Let be $f(x)=\frac{2x}{1+x} $ function and $ x_0 > 0 $. With the help of this, form the $x_{n+1}=f(x_n)$ sequence. Is $x_n$ convergent and if yes what is the limit?
Thank you very much in advance!
Let be $f(x)=\frac{2x}{1+x} $ function and $ x_0 > 0 $. With the help of this, form the $x_{n+1}=f(x_n)$ sequence. Is $x_n$ convergent and if yes what is the limit?
Thank you very much in advance!
This is a different method that you will be able to use in cases where it is not possible to obtain a closed formula for $x_n$.
The function $f$ is increasing in $[0,\infty)$. It cuts the line $y=x$ (the diagonal of he first quadrant) at two points: $(0,0)$ and $(1,1)$. If $0
Assume $0
If $x_0=1$ then $x_n=1$ for all $n$. I leave to you the case $x_0>1$.
It's not difficult to prove by induction that $ x_{n+k} = \frac{2^{k}x_n}{1 + (2^k -1)x_n} $ for any $ k \geq 0 $ and any $ n $.
So we have that $ x_k = \frac{2^k x_0}{1 + (2^k - 1)x_0} $ for $ k \geq 0 $
Does this help?