On Mar 28, 2017, at 9:29 AM, Henry Baker <hbaker1@pipeline.com> wrote:
If f(2*x)=2*f(x)+b*f(x)^3, then we can substitute x=2^(y-1):
f(2*2^(y-1))=2*f(2^(y-1))+b*f(2^(y-1))^3, so
f(2^y) = 2*f(2^(y-1)) + b*f(2^(y-1))^3, so if g(y)=f(2^y) then
g(y) = 2*g(y-1) + b*g(y-1)^3
So how do we solve this non-linear recurrence relation for g(y)?
Failed attempt: --------------- Set g(y) = Sum_{n=0...oo} c_n y^n. The last equation above gives (*) Sum_{n=0...oo} c_n y^n = 2 * Sum_{n=0...oo} c_n (y-1)^n + b * Sum_{n=0...oo} c_n (y-1)^3n or setting z = y-1: Sum_{n=0...oo} c_n (z+1)^n = 2 * Sum_{n=0...oo} c_n z^n + b * Sum_{n=0...oo} c_n z^3n Now setting constants equal gives: Sum_{n=0...oo} c_n = 2 c_0 + b c_0 or c_0 = (1/(b+1)) Sum_{n=1...oo} c_n assuming b <> -1 (not that helpful!) Now setting z^1 powers equal gives: (Sum_{n=0...oo} n c_n) z = 2 c_1 z or c_1 = Sum_{n=2...oo} n c_n Hmm, this doesn't look easily solvable for the coefficients. —Dan