I came across this function as an example of a non-linear function being numerically fit to some data points using the Levenberg-Marquardt algorithm:
$f(t; b_1, b_2, b_3) = b_1 (t+b_2)^{-\frac{1}{b_3}}$
It was labeled bennet5, and it looks like a scaled and shifted power function, but no information about the domain (i.e. what are valid choices of $b_1,b_2,b_3$ and $t$) was provided. I would like to learn more about this family of functions but am not sure what to search for. Could anyone here offer some clues as to what it is called or where I could read more?
$$f(t) = b_1 (t+b_2)^{-\frac{1}{b_3}} \tag 1$$
Inverting leads to : $$t(f)=-b_2+\left(\frac{f}{b_1} \right)^{-b_3} \tag 2$$ This function was considered in the paper referenced below.
In order to make it consistent with the cited paper, let change the notations : $$\begin{cases} y=t \\ a=-b_2 \\ b=b_1^{b_3} \\ c=-b_3 \\ X=f \end{cases} \qquad\text{or}\qquad \begin{cases} t=y \\ b_2=-a \\ b_1=b^{-1/c} \\ b_3=-c \\ f=X \end{cases}$$
The equation is transformed into : $$y(X)=a+bX^c$$ To compute approximate $a$ , $b$ and $c$ , a straightforward method of regression ( not iterative, no initial guess) is provided page 17 in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
Then, the approximate $b_1$ , $b_3$ and $b_3$ are easy to derive thanks to the above relationships.
LATTER EDITION :
Instead of the fitting of the inverse function $(2)$ as proposed above, the fitting of the function $(1)$ can be directly carried out thanks to the same general method, but with an integral equation which $(1)$ is solution : $$f(t)-f_1=\frac{1}{b_2}\left(1-\frac{1}{b_3}\right)\int_{t_1}^t f(t)dt-\frac{1}{b_2}\left(tf(t)-t_1f_1\right)$$ where $f_1=f(t_1)$
At first sight this seems complicated. But the numerical calculus is very simple as shown below.
The initial data is : $$(f_1\:,\:t_1)\:,\:(f_2\:,\:t_2)\:,\:...\:,\:(f_k\:,\:t_k)\:,\:...\:,\:(f_n\:,\:t_n)$$ The data have to be ranked in increasing order of $t_k$ , such as $t_k\leq t_{k+1}$.
Then we compute the approximate values of the integral : $$S_1=0\quad;\quad S_k=S_{k-1}+\frac12(f_k+f_{k-1})(t_k-t_{k-1})$$ Then we compute the coefficients $A$ and $B$ :
Which gives :
$$b_2\simeq -\frac{1}{B} \quad\text{ and }\quad b_3\simeq \frac{B}{A+B}$$
NUMERICAL EXAMPLE :