How would I simplify this function $\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{\dots}}}}$ How do I simplify $\rho(x)$ into simple terms?
$$\rho(x)=x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$$
where the subtracting and the adding follows the Thue–Morse sequence $$+,-,-,+,-,+,+,-,-,+,+,-,+,-,-,+,\dots$$
I tried doing it with $x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\dots}}}}}}}}$ and  got a answer by myself and I did it with $x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{\dots}}}}}}}}$ and found a post here Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$ and I understood how it worked I would like to know how to solve a problem like this? where the adding and the subtracting never repeats.
 A: I wouldn't call this an exact closed form, but a 'close' one indeed. I remember a result in the paper (Page $28$); A chronology of continued square roots and other continued compositions by Dixon J. Jones; he refers another problem reffered in $1899$ by Karl Bochow (Problem 1740. Zeitschrift f¨ur mathematischen und
naturwissenschaftlichen Unterricht) Which asks the reader:

Assuming $0<a<1/2$: $$2\sin(\pi
a)=l_0\sqrt{2+l_{1}\sqrt{2+l_{2}\sqrt{2+l_{3}\sqrt{2+l_{4}\sqrt{...}}}}}$$ For $l_n$ being either $-1$ or $+1$. Then;
$$a=\frac{l_{0}}{2^{2}}+\frac{l_{0}l_{1}}{2^{3}}+\frac{l_{0}l_{1}l_{2}}{2^{4}}+\frac{l_{0}l_{1}l_{2}l_{3}}{2^{5}}+...$$

In your posed problem, we have
$$\small{\begin{align}
ρ\left(x\right) & = x+\sqrt{x-\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x+...}}}}}} \\
 & = x+\left(-1\right)^{0}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{0}\sqrt{x+\left(-1\right)^{1}\sqrt{x+\left(-1\right)^{0}\sqrt{x+...}}}}}} \\ 
 & = x+\left(-1\right)^{m_{0}}\sqrt{x+\left(-1\right)^{m_{1}}\sqrt{x+\left(-1\right)^{m_{2}}\sqrt{x+\left(-1\right)^{m_{2}}\sqrt{x+\left(-1\right)^{m_{3}}\sqrt{x+\left(-1\right)^{m_{4}}\sqrt{x+...}}}}}}
\end{align}}$$
Where $m_n$ is the n-th Thue-Morse element (having $m_0,m_1,m_2... = 0,1,1,0..$). Now applying the first result:
$$\small{\begin{align}
ρ\left(2\right) & = 2+\left(-1\right)^{m_{0}}\sqrt{2+\left(-1\right)^{m_{1}}\sqrt{2+\left(-\right)^{m_{2}}\sqrt{2+\left(-1\right)^{m_{2}}\sqrt{2+\left(-\right)^{m_{3}}\sqrt{2+\left(-1\right)^{m_{4}}\sqrt{2+...}}}}}} \\ 
 & = 2+2\sin\left\{\pi\left(\frac{\left(-1\right)^{m_{0}}}{2^{2}}+\frac{\left(-1\right)^{m_{0}+m_{1}}}{2^{3}}+\frac{\left(-1\right)^{m_{0}+m_{1}+m_{2}}}{2^{4}}+\frac{\left(-1\right)^{m_{0}+m_{1}+m_{2}+m_{3}}}{2^{5}}+...\right)\right\} \\
& = 2+2\sin\left\{\pi\left(\frac{\left(-1\right)^{Sm_{0}}}{2^{2}}+\frac{\left(-1\right)^{Sm_{1}}}{2^{3}}+\frac{\left(-1\right)^{Sm_{2}}}{2^{4}}+\frac{\left(-1\right)^{Sm_{3}}}{2^{5}}+...\right)\right\} \\
& = 2+2\sin\left\{\frac{\pi}{4}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}\right\}\tag{1}
\end{align}}$$
Where $Sm_n=\sum m_k=m_0+m_1+...+m_n$. $Sm_n$ is also called the partial sum of the Theu-Morse sequence (find the sequence of that here). Now I'm not sure whether that sum has a closed form or not. I've seen certain other infinite series using Thue-Morse that has closed forms. There's a similar series to the sum in $(1)$:
$$\sum_{n=0}^{\infty}\frac{\left(-1\right)^{m_{n}}}{2^{n}}=2\left(1-2\tau\right)$$
Where $\tau$ is called the Thue-Morse constant
EDIT: To compute the series in $(1)$, you can use the following asymptote
$$\sum_{n=0}^{\infty}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}\sim\sum_{n=0}^{4x}\frac{\left(-1\right)^{Sm_{n}}}{2^{n}}-\frac{3}{5}2^{-4x}$$
Which stems from the result that:
$$\lim_{n \rightarrow \infty}\frac{Sm_n}{n}=\frac{1}{2}$$
