Question

Let `f(t) = [t+ sqrt(3)]/[1 - tsqrt(3)]`. Prove that `f` is a cyclic function?

Answer 1

See below.

Explanation:

A function is cyclic if it satisty the conditon:

`f^n(x)=x`

`n` is the order (not the nth derivative) that is the number of times the function is applied to itself before the cycle is complete.

`f(t)=(t+sqrt(3))/(1-tsqrt(3))`
`f^1(t)=((t+sqrt(3))/(1-tsqrt(3)) +sqrt(3))/(1-(t+sqrt(3))/(1-tsqrt(3)) sqrt(3))`
`=(t+sqrt(3)+sqrt(3)-3t)/(1-tsqrt(3)-tsqrt(3)-3)`
`=(sqrt(3)-t) /(-1-tsqrt(3))`

`f^2(t)=(sqrt(3)-(t+sqrt(3))/(1-tsqrt(3)))/(-1-(t+sqrt(3))/(1-tsqrt(3))sqrt(3))`
`=(sqrt(3)-3t-t-sqrt(3))/(-1+tsqrt(3)-tsqrt(3)-3)=(-4t)/(-4)=t`

Thus it is proven that the function is cyclic.

QED

An example of a cycle:
`f(1)=(1+sqrt(3))/(1-(1)sqrt(3))=-3.732`

`f(-3.732)=-0.268`

`f(-0.268)=1`

Notice how it circles back to the first value.