Doubts about infinite nested root Find $f(a)=\sqrt{a-\sqrt{a^2-\sqrt{a^4-\cdots}}}$ where $a\in\mathbb{R}$.
My Attempt :
I consider $\frac{f(a)}{a}=\sqrt{1-\sqrt{1-\sqrt{1-\cdots}}}$. Now to finding this limit is easy but I cannot prove the limit exists. Also I have another doubt can we write$\frac{f(a)}{a}$ as that infinite sum ? Since we consider infinitely many terms, I have doubts that this is true. Any help will be welcomed. Thanks.
 A: consider a sequence $t_{n+1} = \sqrt{1 - t_n}$ with $0<t_1 < 1$, you can prove via induction that the limit converges because it's obvious.
also note that $\lim_{n\to\infty}t_{n+1} = \lim_{n\to\infty}t_{n} = t$ from which you get $t = \sqrt{1-t}$.
A: For the expression
$$
\sqrt{1-\sqrt{1-\sqrt{1-\cdots}}}
$$
we can write this $t_0=1$, $t_{k+1}=\sqrt{1-t_k}$ where
$$
t_k=\sqrt{1-\sqrt{1-\cdots\sqrt{1-\sqrt{1}}}}
\quad\text{containing $k+1$ terms ($k$ subtractions)}
$$
and check if the limit of $t_k$ exists. This is the definition of limit of infinitely nested expressions: take a finite number of nestings and let the number of nestings go to infinity and see if it converges.
However, we get $t_0=1$, $t_1=0$, $t_2=1$, $t_3=0$, etc. so it will keep alternating between $0$ and $1$ without converging.
If you start with $t_0\in(0,1)$, it will converge, and the limit $t$ will satisfy $t=\sqrt{1-t}$: i.e. $t=\frac{1}{2}(\sqrt5-1)$.
The only modification when you include the $a$ is that you get
$$
\sqrt{a-\sqrt{a^2-\cdots\sqrt{a^{2^{k-1}}-\sqrt{a^{2^k}}}}}=\sqrt{a}\cdot t_k
$$
as you've already figured out and take the limit of this instead with $a$ a constant.
