# Given domain N and range N of function f(n) with properties f(f(n)=4n-3) and f(2^k)=2^(k+1)-1, k natual number, what is f(993)?

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### 2 Answers

The problem provides two properties of the given function, hence, considering the first property `f(f(n)) = 4n - 3` , yields:

`f(f(n)) = 4n - 3 => f(f(f(n))) = 4f(n) - 3`

Since `f(f(n)) = 4n - 3` , hence, replacing `4n - 3` for `f(f(n))` yields:

`f(f(f(n))) = 4f(n) - 3 => f(4n - 3)=4f(n) - 3`

Using the pattern `f(4n - 3)=4f(n) - 3` , you should notice that `993 = 4*249 - 3` , hence, you may evaluate `f(993)` , such that:

`f(993) = f(4*249 - 3) = 4*f(249) - 3`

Using the same pattern, you may evaluate `f(249)` such that:

`f(249) = f(4*64 - 3) = 4*f(64) - 3`

Since `64 = 2^6` , you may use the next property of the function, provided by the problem, `f(2^k)=2^(k+1)-1` , such that:

`f(2^6)=2^(6+1)-1 = 2^7 - 1= 127`

`f(993) = 4*(4*f(249) - 3) - 3 => f(993) = 4*(4*(4*f(64) - 3) - 3) - 3`

Replacing `127` for `f(64)` yields:

`f(993) = 4*(4*(4*127 - 3) - 3) - 3 = 8065`

**Hence, evaluating `f(993)` using the properties of the function provided by the problem, yields **`f(993) = 8065.`

It is not specify that f(n-3)=f(n)-3 ,so it is inappropriate to solve problem with extra assumption.

We have property

f(f(n))=4n-3

Let f is invertible function

So

`f(n)=f^(-1)(4n-3)` (i)

Also

`f(2^k)=2^(k+1)-1` , k is natural number.

Thus by definition of inverse

`f^(-1)(f(2^k))=2^k`

`=> f^(-1)(x)=(x+3)/2 , AA x inR`

since natural number is smallest subset of real number .

`f(x)=4((x+3)/2)-3`

Thus

`f(993)=4((993+3)/2)-3`

`=1992-3`

`=1989`