Let $f: \mathbb R\to\mathbb R$, $f(x) = \frac{x}{1-x}$. Define $f^{2}(x) = f(f(x))$, $f^{3}(x) = f(f(f(x)))$, ...
Guess the form for $f^{n}(x)$ and prove your answer is correct using induction on n.
My Approach:
$f^{n}(x)$ = $\frac{x}{1 - nx}$ since $f^{2}(x)$ = $\frac{x}{1-x}$ $/$ $1$ - $\frac{x} {1-x}$
= $\frac{x}{1 - 2x}$ = $\frac{x}{1 - nx}$ (Since 2 = n so it must be n * x)
Proof By Induction:
Denote $f^{n}(x)$ = $\frac{x}{1 - nx}$
Base Case:
When $n$ = $1$
$f^{1}(x)$ = $\frac{x}{1-(1)*(x)}$ = $\frac{x}{1 - x}$ = $f(x)$
which is true since $f^{1}(x)$ = $f(x)$
I.H:
$f^{k}(x)$ = $\frac{x}{1 - kx}$ where $n$ = $k$ for any arbitrary $k > 0$ this holds true because of the base case
I.S:
$f^{k+1}(x)$ = $f(f^{k}(x))$
= $f$($\frac{x}{1 - kx}$$)$
= $f$($\frac{x}{(1 - x)k+1}$$)$
= $f$($\frac{1}{(1)(k+1)}$$)$ (Canceling $x$ out)
= $f$($\frac{k+1}{1}$$)$ (Reciprocal of $\frac{1}{k + 1}$ is $\frac{k + 1}{1}$)
= $f$($k + 1$)
Any help or hints would be appreciated
First let's compute $f^2(x)$ to guess the form of the general expression:
$$f^{\color{red}2}(x)=\frac{\frac x{1-x}}{1-\frac x{1-x}}=\frac x{1-\color{red}2x}$$ Now can you prove by induction that
$$f^{\color{red}n}(x)=\frac x{1-\color{red}nx}$$