Is the scheme for generating $\displaystyle p_n=\left(\frac{1}{3}\right)^n$ stable?
$\displaystyle p_{n} = \frac{5}{6} p_{n-1} - \frac{1}{6} p_{n-2}$
Asked
Active
Viewed 37 times
0
-
1How is this different from http://math.stackexchange.com/questions/672727/is-the-following-scheme-for-generating-p-n-1-3n-stable-or-not-p-n-5-6p ? – lhf Feb 11 '14 at 23:50
-
It's not. It was even posted by the same user. – rubberchicken Feb 12 '14 at 01:41
1 Answers
0
What have you tried? Consider writing: $$ (5/6)p_n - (1/6)p_{n-2}$$ and then simplifying. You should get:
$$ (5/6)p_n - (1/6)p_{n-2} = (1/3)^n( (5/6) - (9/6))$$
Does that help? Also -- In general, your question could be re-phrased as: "Does $p_n$ satisfy the following recurrence relation: $p_n = \frac{5}{6} p_n - \frac{1}{6} p_{n-2}$ ?"
pgadey
- 2,226