This is my very first encounter with functional equation of this kind, and methods of series solution and differential equations are of not much help. The solution to this problem, or at least the mathematical prerequisites to understand solution to this problem is asked.

This function is asymptotically zero at both +_ infinity, positive otherwise, and has a peak near zero.

If I am correct, then this is the frequency distribution of numbers fed to the sequence xn+1=a ln (xn2) as long as the sequence generated is chaotic and sufficiently large in number (value of a is usually limited to 0.2 to 1.3, Positive and negative signs of a are essentially immaterial except for the sequence is negated after first term) .The sequence is initiated or seeded with number roughly as the same order of magnitude of 1 in both positive and negative side, except for the values that eventually lead to zero or infinity in the sequence. The sequence is allowed to proceed, and frequency distribution of numbers in the sequence are noted, from which a continuous probability distribution may be numerically guessed but not analytically found. The expression to find out formula of the continuous probability distribution comes to me from the following reasoning

  • Suppose, the probability distribution is given by y=f(x). Now, if I consider a "dx" (infinitesimally) thin strip around x, then I come up with f(x) dx fraction of all points required to construct the probability distribution. When this fraction of all points are passed through yet another sequence of transformation through the recurrence xn+1 =a ln (xn)2 , the fraction of points involved must be unchanged. That is, when x is substituted with a ln x2 , the infinitesimal strip area, which changes to f( a ln x^2) d (a ln x^2), must be numerically equal to f(x) dx, thus the functional equation is postulated
  • I am not entirely sure about this reasoning, and experts are welcome to point out my fault of reasoning and show the correct equation , if I am mistaken.

Please see my related question for further details.

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