(In reply to
Is my reasoning valid? by Jer)
I think you’re right Jer; for integers >=0 your inspired function, f(x), is just x in disguise.
To get round the singularity that Steve mentions, and find f(3), I believe we can use the first derivative as follows:
(x  2) f(x + 1) = f(x)^{2}  x  2 (1) This can be differentiated to give
(x  2) f’(x + 1) + f(x + 1) = 2 f(x) f’(x)  1
Using x = 0, 1 and 2, this equation gives
2 f’(1) + f(1) = 2 f(0) f’(0)  1 (2)
f’(2) +f(2) = 2 f(1) f’(1)  1 (3)
f(3) = 2 f(2) f’(2)  1 (4)
We know that f(0)=0, f(1)=1 and f(2)=2 by direct substitution into your formula, so
(2) gives f’(1) = 1
(3) then gives f’(2) = 1
(4) then gives f(3) = 3
x=3 in (1) then gives f(4) = 3^{2} 3  2 = 4, which solves the problem.
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I think this may pave the way to an inductive proof that f(x) = x for all positive integers. I’m not sure what happens between the integers, but my plotting software suggests something very interesting... (not had time to explore yet).

Posted by Harry
on 20100205 08:57:46 