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
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