<p>We have the recurrence relation</p>

<p>$$x_{k+1} = \frac{x_{k+8}}{x_{k+1}}$$</p>

<p>If the denominator is nonzero, this recurrence relation can be rewritten as follows</p>

<p>$$x_{k+7} = x_k^2$$</p>

<p>Assuming positivity and taking the logarithm of both sides, we obtain a <em>linear</em> recurrence relation</p>

<p>$$\ln (x_{k+7}) = 2 \ln (x_k)$$</p>

<p>Shifting,</p>

<p>$$\ln (x_{k+1}) = 2 \ln (x_{k-6})$$</p>

<p>Let</p>

<p>$$\eta_k := \left( \ln (x_k), \ln (x_{k-1}), \dots, \ln (x_{k-6}) \right)$$</p>

<p>be a $7$-dimensional state vector. In matrix form,</p>

<p>$$\eta_{k+1} = \begin{bmatrix} 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 2\\ 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\end{bmatrix} \eta_k$$</p>

<p>Hence, we need $7$ initial conditions. We recover $x_k$ via $x_k = \exp( \mathrm e_1^{\top}\eta_k )$.</p>

CC BY-SA 3.0
2016-10-30T10:57:17.193Z
2016-10-30T11:18:10.343Z
2016-10-30T11:18:10.343Z
Rodrigo de Azevedo
<p>I'm doing economic modelling where $x_t$ is the intertemporal cash-flow variable. I need to solve the following recurrence relation</p>

<p>$$x_{t+1}=\frac{x_{t+8}}{x_{t+1}}$$</p>

<p>My problem is that I don't exactly know how to linearize this function. Maybe I should introduce a new variable $y_{t} = x_{t+1}$ and get</p>

<p>$$y_{t}^2={x_{t+8}}$$</p>

<p>Could you help me out with the correct linearization technique, please? I'm looking forward to your answer.</p>

CC BY-SA 3.0
2015-12-12T23:24:36.657Z
Linear in what exactly? You have $x_{t+1}^2=x_{t+8}$. Take log on both sides: $2\ln(x_{t+1})=\ln(x_{t+8})$. Is this linear enough? Or do you want the function to be linear in $x_t$?
Herr K.
CC BY-SA 3.0
2015-12-13T11:01:45.850Z
Thanks, I've already thought about loglinearization. But what to do after that? I'm looking for the $x_{t}$=...  form solution.
Übel Yildmar
CC BY-SA 3.0
2015-12-13T11:23:07.057Z
Do you have some boundary condition(s) as part of the definition of the difference equation, e.g. $x_0$ or $x_T$ given? From the log-linearized form, you'll have $x_t=\sqrt{x_{t+7}}$. So if you know $x_0$, then you'll know $x_7$.
Herr K.
CC BY-SA 3.0
2015-12-13T11:30:26.367Z
I have a basic condition like $x_{0} > 0$. Maybe this stands for to make available the loglinearization. Thank you @Herr K.
Übel Yildmar
CC BY-SA 3.0
2015-12-13T11:34:10.850Z
Do you know how could I linearize this function using Taylor-series?
Übel Yildmar
CC BY-SA 3.0
2015-12-13T23:44:17.723Z
Not sure if Taylor expansion is the way to go... Perhaps including more details of your question would help clarify things a bit.
Herr K.
CC BY-SA 3.0
2015-12-14T15:20:18.893Z
There's no more addition :/ I'm just looking for the solution for this simple difference equation. I'm not 100% (80% maybe) sure whether your comment is correct or not.
Übel Yildmar
<time-series><difference-equations>
How to linearize the following difference equation?
Übel Yildmar
# How to linearize the following difference equation?

(Asked by: username_0 on 2015-12-12)

I'm doing economic modelling where $x\_t$ is the intertemporal cash-flow variable. I need to solve the following recurrence relation

$$x\_{t+1}=\frac{x\_{t+8}}{x\_{t+1}}$$

My problem is that I don't exactly know how to linearize this function. Maybe I should introduce a new variable $y\_{t} = x\_{t+1}$ and get

$$y\_{t}^2={x\_{t+8}}$$

Could you help me out with the correct linearization technique, please? I'm looking forward to your answer.

**Tags:** time-series, difference-equations
> **Comments:**
> **username_1**: Linear in what exactly? You have $x_{t+1}^2=x_{t+8}$. Take log on both sides: $2\ln(x_{t+1})=\ln(x_{t+8})$. Is this linear enough? Or do you want the function to be linear in $x_t$?  
> **username_0**: Thanks, I've already thought about loglinearization. But what to do after that? I'm looking for the $x_{t}$=...  form solution.  
> **username_1**: Do you have some boundary condition(s) as part of the definition of the difference equation, e.g. $x_0$ or $x_T$ given? From the log-linearized form, you'll have $x_t=\sqrt{x_{t+7}}$. So if you know $x_0$, then you'll know $x_7$.  
> **username_0**: I have a basic condition like $x_{0} > 0$. Maybe this stands for to make available the loglinearization. Thank you @Herr K.  
> **username_0**: Do you know how could I linearize this function using Taylor-series?  
> **username_1**: Not sure if Taylor expansion is the way to go... Perhaps including more details of your question would help clarify things a bit.  
> **username_0**: There's no more addition :/ I'm just looking for the solution for this simple difference equation. I'm not 100% (80% maybe) sure whether your comment is correct or not.  

---

## 1 answers

### Answer by username_2 on 2016-10-30. Score: 3

We have the recurrence relation

$$x\_{k+1} = \frac{x\_{k+8}}{x\_{k+1}}$$

If the denominator is nonzero, this recurrence relation can be rewritten as follows

$$x\_{k+7} = x\_k^2$$

Assuming positivity and taking the logarithm of both sides, we obtain a *linear* recurrence relation

$$\ln (x\_{k+7}) = 2 \ln (x\_k)$$

Shifting,

$$\ln (x\_{k+1}) = 2 \ln (x\_{k-6})$$

Let

$$\eta\_k := \left( \ln (x\_k), \ln (x\_{k-1}), \dots, \ln (x\_{k-6}) \right)$$

be a $7$-dimensional state vector. In matrix form,

$$\eta\_{k+1} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 2\\ 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix} \eta\_k$$

Hence, we need $7$ initial conditions. We recover $x\_k$ via $x\_k = \exp( \mathrm e\_1^{\top}\eta\_k )$.


---

