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Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical $$ \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} $$ Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that $$ f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} $$ We can see that $$\begin{align} f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\ &= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} \end{align}$$ And we begin solving by $$\begin{align} f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\ f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\ &= 2^x + f(x + 1) \\ f(x + 1) &= f(x)^2 - 2^x \end{align}$$ At this point I find myself stuck, as I have little experience with recurrence relations.

How would this recurrence relation be solved? Would the method extend easily to $$\begin{align} f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\ f_n(x)^2 &= n^x + f_n(x + 1)~\text ? \end{align}$$

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Before even trying to evaluate it, can you prove even that it converges at all? –  Thomas Andrews Feb 11 '13 at 15:33
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I'm not sure the question of whether it converges in the usual sense must absolutely be settled before working out the value of the expression. If it diverges, then one might wonder whether it converges in some other sense. –  Michael Hardy Feb 11 '13 at 15:46
    
It does converge, to some limit between 1.78 and 1.94. –  Did Feb 11 '13 at 16:03
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Computation shows that $f(0)=1.7831658\dots$. Likely there is no exact form for the radical. –  Eric Naslund Feb 11 '13 at 16:09
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If you set $y = 2^x$ and $z = 2^y$ and $f(x) = g(y) = h(z)$, the recursion "simplifies" to $$g(2y) = g(y)^2 - y \qquad \qquad h(z^2) = h(z)^2 - \log_2 z$$ I don't know either helps. –  Hurkyl Feb 11 '13 at 16:50

4 Answers 4

Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists: $$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$ Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-form solution: $$ [x,x,\ldots]^2=x+[x,x,\ldots] \implies [x,x,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{1+4x}. $$ If $b_i \le a_i$ for each $i$, then clearly $[b_0,b_1,\ldots]\le[a_0,a_1,\ldots]$. What happens when a multiplicative factor is introduced? You have $$k[a_0,a_1,a_2,...]=\sqrt{k^2 a_0 + k^2[a_1,a_2,...]}=[k^2a_0,k^4a_1,k^8a_2,\ldots]$$ In your case, $[1,2,4,\ldots]=\sqrt{1+[2,4,8,\ldots]}=\sqrt{1+\sqrt{2}[1,1,1/2,1/16,\ldots]}$. Using the bounds $$ \sqrt{2}=[1,1]\le[1,1,1/2,1/16,\ldots]\le[1,1,1,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{5}, $$ you have $$ 1.732 \approx \sqrt{3} \le [1,2,4,\ldots] \le \sqrt{1+\frac{1+\sqrt{5}}{\sqrt{2}}}\approx 1.813. $$ Tighter bounds can be provided, of course, but this suffices to show that the limit exists.

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Just to add - the actual answer is around: $1.78316580926410$ –  nbubis Feb 11 '13 at 21:24

Let $x_0 = \sqrt{1}$, $x_1 = \sqrt{1 + \sqrt{2}}$, $x_3 = \sqrt{1 + \sqrt{2 + \sqrt{4}}}$, and so on. Then we have:

$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \cdots}}} = \lim_{n \to \infty} x_n$$

Clearly, this sequence is monotonically increasing. It also converges, since we can see each new term appears under $n$ square roots, and hence $| x_n - x_{n - 1} | \propto 2^{-n}$ which should be enough.


From observation, the value of $x_n - x_{n - 1}$ is a root of a polynomial of order $2^{2n - 3}$. In this sense, a closed-form solution is very unlikely to exist. Not a full answer, but it looks complicated.

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I'd rather try this: If $$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$ then $$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\ \end{align}$$ Thus if $u=\frac1{\sqrt 2}$ then $$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$ The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.

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The $n$term in the radicals is $2^{n-1}u^{2^n}$. This does not simplify when $u=1/\sqrt2$ (or for any other value of $u$), does it? –  Did Feb 11 '13 at 15:46
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@Hagen Your method, once corrected, shows that what the OP calls the infinite radical does converge, to a limit which is at most $y\sqrt2\lt2$. –  Did Feb 11 '13 at 15:57
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This method doesn't work because $4u^8 = \frac{4}{\sqrt{2}^8} = \frac{4}{2^4} = \frac{4}{16} = \frac{1}{4}$. I tried something like this on my first go, and the problem with it is there is no way to bring things in evenly through all the radicals: the coefficients increase by powers of two, while the factor you bring in gets squared every time. –  algorithmshark Feb 11 '13 at 16:05

Since you have recurrence relation $(f(x))^2=2^x+f(x+1)$ you could find an approximate solution by approximating $f(x+1)\approx f(x)$ and then you get quadratic equation for the function $f$ and in doing so you can find an approximate value $f(x)$ for every value of $x$. It is clear from the recurrence relation that the exact expression for $f(x)$ may not be expressible as some combination of the functions that have been studied to this day, but maybe I am wrong.

If you still want to search for closed forms for certain values of $f$ maybe it is better to study expressions of this type that are finite, such as $f(2,x,n) =\sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} +\sqrt {\ldots+\sqrt{2^{x+n}}}}}}}$ and then let $n\to\infty$ to get your $f$.

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These finite expressions are in general much harder to study because fixed point techniques are not available. A simplified closed form is even more unlikely to exist for your $f(2,x,n)$ than for the OP's original expression. –  Eckhard Feb 12 '13 at 17:10

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