Help me find a fixed point.

Question

I want to find a fixed point of $1/(x+1)$.

I set:

$p = 1/(p+1)$

$p^2+p-1=0$

$p= (-1\pm\sqrt(5))/2$

But I plug in

$f((-1 + \sqrt(5))/2)$ and $(-1 - \sqrt(5))/2$ but I don't get the same output to make it a fixed point. Please help me.

Answer 1

Let $p = \frac{-1\pm\sqrt{5}}{2}$, then $$p = \frac{1}{p+1}=\frac{1}{\frac{-1\pm\sqrt{5}}{2}+1}=\frac{1}{\frac{-1\pm\sqrt{5}}{2}+\frac{2}{2}}=\frac{1}{\frac{1\pm\sqrt{5}}{2}}=\frac{2}{1\pm\sqrt{5}}$$ $$=\frac{2(1\mp\sqrt{5})}{(1\pm\sqrt{5})(1\mp\sqrt{5})}=\frac{2(1\mp\sqrt{5})}{1\pm\sqrt{5}\mp\sqrt{5}-5}=\frac{2(1\mp\sqrt{5})}{1\pm\sqrt{5}\mp\sqrt{5}-5}=-\frac{1\mp\sqrt{5}}{2}=\frac{-1\pm\sqrt{5}}{2}$$