# Can this nonlinear model be transformed to a linear one?

Consider the model

$$y=\frac{1}{1-\theta_1x^{\theta_2}}+\theta_3+\epsilon$$

(assume Gaussian iid errors, even though this is not relevant for the question). Is it possible to transform it into a linear one? I fiddled with it a bit, without success. Am I missing something, or is it just impossible to convert it to a linear model?

• Do you mean linear in some function of $x$, linear in some function of $(\theta_1,\theta_2,\theta_3)$, or both?
– whuber
Commented Nov 10, 2017 at 18:08
• @whuber I mean linear in the parameters, thus according to this definition $y=\beta_0+\beta_1\sin(x)$ would be linear but $y=\beta_0+\sin(\beta_1x)$ wouldn't. Transforming $x$ and $y$ to get to a linear model is ok. Ideally, I would also like to understand how the errors transform, since I'll have to fit the model to data, but let's say that for now the most urgent point is to understand whether this model is equivalent to a linear one or not. Commented Nov 10, 2017 at 19:28

$$f(x) = f(0) + f'(0)x + \frac{f''(\xi)}{2}x^2, (\xi \in [0, x]).$$
If you apply this to the geometric-series expansion of $\frac{1}{1 - \theta_1x^{\theta_2}}$, and then apply on the result the exponential function expansion, it is easy to see that there is no possible form that is linear in $\theta_1$ and $\theta_2$. Otherwise, you would have two equivalent polynomials, where one has a nonzero $\theta_1 \theta_2$ term, but the other doesn't, which is impossible.
• ...because $\xi$ depends on $x$, so I don't understand why your argument holds. 2) You propose to use the geometric series expansion: does this mean you're making the substitution $z=\theta_1 x^{\theta_2}$ and then expanding $\frac{1}{1-z}$? Commented Nov 11, 2017 at 14:00