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folks.

I'm a begginer with Mathematica and all I know is just a few basics commands and i'm trying to improve it.

I have this list of experimental data taken from a paper which describes the magnetization of a sample (actually a rock) toward the equilibrium when subject to an external magnetic field (y axis is magnetization, x axis is time in log scale):

data={{0.995, 0.142}, {3.003, 0.2}, {5.908, 0.25}, {10.525, 0.36}, {13.617,
0.498}, {24.321, 0.616}, {33.917, 0.599}, {47.843, 0.7}, {64.172, 
0.835}, {91.353, 1.102}, {126.745, 1.083}, {174.118, 
1.225}, {225.059, 1.133}, {292.998, 1.165}, {369.133, 
1.298}, {640.295, 1.365}, {828.169, 1.298}, {1255.39, 
1.373}, {1496.61, 1.409}, {1942.79, 1.538}}

This data should be ploted using LogLinearPlot[data] which yields:

enter image description here

There's a integral equation that should fit this data:

$ m(t) = \int N V M_{eq} (1-e^{-t/\tau})f(V) dV$

where $N$ is the total number of particle, $t$ is the elapsed time,$V$ the volume of the particles, $M_{eq}$ the magnetization at the equilibrium state,$\tau$ is a parameter that dependes on $V$ and $f(V)$ is the distribution of volume of the particles.

I'd like to know:

i)if it's possible (and if so, how to do it) to fit this data using this integral equation with, for instance, a log gaussian distribution $f(V)$.

ii) Given this equation, is there a way to create a routine which calculates the best-fitting of the data and, as an output, gives the $f(V)$ distribution which provides this best fitting?

Thanks in advance

|improve this question
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  • $\begingroup$ Statistics begins from 30 (google.com.ua/…). The size of your data is too small for reliable conclusions. $\endgroup$ – user64494 May 2 at 18:17
  • $\begingroup$ Hi, user64494. Thanks for your feedback. So let's suppose that I have a log-normal distribution $f(V)$. Is there a way to numerically solve the equation using the given distribution and plot this solution as a function of time $t$? $\endgroup$ – José Augusto Devienne May 2 at 18:36
  • $\begingroup$ From your previous question: Please don't refer to these curves with the word "distribution" as there is no probabilistic interpretation of the curves. You are performing a regression with curves having a similar form as a probability distribution. $\endgroup$ – JimB May 2 at 18:45

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