_n2)/3 ?
Investigate the k th order recurrence relation.