On the rigour of a proof.

Question

I've been studying the following function $f$ defined on $\mathbb{R}^+$ as follows: $$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$ Which I have found a closed form for on the same domain of definition in the following manner: Because of the recursive nature of the series, one can write the following: $$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}\Longleftrightarrow f(x)=\sqrt{x+f(x)}$$ Hence, treating $f(x)$ as the unknown element of the equation yields: $$f^2(x)-f(x)-x=0$$ And consequently solving for $f(x)$: $$\Delta=1^2+4x\Longrightarrow f_1(x)=\frac{1+\sqrt{1+4x}}{2}\wedge f_2(x)=\frac{1-\sqrt{1+4x}}{2}$$ Only one of these definitions is in fact correct, considering all outputs of the function have to be positive and remarking that, on $\mathbb{R}^+$: $$1\leq1+4x\Longleftrightarrow f_2(x)\leq0\ \ \forall x\in\mathbb{R}^+$$ Leaving us with the only possible conclusion that: $$f(x)=\frac{1+\sqrt{1+4x}}{2}$$ My question is as to how rigorous this proof is. Does it truly hold for all values of $x\in\mathbb{R}^+$? The two functions certainly seem to match up on graphing calculators like Desmos. Is this rigorous enough, or still too hand-wavy?

Answer 1

To make this more rigorous, you would first need to make it clear what $\sqrt{x + \sqrt{x + \ldots}}$ actually means, and typically an infinite operation like this is defined as the limit of a sequence of finite operations. So we might start by defining:

$f_n(x) := \begin{cases} \sqrt{x} & \mbox{for } n = 0 \\ \sqrt{x + f_{n-1}(x)} & \mbox{for } n > 0 \end{cases}$

So $f_0(x) = \sqrt{x}$, $f_1(x) = \sqrt{x + \sqrt{x}}$, and so forth. Then we can say that $f(x) = \lim_{n \rightarrow \infty} f_n(x)$, assuming that such a limit exists.

To prove that $f(x)$ is well-defined for $x > 0$, we could then apply something like the monotone convergence theorem - for any given $x$, if you can show that $f_n(x)$ increases as $n$ increases, and also that $f_n(x) \leq M$ for some real number $M$, then $f_n(x)$ must converge to a limit.

Once you do that, then knowing $f(x)$ is the limit of $f_n(x)$, you can essentially perform the calculations you've done and show that the closed form is indeed correct. It would also help to show that the resultant formula for $f(x)$ is not dependent on the choice of $f_0(x)$, which helps justify the somewhat arbitrary choice I made there.