Compositional Functions

[r.jereen]

Given: f_o(x)= 1/(1-x) & f_n+1= f_o o f_n

Find: f₁₀₀(3) =

I know the answer is 2/3 , i just need to find out how to reach the answer , i dont really know how to start this problem.

 

[pka]

Given: f_o(x)= 1/(1-x) & f_n+1= f_o o f_n
Find: f₁₀₀(3) =
I know the answer is 2/3 , i just need to find out how to reach the answer , i dont really know how to start this problem.

\(\displaystyle \[\begin{gathered}
f_0 (3) = \frac{1}
{{1 - 3}} = - \frac{1}
{2} \hfill \\
f_1 \left( 3 \right) = \frac{1}
{{1 + \tfrac{1}
{2}}} = \frac{2}
{3} \hfill \\
f_2 (3) = \frac{1}
{{1 - \tfrac{2}
{3}}} = 3 \hfill \\
\end{gathered} \)

Now you find \(\displaystyle f_3(3),~f_4(3),~\&~f_5(3)\).
You should see a pattern.

 

[soroban]

Hello, r.jereen!

A slightly different approach . . .

\(\displaystyle \text{Given: }\:f_0(x) \,=\,\dfrac{1}{1-x}\,\text{ and }\,f_{n+1} \:=\:f_0 \circ f_n \)

\(\displaystyle \text{Find }\,f_{100}(3).\)

\(\displaystyle f_0 \;=\;\dfrac{1}{1-x}\)

\(\displaystyle f_1 \;=\;f_0(f_1) \;=\;f_0\left(\frac{1}{1-x}\right) \;=\;\dfrac{1}{1-\frac{1}{1-x}} \;=\;\dfrac{x-1}{x} \)

\(\displaystyle f_2 \;=\;f_0(f_2) \;=\;f_0\left(\frac{x-1}{x}\right) \;=\;\dfrac{1}{1 - \frac{x-1}{x}} \;=\;x\)

\(\displaystyle f_3 \;=\;f_0(f_2) \;=\;f_0(x) \;=\;\dfrac{1}{1-x}\)


We see that the sequence repeats in a 3-stage cycle.

\(\displaystyle \text{We can write: }\:f_n \;=\;\begin{Bmatrix} \dfrac{1}{1-x} && \text{for }n \equiv 0\text{ (mod 3)} \\ \\ \dfrac{x-1}{x} && \text{for }n \equiv 1\text{ (mod 3)} \\ \\ x && \text{for }n \equiv 2\text{ (mod 3)} \end{array}\)


\(\displaystyle \text{Therefore: }\:f_{100}(3) \:\equiv\:f_1(3) \;=\;\dfrac{3-1}{3} \;=\;\dfrac{2}{3}\)