# Solving a recursive relation

Let $$\{c_t\}_{t = 1}^k$$ be a (non-monotone) sequence of real numbers such that $$c_t \in (0, 1]$$ for all $$t = 1, \dots, k$$. Consider the recursive sequence $$\left \{ \begin{array}{ll} x_1 & = c_1 & \\ x_{t + 1} & = c_{k + 1} \left (1 - \alpha \sum_{j = 1}^t x_{j} \right ) & \mbox{for all } 1 \leq t < k\\ \end{array} \right .$$ with $$\alpha > 0$$ a constant indipendent of $$t$$. Find a closed-form formula for the sequence $$\{x_t\}_{t = 1}^k$$.

As

$$\frac{x_{k+1}}{c_{k+1}} = 1-\alpha\sum_{j=1}^{j=k}x_j$$

calling $$y_k = \frac{x_k}{c_k}$$ we have

$$y_{k+1}-y_k = -\alpha x_k = -\alpha c_k y_k$$

so we have

$$y_{k+1}+(\alpha c_k - 1)y_k = 0$$

this is a recurrence linear equation with solution

$$y_k = C_0 \prod_{j=1}^{j=k-1}(1-\alpha c_j)$$

and finally

$$x_k = C_0 c_k\prod_{j=1}^{j=k-1}(1-\alpha c_j)$$

but as $$x_{1} = c_1$$ follows $$C_0 = 1$$

Consider $$\frac{x_{t+1}}{c_{t+1}}-\frac{x_{t}}{c_{t}}=-\alpha x_t$$ Can you take it from here?