$z_t=x_{t+7}/x_t$. Solve back for x. model is $z_t=alpha*z_{t-1}$ I want to create a model of x, 
Now my issue is that to get this fit I need to transform the original data such that $z=x_{t+7}/x_t$ the absolutely best fit I could get is 
by regressing $z_t=alpha*z_{t-1}$. 
This gives me an R^2 of 0.92. Now I thought by having the parameter from this regression I could backsolve for $x_{t+7}=x_t*alpha*(x_{t+6}/x_{t-1}))$, 
so I did this and then I looped though over x and I expected that the fit would be at the same level as the transformed equation, however, the fit is absolutely terrible. 
This means that I am either doing something wrong or one can not simply transform a time series by dividing by it's lag. Would anyone be able to explain what I've done wrong, why the fit becomes so much worse after transforming back, and what I should be doing?
 A: A simple solution
You can solve this as following:

*

*The function $z(t)$ is an exponetial function
$$z_t = \beta \cdot\alpha^t$$


*For $x_t$ you could reparameterize by using the exponent of a function $f(t)$
$$x_t = e^{f(t)}$$
such that
$$\frac{x_{t+7}}{x_{t}} = \frac{e^{f(t+7)}}{e^{f(t)}} = e^{f(t+7)- f(t)} = z(t)$$
or
$$f(t+7)- f(t) = \log( z(t)) = \log \beta +  \log(\alpha) t$$
a function $f(t)$ that satisfies this is a polynomial function
$$f(t) = a + bt + ct^2$$
such that
$$\begin{array}{}
f(t+7)- f(t) &=& a + b(t+7) + c(t+7)^2 - a - bt - ct^2 \\ &=& \underbrace{49 c}_{\log \beta} + \underbrace{(7b+14 c)}_{\log(\alpha)}t \end{array}$$

More general solutions
The polynomial function is not the only solution.
I have not solid theoretic/mathematical proof for the entire space of solutions, but there are many functions that will satisfy.
Intuitively, what you have is
$$\frac{x(t + \Delta)}{x(t)}  = c \cdot \alpha^t$$
or (taking the logarithm of both sides)
$$ \log({x(t + \Delta)}) - \log({x(t)})  =  \log(c) + \log(\alpha ) t$$
and in terms of $x_t = e^{f(t)}$
$$ f(t + \Delta) - f(t)  =  \log(c) + \log(\alpha ) t$$
and any function $x(t)$ where the difference of the logarithm ($f(t + \Delta) - f(t)$) is a linear function of time will work.
So alternative functions to
$$x(t) = e^{a+bt+ct^2}$$
will be
$$x(t) = e^{a+bt+ct^2 + g(t)}$$
where $g(t)$ is a cyclic function such that $g(t+7) = g(t)$

Why your fitting/backsolving may be problematic
I believe that is it is better to fit the equation $x(t) = exp(a+bt+ct^2)$ directly to your data.

*

*In your current situation, fitting $x_{t+7}/x_{t}$ (which is sort of like fitting the derivative) you only determine the parameters $\alpha$ and $\beta$, which relates to the parameters $b$ and $c$ of the polynomial, but there is a scaling/integration parameter $a$ that is not determined by this.


*You are computing/backsolving your estimates of the data based on the values $z_t$ determined from the data $x_{t+7}/x_{t}$. However, these values $z_t = x_{t+7}/x_{t}$ have random errors (assuming you are fitting a model for values $x(t)$ that has additional noise and is not exactly like the model), and this will make you modelled curve/estimate change a lot.
I am not sure what your are doing with your backsolving method (you didn't describe the alogrithm completely), but it can go wrong in many ways, and I imagine that the errors in the data are building up in your backsolved solution. Anyway, your algorithm for backsolving the solution might not need to be the same as the function $x(t) = e^{a+bt+c t^2 + g(t)}$ and neither the function with additional errors $x(t) = e^{a+bt+c t^2 + g(t)} + \epsilon(t)$ and it may include the errors $\epsilon(t)$ in a wrong way, such that your result starts to strongly deviate from the data.

Analyzing your problem
possibly you could gain some insight into your time series by plotting the logarithm of your data series
$$\log(x(t)) = a+bt+c t^2 + g(t)$$
How does that look like? Do you have a polynomial function? Do you find autocorrelation (the function $g(t)$) in the residuals of a polynomial fit?
