# domain function N natural numbers range Q without -1, 0 f(0)=7/9 f(n+2)=f(n)-1/f(n)+1calculate f(n+4),f(n+8) and show that f is periodic function

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You need to evaluate `f(n+4) ` using the given reccurence relation such that:

`f(n+4) = (f(n+2) - 1)/(f(n+2) + 1)`

Using the reccurence relation, you need to substitute `(f(n)-1)/(f(n)+1)` for `f(n+2) ` such that:

`f(n+4) = ((f(n)-1)/(f(n)+1) - 1)/((f(n)-1)/(f(n)+1) + 1)`

`f(n+4) = (f(n) - 1- f(n) - 1)/(f(n) - 1 + f(n) + 1)`

`f(n+4) = -2/(2f(n)) => f(n+4) = -1/f(n)`

You need to evaluate `f(n+8)` such that:

`f(n+8) = (f(n+6) - 1)/(f(n+6) + 1)`

`f(n+6) = (f(n+4)-1)/(f(n+4)+1) `

Substituting -`1/f(n)` for `f(n+4)` yields:

`f(n+6) = (-1/f(n) - 1)/(-1/f(n) + 1)`

`f(n+6) = (-1 - f(n))/(-1 + f(n))`

`f(n+8) = ((-1 - f(n))/(-1 + f(n)) - 1)/((-1 - f(n))/(-1 + f(n)) + 1)`

`f(n+8) = (-1 - f(n) + 1 - f(n))/(-1 - f(n) - 1 + f(n))`

`f(n+8) = (-2f(n))/-2 = f(n)`

**Notice that `f(n+4) = -1/f(n)` and `f(n+8) = f(n)` , hence, f(n) is a periodic function and evaluating its period yields `p = 8.` **