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# Recursion

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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 $$|f^7(2)|$$

Dec 23, 2021

#1
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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

=^._.^=

Dec 23, 2021
#2
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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. :)

MathProblemSolver101  Dec 23, 2021