Wow. Experimentally, it's also x+1 for -1≤x≤0, -x-1 for -2≤x≤-1, and √(1-2x-x²) for -3≤x≤-2, wherein it goes imaginary. gosper.org/puzfn.png It can probably be guessed anywhere. Where has it been proven? Note the tiny flat spot in the imaginary part near x = -3.5. --rwg On 2018-05-12 03:54, James Propp wrote:
Could f be a polynomial? Assuming deg f is at least 1, Joerg’s equation implies 2 deg f = 1 + def f, so we’ll need def f = 1. Putting f(x) = ax+b in Joerg’s equation, we get (ax+b)^2 - 1 = x (ax+a+b), or
(a^2) x^2 + (2ab) x + (b^2 - 1) = (a) x^2 + (a+b) x,
with unique solution a=b=1.
So I’m guessing f(x) = x+1.
Jim Propp
On Saturday, May 12, 2018, Joerg Arndt <arndt@jjj.de> wrote:
f(x)^2 - 1 = x * f(x+1) No idea how to proceed from there, though.
Best regards, jj
* Dan Asimov <dasimov@earthlink.net> [May 05. 2018 08:20]:
Define a function f : [0,oo) —> R via
f(x) = sqrt(1 + x*sqrt(1 + (x+1)*sqrt(1 + (x+2)*sqrt(1+...
Evaluate f(x) in closed form.
—Dan