We have a one-variable equation $\rho(R)$ where

ρ = (14656.4+277.526*R^2)/(45.9225+R^2)^{5/2} + 0.370036/(R*(0.25+R)^3)

This equations describes the evolution of a density distribution, and as the following plot shows, the value of $\rho$ decreases rapidly with increasing $R$.

enter image description here

Usually density distributions drop following a law $R^{-k}$. So, is it possible to use Mathematica in order to determine the falling law at large radii $R$ in this case? In other words, to determine the values of $k$.


Following @Vitaliy Kaurov solution I got the desired value of $k$. At the following plot, the red line corresponds to $277.526/R^3$ and as we can see it fully coincides with $\rho(R)$.

enter image description here

  • $\begingroup$ One approach is to approximate your function for large R (say 10<R<50) by a straight line. The slope of this line will be k. Or do you have something else in mind? $\endgroup$ – bill s Aug 23 '13 at 15:50
  • $\begingroup$ @bills Let's try this approach! How could we compute the slope of this line? Maybe posting a detailed answer rather than a comment. $\endgroup$ – Vaggelis_Z Aug 23 '13 at 16:14

==== Asymptotic behavior ====

Your function is:

\[Rho][R_] := 0.370036`/(R (0.25` + R)^3) + 
 (14656.4` + 277.526` R^2)/(45.9225` + R^2)^(5/2)

Wouldn't the leading term be answer to your question as k = 3?

Series[\[Rho][R], {R, Infinity, 5}]

enter image description here

This is obvious from the equation, but you can also see correction terms from the next orders of expansion.

==== Finite range behavior ====

If you are curious in a finite range - say from 20 to 50 - what would be the best law to describe the behavior? Then you could to a simple trick of fitting:

data = Table[{R, \[Rho][R]}, {R, 20, 50}];
model = a /x^k;
fit = FindFit[data, model, {a, k}, x]

{a -> 145.639, k -> 2.83169}

And to see how well this law fits:

modelf = Function[{t}, Evaluate[model /. fit]];
Plot[modelf[x], {x, 20, 50}, Epilog -> {Red, Point[data]}]

enter image description here

  • $\begingroup$ Yes, it's working fine; see my EDIT. BTW, how did you enclose the first term with red dashed line? $\endgroup$ – Vaggelis_Z Aug 23 '13 at 17:15
  • $\begingroup$ I added some other approach. For dashed line... Select cell of expression and go Top Menu >> Cell >> Convert to >> Bitmap. Then press CTRL-t for drawing tools. $\endgroup$ – Vitaliy Kaurov Aug 23 '13 at 17:26
  • $\begingroup$ ctrl-t ==> ctrl-D ?? $\endgroup$ – Sjoerd C. de Vries Aug 24 '13 at 17:23
  • $\begingroup$ @SjoerdC.deVries I am on Mac ;) And I think yes on Win it is CTRL-D - thanks for catching this. $\endgroup$ – Vitaliy Kaurov Aug 24 '13 at 17:57

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