Problems finding fixed point.

Question

$x_n=f(x_{n-1})={ax_{n-1} +e^{-x_{n-1}}\over 1+a}$ for $a>0$

Setting $f(x)=x$. The problem is while trying to find the fixed point of $x={ax +e^{-x}\over 1+a}$ I only get $x=e^{-x}$. What's the solution of this?

Answer 1

In case what you're trying to do is to prove the existence of a fixed point, you might want to try the following:

Consider the function $$g(x)=x-e^{-x}.$$

This function is easily shown to be continuous. Moreover, when $x$ tends to $\infty$, $e^{-x}$ tends to $0$, so in particular $g(x)$ obtains positive values at some $x\gg 0$. Likewise, it isn't very hard to show that $$\lim_{x\to-\infty}x-e^{-x}=\lim_{x\to\infty}-x-e^x=-\infty.$$ Which means that $g(x)$ is negative for $x\ll 0$. Consequently (by what theorem?), there is some point where $g(x)=0$, and this will be your fixed point.