First what if we go backwards?
Using my function f(x) = sqrt((x+2)+(x-2)f(x+1))
f(0) = sqrt(2+-2*f(1)) = sqrt(2+-2) = 0
f(-1) = sqrt(1+-3*f(0)) = sqrt(1+0) = 1
which is promising but
f(-2) = sqrt(0+-4*f(-1)) = sqrt(-4) = 2i
which is less so
f(-3) =sqrt(-1+-5*f(-2)) = sqrt(-1-10i) = 2.12519-2.35052i
and they seem to continue as irrational complex numbers without too much pattern.
Second the recursive proof that f(x)=x for x a positive integer:
We now know f(3)=3
assume f(x)=x to prove f(x+1)=x+1
f(x+1) = (f(x)^2 - (x+2)) / (x-2)
= ((x)^2 - x - 2) / (x-2)
=(x+1)(x-2) / (x-2)
=x+1
[I was very close to this in my other post but it requires x>2 which we didn't have at the time.]
Finally to show this formula works for any real number > 0
We prove f(x+a) = x+a where x is a positive integer and a is a positive real.
x+a = f(x)+a
so we have f(x+a)=f(x)+a and we are done.
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Posted by Jer
on 2010-02-06 15:20:25 |
A couple of things
