Re: " How do you solve f(3x + 2f(x) + 1) = f^[2](x -1) "

alainverghote@xxxxxxxx nous a récemment amicalement signifié :
Well, it would certainly be true if
3 x + 2 f(x) + 1 = f(x-1).
That linear recursion has general solution
f(x) = 2 - 3 x + a(x) 2^(-x) where a(x) is any
function that is periodic with period 1.

Another family of solutions is f(x) = constant.

It does not seem your general solution
f(x) =2 -3x + a(x)*2^(- x)
works in the given equation
f(3x + 2f(x) +1 ) = f(x -1) , order f^[2]


No, Robert's solution doesn't work with f(3x + 2f(x) +1 ) = f(x -1)
But it works with 3 x + 2 f(x) + 1 = f(x-1).
So it works with f(3x + 2f(x) +1 ) = f(f(x -1))
Which was your given equation


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