Let $x$ be a real number such that $x\ge{0}$, then
$$1=\sqrt{\frac{\sqrt{\frac{\sqrt{\frac{\sqrt{\frac{x+1}{2}}+1}{2}}+1}{2}}+1}{2}}+...$$
At least I haven't seen it on the internet.
Questions: a) Is this known?, b) Does this formula generalize any well-known nested radical?
Consider the function $f:[-b/a,\infty)$ with $a\gt0,$ $b\in\mathbb{R}$ such that $f(x)=\sqrt{ax+b}.$ We can define the sequence family $s[f]$ such that $s[f]_{n+1}=f(s[f]_n).$ Define $f^n$ such that $f^{n+1}=f\circ{f^n}.$ Then $s[f]_n=f^n(s[f]_0).$ The nested radical in question is then equal to $\lim_{n\to\infty}s[f]_n.$ The fixed points of $f$ are given by the solutions of the equation $x=\sqrt{ax+b}.$ $x=\sqrt{ax+b}$ is equivalent to $ax=a\sqrt{ax+b}=(ax+b)-b=\sqrt{ax+b}^2-b=a\sqrt{ax+b},$ which, with the caveat that $\left(\frac{a}2\right)^2+b\gt0,$ is equivalent to $\sqrt{ax+b}^2-a\sqrt{ax+b}-b=0=\left(\sqrt{ax+b}-\frac{a}2\right)^2-\left[\left(\frac{a}2\right)^2+b\right]=\left[\sqrt{ax+b}-\frac{a}2-\sqrt{\left(\frac{a}2\right)^2+b}\right]\left[\sqrt{ax+b}-\frac{a}2+\sqrt{\left(\frac{a}2\right)^2+b}\right].$
In the case that $\sqrt{\left(\frac{a}2\right)^2+b}-\frac{a}2\gt0,$ $\frac{a}2-\sqrt{\left(\frac{a}2\right)^2+b}$ is not a fixed point, so only $\frac{a}2+\sqrt{\left(\frac{a}2\right)^2+b}$ is a fixed point. This is the case when $b\gt0.$ In that case, you would use the monotone convergence theorem to show that $$\lim_{n\to\infty}s[f]_n=\frac{a}2+\sqrt{\left(\frac{a}2\right)^2+b}.$$ So this indeed generalize other well-known results. For example, if $a=1$ and $b=1,$ then $$\frac{1+\sqrt{5}}2=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}.$$