0
$\begingroup$

I have a problem that I have been struggling with for several days. I have a dataset and I know the left part is from a curve like y = a/x and the left part of y = bx^c, where a, b and c are the variables I need to find. I have to fit one curve (my case is exactly like the picture below) but this curve consists of two different "regimes". How can I do this to get a smooth transition between the left and right parts? I tried to do it in Origin but to no avail, I am trying to do it with Mathematica but I don't know the language very well. I will be very grateful for any help!

My data: {1., 200}, {2., 300}, {5.,700}, {0.1,100}, {0.04,1000}, {0.5,130}, {0.05,200}, {0.03,1200}

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ There are many ways to accomplish this. Consider selecting a method based on a reasonable physical model of your system that also accounts for errors and uncertainties. I describe one method of fitting this shape at stats.stackexchange.com/a/148166/919. Possibly some of those ideas might apply in your case. $\endgroup$
    – whuber
    May 15, 2021 at 15:03

1 Answer 1

0
$\begingroup$

The shape of the curve looks like an hyperbola. So, we can try to fit an hyperbola to the data.

But this is probably not the best choice. I agree with the whuber's comment: A better choice of model should be derived from physical considerations.

The numerical data provided in the question apparently doesn't match to the figure. Since it is better to have more points for the example of calculus, I used the figure to generate the data below. Of course this is not very accurate due to the uncertain scanning of the pixels of the graph.

Notations :

The coordinates of the points on the graph are noted $(X_1,Y_1),(X_2,Y_2) ,..., (X_k,Y_k) ,..., (X_n,Y_n)$

Since the scales are logarithmic they correspond to the "experimental" values : $$(x_1,y_1),(x_2,y_2) ,..., (x_k,y_k) ,...,(x_n,y_n) \quad \text{with}\quad \begin{cases} X_k=\log_{10}(x_k)\\ Y_k=\log_{10}(y_k)\end{cases}$$

Asymptotically we want $\begin{cases} y=\frac{a}{x}\\ y=bx^c \end{cases}\quad\implies\quad\begin{cases} Y=A-X\\ Y=B+cX \end{cases}$

Equation of the asymptotes : $\big(Y-(A-X)\big)\big(Y-(B+cX)\big)=0$

Equation of the hyperbola : $\big(Y-(A-X)\big)\big(Y-(B+cX)\big)=G$

$G$ is a constant to be optimiezed to fit the hyperbola to the data.

Equation of the hyperbola : $$Y^2+(1-c)XY-cX^2-(A+B)Y+(Ac-B)X+AB-G=0$$

The optimized coefficients can be computed thanks to linear regression : $$Y^2+XY= c(X^2+XY)+(A+B)Y+(B-Ac)X-(AB+G)$$ $$V=cU+DY+EX+F \qquad\begin{cases} V=Y^2+XY\\ U=X^2+XY\\ D=A+B\\ E=B-Ac\\ F=-(AB+G) \end{cases}$$ The linear regression gives $c, D,E,F\quad$ then $\quad\begin{cases} A=\frac{D-E}{1+c}\\ B=D-A\\ \end{cases}\quad\text{and}\quad\begin{cases} a=10^{A}\\ b=10^{B}\end{cases}$

enter image description here

The equation of the fitted curve comes from solving for $Y$ the next equation : $$Y^2+(1-c)XY-cX^2-(A+B)Y+EX-F=0$$ $$Y=\frac{1}{2}\left((c-1)X+A+B+\sqrt{((c-1)X+A+B)^2+4(cX^2+EX+F}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.