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.
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.