+876 There is a function that is defined the following
\(\begin{align*} f^1(x) &= (x-1)^2 \\ f^n(x) &=f^1(f^{n-1}(x)) \end{align*}\)
What is \(|f^7(2)|\)
+2407 f^1(1) = (2 - 1)^2 = 1
f^2(2) = f^1(f^1(2)) = f^1(1) = 0
f^3(2) = f^1(f^2(2)) = f^1(0) = 1
f^4(2) = f^1(f^3(2)) = f^1(1) = 0
There's a pattern. :))
It seems like when f(a) = 0, f(a+1) = 1 and when f(a) = 1, f(a+1) = 0.
So when n is odd, then the answer is 1.
If n is even, the answer is 0.
f^7(2) = 1
=^._.^=
+876 There is a reason for the pattern - can you see why? Do you see anything intesting if we generalize \(x\) for not just 2, but for all integral \(x\)? Try finding a general formula. Also notice - does your 1, 0, etc. pattern have anything to do with \(x=2\)? Maybe there is, and maybe not. Try to find out. :)
