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

The logistic map comes from concatenations of the function f(x)=λx(1x) f(x) = \lambda x(1-x), where x=12x=\frac12 andλ\lambda is an argument within the domain [0,4][0,4]. We'll begin by defining a function fn(x)f_n (x) as nn instances of f(x)f(x) concatenated together, observing the fractional form of each incrementation of nn: f1(12)=λ12(112)=λ4 f_1 \left(\frac12\right) = \lambda \frac12 \left(1-\frac12\right) = \frac{\lambda}4 f2(12)=λλ4(1λ4)=λ2(4λ)16 f_2 \left(\frac12\right) = \lambda \frac{\lambda}4 \left(1-\frac{\lambda}4\right) = \frac{\lambda^2 (4-\lambda)}{16} f3(12)=λλ2(4λ)16(1λ2(4λ)16)=λ3(4λ)(16λ2(4λ))256 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 44 is 2212^{2^1}, we can rewrite fn(12)=fn(12)22nf_n \left(\frac12\right) = \frac{f_n^{'}\left(\frac12\right)}{2^{2^n}} (here fn(x)f_n^{'}(x) just denotes the remaining sections of fn(x)f_n (x).)
Note that the leading power of λ\lambda in each term cannot be taken out in this state, due to it being concentated into fn+1f_{n+1}. However, this option may be feasible within sums of λnk\lambda^{n-k}. To derive bounds for each power, we can: f1(12)=λ f_1^{'} \left(\frac12\right) = \lambda f2(12)=λ2(4λ)=4λ2λ3 f_2^{'} \left(\frac12\right) = \lambda^2 (4-\lambda) = 4\lambda^2-\lambda^3 f3(12)=λ(4λ2λ3)(164λ2+λ3)=64λ316λ5+4λ616λ4+4λ6λ7 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=64λ316λ416λ5+8λ6λ7= 64\lambda^3-16\lambda^4-16\lambda^5+8\lambda^6-\lambda^7 The lowest order kk of λk\lambda^k is nn, as a constant times λ\lambda times λn1\lambda^{n-1}, and the highest order is 2n12^n-1, as (λ2n11)2\left(\lambda^{2^{n-1}-1}\right)^2 times λ\lambda produces λ2(2n11)+1=λ2n1\lambda^{2(2^{n-1}-1)+1}=\lambda^{2^n-1}. Given the coefficients cn,kc_{n,k}, we can express fn(x)f_n^{'}(x) as k=02nn1cn,kxk+n\sum_{k=0}^{2^n-n-1} c_{n,k} x^{k+n} More iterations of nn will be needed to figure out a general formula for these coefficients.

Note by William Crabbe
1 year, 4 months ago

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