Using fixed point iteration find the positive root of $f(x)=e^{-x}-x^2$

Question

Consider $f(x)=e^{-x}-x^2$.

I'm suppose to find the positive root using fixed point iteration.

after drawing the graph, it's safe to set the interval from [0.25,1]. (I actually want to set it from [0.5,1] but i'm unsure as i can't zoom into the graph).

The problem i'm having is deriving the function $g$.

I got $g(x)=-2\ln(x)$.

I can't seem to prove the uniqueness of this root as $g'(x)=\frac{-2}{x}$ but i can bound $|g'(x)|<K$ such that K<1 under this interval. Any help is appreciated.

I've tried to think of other $g(x)$ but none of them seem convergent to me

Answer 1

I think $g(x)=e^{-x/2}$ works.

Answer 2

Hint- Try $x=e^{-x/2}=g(x)$ such that $|g'(x)|<1$ for $x\in(0,1)$

Answer 3

Since you've proved $|g'(x)| < K$ with $ K < 1 $ You can use the mean value theorem that states that for all $x,y$ in $[0.25,1]$ $|g(x)-g(y)|\le K|x-y]$

Let $l_1$ and $l_2$ be fixed points of g in $[0.25,1]$ $$ |g(l_1)-g(l_2)| \le K|l_1-l_2| $$ Since $g(l_1) = l_1$ and $g(l_2) = l_2$ $$ |l_1-l_2| \le K|l_1-l_2|$$ $$ (|l_1-l_2|)(1-K) \le 0 $$ And since $K \lt 1$ then $1-K \gt 0$ which leads us to $|l_1 - l_2| \le 0$ however we know $|l_1 - l_2| \ge 0$ We can conclude that $ l_1 = l_2 $ Which proves the uniqueness of the fixed point of $g$ and thus the uniqueness of the of $f$

And as others suggested, i think $g \mapsto e^{-x/2}$ works better