Arithmetico–geometric recursive sequence and the fixed point distance from the first term

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Consider the arithmetico–geometric recursive sequence defined by $\begin{cases} u_0 = 1\\ u_{n+1} = \frac{3}{2} u_n + 1\\ \end{cases}$

The fixed point $\alpha$ is the intersection between the affine representation of $(u_n)$ and the first bisector $y = x$. The unique solution of the equation is $\alpha = \frac{3}{2} \alpha + 1 \iff \alpha = -2$.

sequence plot

The geometric interpretation shows that the distance $u_0 – \alpha$ is proportional to the distance $u_1 – \alpha$ by a real number $k$. We can generalize this to any term of the given sequence by

$$u_{n+1} – \alpha = k(u_n – \alpha)$$

so that each distance between the current term and the fixed point is directly proportional to the distance between the previous term and the fixed point.

For a geometric sequence, the fixed point is at the origin ($\alpha = 0$).

Any recursive geometric application is defined by the ratio $k$ between the distance $v_n$ of any term $u_n $from the fixed point $\alpha$ over the distance $v_{n+1}$ of the immediate following term $u_{n+1}$ from the fixed point $\alpha$. The fixed point is the lowest (respectively highest) possible value for an increasing (respectively decreasing) geometric (or arithmetico–geometric) sequence.

This is confirmed by a strong intuition feeling, however, is there any specific reason for which the fixed point is the reference for this proportional distance?

Best Answer

Drop the particular condition $u_0 = 1$ and consider the set of all possible sequences satisfying $u_{n+1} = \frac 32u_n + 1$. Note that if $u_n, v_n$ are any two such sequences, then their difference $c_n = u_n - v_n$ satisfies $$c_{n+1} = u_{n+1} - v_{n+1} = \left(\frac 32u_n + 1\right) - \left(\frac 32v_n + 1\right) = \frac 32c_n$$ Thus $c_n = \left(\dfrac 32\right)^nc_0$.

Now $\alpha = \dfrac 32\alpha + 1$, so if $v_0 = \alpha$, if follows that $v_n = \alpha$ for all $n$. And therefore $c_0 = u_0 - \alpha$, and $$(u_n - \alpha) = \left(\dfrac 32\right)^n(u_0 - \alpha)$$

So the reason $\alpha$ is the reference is that $\alpha$ gives a constant sequence, and the difference between any two sequences is geometric.

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