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?

  • $\begingroup$ Sure. Holds for all $x \in \mathbb{R}^+$. $\endgroup$ Apr 15, 2023 at 22:55
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    $\begingroup$ The only issue is the first step, where you definite $f$ using an infinite formula. I suggest defining $f_n$ which has $n$ square roots in its formula and establishing the domain on which the sequence converges. You can then say that the limiting function satisfies the functional equation stated in your second displayed equation. $\endgroup$
    – Deane
    Apr 15, 2023 at 22:59
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    $\begingroup$ This is not a proof. It is a (very reasonable) interpretation of a weird pseudo-definition. $\endgroup$ Apr 15, 2023 at 23:06
  • $\begingroup$ @Deane I had the same idea but quickly dropped it for 2 (tightly linked) reasons: 1) the (pseudo)definition of $f$ is not written in that spirit (the dots inside do not denote a limiting process but an everlasting one in the past, since eternity, and which just stopped); 2) The initialization of the sequence seems arbitrary: $\sqrt x$? $x$? $0$? any other $a\ge0$? $\endgroup$ Apr 15, 2023 at 23:18
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    $\begingroup$ To this I fully agree. This was the meaning of my short first comment but you formulate it better. @Deane $\endgroup$ Apr 15, 2023 at 23:47

1 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.


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