I want to use the least squares adjustment to get the parameter $c$ and $a$ of the following formula:
$$f(a,c) = c \cdot e^{-a^2 \cdot r^2}$$
For the design matrix, I used the Taylor series:
$$f(x) = c \cdot \sum_{k=0}^{\infty} \frac{(-a^2r^2)^k}{k!}.$$ As an approximation the first 3 elements would be $$f(x) \approx c - \frac{a^2r^2}{1} + \frac{(a^2r^2)^2}{2} - \frac{(a^2r^2)^3}{6}.$$
But now I am not sure how to linearize this term to get the design matrix $$\mathbf{D} = \left[\frac{\partial f(x)}{\partial c} \;\; \frac{\partial f(x)}{\partial a}\right].$$
Guess, the problem can be solved easy but at the moment I can't see the solution.
Does someone has an idea?
Thanks a lot for you help!