Not too Complicated

Algebra Level 4

Let f ( x ) = x − 1 x + 1 f(x) = \frac{x-1}{x+1} and f n ( x ) f^{n}(x) denote the n − fold n-\text{fold} composition of f f with itself. That is, f 1 ( x ) = f ( x ) f^1(x) =f(x) and f n ( x ) = f ( f n − 1 ( x ) ) f^n(x) = f(f^{n-1}(x)) . Find f 2007 ( 2 ) f^{2007}(2) .


The answer is -3.

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Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Note that f 2 ( x ) = − 1 x f^2(x) = -\frac{1}{x} f 3 ( x ) = − x + 1 x − 1 f^3(x) = -\frac{x+1}{x-1} f 4 ( x ) = x f^4(x) = x f 5 ( x ) = f ( x ) f^5(x) = f(x) Since the remainder r r when 2007 2007 is divided by 4 4 is 3 3 ( 2004 = 4 ∗ 501 + 3 ) (2004 = 4*501+3) , it follows f 2007 ( x ) = f 3 ( x ) = − x + 1 x − 1 f^{2007}(x) = f^3(x) = -\frac{x+1}{x-1} .

Therefore, f 2007 ( 2 ) = − 3 f^{2007}(2) = -3

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Please don't post solutions, spare us the trouble of posting solutions. 😀

Hem Shailabh Sahu - 6 years, 2 months ago

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