Mean Sequence
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 $u$1=2 and $u$2=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 $u$1, 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 = $u$1+2($u$2$u$1)/3 = ($u$1+2$u$2)/3.
If $u$1 $u$2 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 = $u$12($u$1$u$2)/3 = ($u$1+$u$2)/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.