Solve recurrence equation.

Question

Solve recurrence equation: $$\frac{y_{t+2}-y_{t+1}}{y_{t+1}} = 0.995 \frac{y_{t+1}-y_{t}}{y_{t}} $$

A friend gave me this question, but I'm not familiar with recurrence equations.

I can solve it for $$k_t := \frac{y_{t+1}-y_t}{y_t} = \frac{y_{t+1}}{y_t}-1$$ but this doesn't give me a closed form for $y_t.$

How do I proceed?

Answer 1

Let $c=0.995$. Since you've already no doubt obtained $k_t=k_0 c^t$ so $\frac{y_{t+1}}{y_t}=1+k_0 c^t$, we get $y_t=y_0\prod_{i=0}^{t-1}(1+k_0 c^{i-1})$, $k_0$ of course being $\frac{y_1}{y_0}-1$. (Unsurprisingly, the original recursion relation's general solution requires the first two terms to become particular.)

Answer 2

If $\dfrac{y_{t+2}-y_{t+1}}{y_{t+1}} = a \dfrac{y_{t+1}-y_{t}}{y_{t}} $ then $\dfrac{y_{t+2}}{y_{t+1}}-1 = a (\dfrac{y_{t+1}}{y_{t}}-1) $ or $\dfrac{y_{t+2}}{y_{t+1}}- a \dfrac{y_{t+1}}{y_{t}} =1-a $.

Let $x_t = \dfrac{y_{t+1}}{y_{t}} $. Then $x_{t+1}-ax_t = 1-a$.

Now apply the usual trick of dividing by $a^{t+1}$. This becomes $\dfrac{x_{t+1}}{a^{t+1}}-\dfrac{x_t}{a^t} = \dfrac{1-a}{a^{t+1}} $.

Letting $z_t = \dfrac{x_t}{a^t}$, this becomes $z_{t+1}-z_t = \dfrac{1-a}{a^{t+1}} $.

Summing from $0$ to $n-1$,

$\begin{array}\\ z_n-z_0 &=\sum_{t=0}^{n-1}(z_{t+1}-z_t)\\ &=\sum_{t=0}^{n-1}\dfrac{1-a}{a^{t+1}}\\ &=\dfrac{1-a}{a}\sum_{t=0}^{n-1}(1/a)^t\\ &=\dfrac{1-a}{a}\dfrac{1-(1/a)^n}{1-1/a}\\ &=-1+(1/a)^n\\ \end{array} $

Working back, since $z_0 = x_0$,

$\dfrac{x_n}{a^n} =z_n =z_0-1+(1/a)^n =x_0-1+(1/a)^n $ so $x_n =a^n(x_0-1)+1 $.

We can now get $y_m$.

$\begin{array}\\ \dfrac{y_m}{y_0} &=\prod_{n=0}^{m-1} \dfrac{y_{n+1}}{y_n}\\ &=\prod_{n=0}^{m-1} (a^n(x_0-1)+1)\\ &=\prod_{n=0}^{m-1} (a^n((y_1/y_0)-1)+1)\\ \end{array} $

so $y_m =y_0\prod_{n=0}^{m-1} (a^n((y_1/y_0)-1)+1) $.

Note: Nothing original here.

Answer 3

Let $w_t:=\dfrac{y_{t+1}-y_t}{y_t}$. The recurrence says

$$w_{t+1}=aw_t$$ and $$w_t=a^tw_0.$$

Now we solve the second recurrence

$$y_{t+1}=(w_t+1)y_t=(a^tw_0+1)y_t$$

with

$$y_t=y_0\prod_{s=1}^t(a^sw_0+1).$$

As far as I know, there is no simpler form.