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I recently became fascinated by infinite nested radicals, first drawn attention to me from a question in my textbook about the value of $\sqrt{1+\sqrt{{1}+\sqrt{{1}+\sqrt{{1}...}}}}$ which turned out to be $\phi$ when I worked it out, a rather beautiful result.

I then tried to find a formula to evaluate the general case $$\sqrt{x+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}...}}}}$$ which I succeeded in; it can be evaluated as $$\frac{1+\sqrt{1+4x}}{2}$$

Multiplying the nested radical which was equal to $\phi$ by $x$ produces the following nested radical:

$$\sqrt{{x^2}+\sqrt{{x^4}+\sqrt{{x^8}+\sqrt{{x^{16}}...}}}}$$

so this is equal to $x\left(\frac{1+\sqrt5}{2}\right)$.

However, I have tried and failed to find the value of the following infinite square root: $$\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$$

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    $\begingroup$ There's an interesting page on Wikipedia which has your upper example with a solution but not the question you specifically ask... Link : en.wikipedia.org/wiki/Nested_radical $\endgroup$ Commented Jul 6, 2020 at 19:13
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    $\begingroup$ You may find either this, this, or this question helpful. $\endgroup$
    – Decaf-Math
    Commented Jul 6, 2020 at 19:25
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    $\begingroup$ Thank you for taking the time to find the suggestions and interesting links everyone :) I have already found formulae myself to evaluate the first 2 square roots in terms of $x$ though ; only the third eludes me. $\endgroup$ Commented Jul 6, 2020 at 21:05
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    $\begingroup$ Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ Commented Jul 11, 2020 at 13:33
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    $\begingroup$ Here is some data I've gathered: $f(-1)=0$, $f(1)=\phi$, $f(\epsilon)=1$ for some $0<\epsilon<<1$ and that $f(x)-\sqrt{2x}$ approaches $\frac{1}{\sqrt{32}}$. Also $f'(\epsilon)=\frac{1}{2}$ $\endgroup$
    – Graviton
    Commented Jul 15, 2020 at 5:16

2 Answers 2

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Partial answers: power series expansions around 0 and $\infty$

Define for $k\in\mathbb{Z}, k\ge 0$ $$ f_k(x) = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots}}} = \sqrt{x^k+f_{k+1}(x)}$$ where the meaning of the infinite square root iteration will be made precise below. The function of the OP is then $f_1(x)$.

For the power series expansion at 0, let us assume in all what follows that $0\lt x\ll 1$. Then for $k>1$, the term $f_{k+1}(x)=\sqrt{x^{k+1}+f_{k+2}(x)}$ dominates $x^k$ in $\sqrt{x^k+f_{k+1}(x)}$, simply because $x^k\ll x^\frac{k+1}{2}$, but $x^{k+1}$ is again dominated by $f_{k+2}(x)$ and so on, so we have approximately $f_k(x)\approx\sqrt{f_{k+1}(x)}\approx\sqrt{\sqrt{f_{k+2}(x)}}\ldots$ But taking square roots again and again on a positive value will aproach 1. Thus it is sensible to conjecture that for $0\lt x\ll 1$, we have $f_k(x)\approx 1$.

Replacing $f_{m+1}(x)$ with 1 somewhere inside the infinitely iterated expressions leads to the following partial expressions ($k,m\in\mathbb{Z}, 0\le k\le m$): $$f_{k,m}(x) := \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots\sqrt{x^m+1}}}},$$ and we can define $$f_k(x):=\lim_{m\to\infty}f_{k,m}(x)$$ if that limit exists (which I won't elaborate). Now it is easily seen that $f_{k,m}(x) = 1 + \mathcal{O}(x^k)$ (just start from $f_{m,m}(x) =\sqrt{x^m+1}= 1 + \mathcal{O}(x^m)$ and proceed by induction on decending $k$: $f_{k,m}(x)=\sqrt{x^k+f_{k+1,m}(x)}=\sqrt{x^k+1+\mathcal{O}(x^{k+1})}=1 + \mathcal{O}(x^k)$).

But this means that for any $m'>m$ \begin{align} f_{k,m'}(x) & = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots\sqrt{x^m+f_{m+1,m'}(x)}}}}\\ & = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots\sqrt{x^m+1+\mathcal{O}(x^{m+1})}}}}\\ & = f_{k,m}(x) + \mathcal{O}(x^{m+1}) \end{align} and so $$f_k(x)=\lim_{m'\to\infty}f_{k,m'}(x)=f_{k,m}(x) + \mathcal{O}(x^{m+1}).$$ Hence, the terms of the power series expansion of $f_k(x)$ up to order $x^m$ are determined by the power series expansion of $f_{k,m}(x)$. For example, the result for $f_1(x)$ up to order $x^{20}$ reads $$f_1(x)= 1 + \frac{1}{2}x + \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128} x^4 - \frac{5}{256} x^5 - \frac{19}{1024} x^6 + \frac{13}{2048} x^7 - \frac{397}{32768} x^8 + \frac{243}{65536} x^9 + \frac{79}{262144} x^{10} + \frac{6415}{524288} x^{11} + \frac{10959}{4194304} x^{12} - \frac{6321}{8388608} x^{13} - \frac{283323}{33554432} x^{14} + \frac{171429}{67108864} x^{15} + \frac{4224323}{2147483648} x^{16} + \frac{22138947}{4294967296} x^{17} - \frac{25215333}{17179869184}x^{18} - \frac{83594725}{34359738368}x^{19} - \frac{1538702507}{274877906944}x^{20} + \mathcal{O}(x^{21}).$$

For analyzing the behavior of $f_k(x)$ for large positive $x$, consider \begin{align} f_k(x^{-2})&=\sqrt{x^{-2k}+\sqrt{x^{-2k-2}+\sqrt{x^{-2k-4}+\ldots}}}\\ &=x^{-k}\sqrt{1+x^{k-1}\sqrt{1+x^k\sqrt{1+x^{k+1}\sqrt{1+\ldots}}}}\\ &=x^{-k}g_{k-1}(x), \end{align} again for $0<x\ll 1$, where $$g_k(x)= \sqrt{1+x^{k}\sqrt{1+x^{k+1}\sqrt{1+x^{k+2}\sqrt{1+\ldots}}}}.$$

Setting for $k\le m$ $$g_{k,m}(x):= \sqrt{1+x^{k}\sqrt{1+x^{k+1}\sqrt{1+\ldots x^{m-1}\sqrt{1+x^m}}}},$$ and we can define $$g_k(x):=\lim_{m\to\infty}g_{k,m}(x)$$ and we find $g_{k,m}(x)=1+\mathcal{O}(x^k)$ and for any $m'> m$ \begin{align} g_{k,m'}(x) & = \sqrt{1+x^k\sqrt{1+x^{k+1}\sqrt{1+\ldots x^{m-1}\sqrt{1+x^m g_{m+1,m'}(x)}}}}\\ & = \sqrt{1+x^k\sqrt{1+x^{k+1}\sqrt{1+\ldots x^{m-1}\sqrt{1+x^m(1+\mathcal{O}(x^{m+1})}}}}\\ & = g_{k,m}(x) + \mathcal{O}(x^{(k+m+1)(m+2-k)/2}), \end{align} so again, the power series expansion of $g_k(x)$ up to any given order can be determined by the power series expansion of $g_{k,m}(x)$ for sufficiently large $m$. For instance, for determining $g_1(x)$ up to order $x^{20}$, expanding $g_{1,5}(x)$ is sufficient, yielding: $$g_1(x)=1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{5}{16} x^3 - \frac{21}{128} x^4 + \frac{15}{256} x^5 + \frac{27}{1024} x^6 + \frac{157}{2048} x^7 - \frac{4237}{32768} x^8 + \frac{1627}{65536} x^9 + \frac{15585}{262144} x^{10} + \frac{20179}{524288} x^{11} - \frac{420737}{4194304} x^{12} + \frac{136155}{8388608} x^{13} + \frac{606675}{33554432} x^{14} + \frac{3116173}{67108864} x^{15} - \frac{166576957}{2147483648} x^{16} + \frac{258982675}{4294967296} x^{17} - \frac{117088187}{17179869184} x^{18} - \frac{516645801}{34359738368} x^{19} - \frac{23704687899}{274877906944} x^{20} + \mathcal{O}(x^{21})$$ and $$g_0(x)=\sqrt{1+g_1(x)}=\sqrt{2}\left(1 + \frac{1}{8} x - \frac{5}{128} x^2 + \frac{85}{1024} x^3 - \frac{1709}{32768} x^4 + \frac{6399}{262144} x^5 - \frac{8145}{4194304} x^6 + \frac{828477}{33554432} x^7 - \frac{83481725}{2147483648} x^8 + \frac{231319419}{17179869184} x^9 + \frac{2532368405}{274877906944} x^{10} + \frac{29815364515}{2199023255552} x^{11} - \frac{2122499603177}{70368744177664} x^{12} + \frac{5230968689963}{562949953421312} x^{13} + \frac{7443547207831}{9007199254740992} x^{14} + \frac{1141411701025037}{72057594037927936} x^{15} - \frac{231372106336231965}{9223372036854775808} x^{16} + \frac{1498156069006490195}{73786976294838206464} x^{17} - \frac{8082528897875176135}{1180591620717411303424} x^{18} + \frac{18359172053830212871}{9444732965739290427392} x^{19} - \frac{8183042653064552822819}{302231454903657293676544} x^{20} + \mathcal{O}(x^{21})\right).$$ This yields immediately the behavior of $f_1(x)=\sqrt{x}g_0(1/\sqrt{x})$ for large $x$: $$f_1(x)=\sqrt{2x}\left(1 + \frac{1}{8\sqrt{x}}- \frac{5}{128x} + \frac{85}{1024 \sqrt{x^3}} - \frac{1709}{32768 x^2} + \frac{6399}{262144\sqrt{x^5}} - \ldots\right).$$ Interestingly, the recursion $$g_k(x)^r=\left(1+x^kg_{k+1}(x)\right)^{r/2}=\sum_{a=0}^\infty{\frac{r}{2}\choose a}x^{ak}g_{k+1}(x)^a$$ can be used to get for $k\ge 1, r\ge 0$ the expression $$g_k(x)^r=\sum_{a_1=0}^\infty\sum_{a_2=0}^\infty\sum_{a_3=0}^\infty\ldots {\frac{r}{2}\choose a_1}{\frac{a_1}{2}\choose a_2}{\frac{a_2}{2}\choose a_3}\ldots x^{a_1 k + a_2(k+1)+a_3(k+2)+\ldots},$$ such that the coefficients of $$g_1(x)=\sum_{r=0}^\infty c_rx^r$$ can be written as $$c_r=\sum_{a_1}\sum_{a_2}\sum_{a_3}\ldots{\frac{1}{2}\choose a_1}{\frac{a_1}{2}\choose a_2}{\frac{a_2}{2}\choose a_3}\ldots,$$ where, for fixed $r$, the summation variables are restricted to $a_i\ge 0$ and $\sum_iia_i=r$, such that the sum is in fact finite. And because the binomial coefficients ${0\choose a}$ are zero for $a>0$, and more generally ${a_i/2\choose a_{i+1}}=0$ for $a_i$ even and $a_{i+1}>a_i/2$, the terms of the sum are non-zero only for those values $(a_1, a_2,\ldots)$ where for each even $a_i$ holds $a_{i+1}\le\frac{a_i}{2}$.

The power expansions suggest that there is no simple expression for the function of the OP. But this does not exclude the possibility that it might be an algebraic function in the sense that there might be a polynomial $p(x,y)$ in two variables $x$ and $y$, such that $p(x,f_1(x))=0$.

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Not an answer, but something I find quite interesting is that when $x = 4$, this converges to $3$.

For the comment regarding my code, it's quite simple. Here it is:

import numpy as np

x = 2 # Or whatever you want
iterations = 10 # Or whatever you want

value = np.sqrt(x ** iterations)
for i in reversed(range(1, iterations)):
  value = np.sqrt(value + x ** i)

print (value)

Another interesting aspect is this:

Define a function as the expression in the title:

$f(x) = \sqrt{x + \sqrt{x^2 + \sqrt{x^3 + \sqrt{x^4 ...}}}}$

For any positive x, $f(f(f(f(...x)))) \approx 2.340649036282968$

This is the intersection between $y=f(x)$ and $y=x$.

Edit: Just another neat fact: this function can be approximated very closely by the function $f(x) = \sqrt{2x} + 0.17555$ for most relatively small values ($x \in (0, 10^{10}]$)

Edit: The case for x = 1 isn’t too difficult.

$a = \sqrt{1 + \sqrt{1^2 + \sqrt{1^3 + \sqrt{1^4 ...}}}}$

$a = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 ...}}}}$

$a^2 - 1 = a$

$a^2 - a - 1 = 0$

Now you have a quadratic that you can easily solve.

I’m still unsure how to do the other cases.

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  • $\begingroup$ Do you have a reference for this claim? $\endgroup$ Commented Jul 6, 2020 at 19:18
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    $\begingroup$ I see, but it would be really cool if this is true. I'd like to see a proper proof if that is possible. $\endgroup$ Commented Jul 6, 2020 at 19:22
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    $\begingroup$ For the sake of having a more complete answer, would you mind sharing your python code? I know it is not difficult to write, but people here may downvote your answer because it doesn't exactly answer the question. I understand why you didn't make a comment instead (as you couldn't due to the lack of reputations). And I think your discovery is worthy of being kept as an answer. $\endgroup$ Commented Jul 6, 2020 at 19:23
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    $\begingroup$ Interesting similarities and differences with this popular MathStack question from 2015 : Link math.stackexchange.com/questions/1588720/… $\endgroup$ Commented Jul 6, 2020 at 19:37
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    $\begingroup$ Here is some sort of proof for $x=4$: math.stackexchange.com/questions/1659502/… $\endgroup$
    – Zeekless
    Commented Jul 6, 2020 at 19:49

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