Fitting Functions

Algebra Level 2

Let f f be a function defined on integers such that f ( 0 ) = 0 f\left( 0 \right) =0 , f ( 2 n + 1 ) = f ( n ) − 1 f\left( 2n+1 \right) =f\left( n \right) -1 and f ( 2 n ) = − 2 f ( n ) f\left( 2n \right) =-2f\left( n \right) . Find the value of f ( 10 ) f\left( 10 \right) .


The answer is -2.

Lee Care Gene by Lee Care Gene , 20, Malaysia

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1 solution

Ralph James
Jun 13, 2016

f ( 2 n + 1 ) = f ( n ) − 1 ⟹ f ( 1 ) = 0 − 1 = − 1 f(2n+1) = f(n)-1 \implies f(1)=0-1=-1 .

⟹ f ( 2 n ) = − 2 f ( n ) \implies f(2n)=-2f(n) , we have f ( 2 ⋅ 1 ) = − 2 ⋅ − 1 = 2 f(2 \cdot 1) = -2 \cdot-1 = 2 .

⟹ f ( 2 ) = 2 → f ( 5 ) = f ( 2 ⋅ 2 + 1 ) = f ( 2 ) − 1 = 1 \implies f(2)=2 \rightarrow f(5) = f(2 \cdot 2 + 1) = f(2)-1 = 1 .

⟹ f ( 10 ) = f ( 2 ⋅ 5 ) = − 2 ⋅ f ( 5 ) = − 2 ⋅ 1 = − 2 \implies f(10) = f(2 \cdot 5) = -2 \cdot f(5) = -2 \cdot 1 = \boxed{-2} .

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