Fixed Point $f : \mathbb{R} \setminus \{ \tfrac{\pi}{2} +k\pi : k \in \mathbb{Z} \} \to \mathbb{R} : x \mapsto \sqrt{x^{2}+1}-\tan{x} $

Question

I have to show that $f : \mathbb{R} \setminus \{ \tfrac{\pi}{2} +k\pi : k \in \mathbb{Z} \} \to \mathbb{R} : x \mapsto \sqrt{x^{2}+1}-\tan{x} $ has a fixed point in the interval $[0,1]$.

Meaning $\exists \xi \in [0,1]: f(\xi) = \xi$

I know how to show the fixed-point-theorem for any continuous function but i don't know how to show:
$\exists\xi \in [0,1] : \varphi (\xi):= f(\xi)-\xi =0 $

score 2 · Accepted Answer · answered Nov 19 '22 at 11:08

Okay so we set

$$\varphi(x)=\sqrt{x^2+1}-\tan x-x.$$

The goal is to show that $\varphi$ has a zero in $[0,1]$. Notice first that

$$\varphi(0)=1.$$

Secondly, notice that

$$\varphi\left(\frac{\pi}{4}\right)=\sqrt{\frac{\pi^2}{16}+1}-1-\frac{\pi}{4}=\frac{\sqrt{\pi^2+16}-\pi-4}{4}<0.$$

Since $\varphi(0)>0$ and $\varphi\left(\frac{\pi}{4}\right)<0$, it follows by the intermediate value theorem that $\varphi(\xi)=0$ for some $\xi\in\left(0,\frac{\pi}{4}\right)\subseteq[0,1]$.

Fixed Point $f : \mathbb{R} \setminus \{ \tfrac{\pi}{2} +k\pi : k \in \mathbb{Z} \} \to \mathbb{R} : x \mapsto \sqrt{x^{2}+1}-\tan{x} $

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