Attempting to derive a formula for logistic map (part 1)

The logistic map comes from concatenations of the function \( f(x) = \lambda x(1-x)\), where \(x=\frac12\) and\(\lambda\) is an argument within the domain \([0,4]\). We'll begin by defining a function \(f_n (x) \) as \(n\) instances of \(f(x)\) concatenated together, observing the fractional form of each incrementation of \(n\): \[ f_1 \left(\frac12\right) = \lambda \frac12 \left(1-\frac12\right) = \frac{\lambda}4\] \[ f_2 \left(\frac12\right) = \lambda \frac{\lambda}4 \left(1-\frac{\lambda}4\right) = \frac{\lambda^2 (4-\lambda)}{16} \] \[ f_3 \left(\frac12\right) = \lambda \frac{\lambda^2 (4-\lambda)}{16} \left(1-\frac{\lambda^2 (4-\lambda)}{16} \right) = \frac{\lambda^3(4-\lambda)(16-\lambda^2(4-\lambda))}{256}\] The denominator of each iteration is the square of the previous denominator. Using the fact that \(4\) is \(2^{2^1}\), we can rewrite \[f_n \left(\frac12\right) = \frac{f_n^{'}\left(\frac12\right)}{2^{2^n}}\] (here \(f_n^{'}(x)\) just denotes the remaining sections of \(f_n (x)\).)
Note that the leading power of \(\lambda\) in each term cannot be taken out in this state, due to it being concentated into \(f_{n+1}\). However, this option may be feasible within sums of \(\lambda^{n-k}\). To derive bounds for each power, we can: \[ f_1^{'} \left(\frac12\right) = \lambda\] \[ f_2^{'} \left(\frac12\right) = \lambda^2 (4-\lambda) = 4\lambda^2-\lambda^3\] \[ f_3^{'} \left(\frac12\right) =\lambda(4\lambda^2-\lambda^3)(16-4\lambda^2+\lambda^3)= 64\lambda^3-16\lambda^5+4\lambda^6-16\lambda^4+4\lambda^6-\lambda^7\]\[= 64\lambda^3-16\lambda^4-16\lambda^5+8\lambda^6-\lambda^7\] The lowest order \(k\) of \(\lambda^k\) is \(n\), as a constant times \(\lambda\) times \(\lambda^{n-1}\), and the highest order is \(2^n-1\), as \(\left(\lambda^{2^{n-1}-1}\right)^2 \) times \(\lambda\) produces \(\lambda^{2(2^{n-1}-1)+1}=\lambda^{2^n-1}\). Given the coefficients \(c_{n,k}\), we can express \(f_n^{'}(x)\) as \[\sum_{k=0}^{2^n-n-1} c_{n,k} x^{k+n}\] More iterations of \(n\) will be needed to figure out a general formula for these coefficients.

Note by William Crabbe
1 week, 5 days ago

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