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

1 Answer
Altrigeos
Jun 4, 2017

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.

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