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I was messing around with infinite square root nesting problems like $$w_1=\sqrt{1+\sqrt{1+\sqrt{...}}}$$ which is an easy example. I decided to try one where the terms inside of the square roots form a geometric sequence: $$w_2=\sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-3}+\sqrt{...}}}}$$ but I can't figure it out. I figured that I probably have to set it equal to $x$ and define $x$ as a function of itself, but I'm not sure how to do that. A similar problem that I did figure out how to do was $$w_3=\sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-4}+\sqrt{2^{-8}+\sqrt{...}}}}}$$ because if you multiply it by $\sqrt{2}$, it collapses to a problem of the first form that I mentioned. Any ideas about that second problem?

Gottfried Helms
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Franklin Pezzuti Dyer
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  • The value appears to be approximately $1.285736763357$, for what it's worth. – Brian Tung Apr 27 '17 at 22:07
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    In Mathematica, the function R[n_] := Fold[Sqrt[#1 + #2] &, 0, 2^-Reverse[Range[n]]] followed by N[R[1000],50] gives the result to 50 decimal places of precision. A search of this number via the inverse symbolic calculator yields no results. – heropup Apr 27 '17 at 22:14
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    Interestingly, when I expand the iterated fraction to the left, meaning I let the iteration begin not at $2^{-1}$ but at $2^0$ or $2^1$ or even $2^{64}$ (or $2^{2k}$) I arrive at a value near $2^{k}+0.35355$ whose right summand -by W|A- is near ${1 \over 2 \sqrt 2}$ ... so that we might write $ \lim_{k \to \infty} \sqrt {2^{2k}+\cdots + \sqrt {2^1 + \sqrt{2^0 + \sqrt{2^{-1} + \sqrt{2^{-2} + \cdots }}}}} \to 2^k + \sqrt{ \frac 18 } $ – Gottfried Helms Apr 28 '17 at 08:23
  • problem 2 can be reformulated . Let $b=\sqrt2$ then we can write: $$w_2 \cdot b = \sqrt{1+b^0\sqrt{1+b^1 \sqrt{1+b^2 \sqrt{1+b^3 \sqrt{...}}}}} $$ However I'm not nearer with this... – Gottfried Helms Jul 06 '17 at 14:13
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    @GottfriedHelms Yeah, we can actually prove that equality. If we let $$r_1=\lim_{k\to\infty}\Big(\sqrt{2^{2k}+\sqrt{...}}-2^k\Big)$$ $$r_2=\lim_{k\to\infty}\frac{\sqrt{2^{2k}+\sqrt{...}}+2^k}{2^k}$$ it is easy to prove (using the difference of squares identity) that $r_1\cdot r_2=1/\sqrt{2}$. Since the limit $r_2=2$ is almost trivial, we have $r_1=1/2\sqrt{2}$ as you calculated. – Franklin Pezzuti Dyer Mar 24 '20 at 13:15

2 Answers2

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If $$G_k(x) = (2x)^{k+1} + \sqrt{G_{k+1}(x)},$$ then $G_0(x)$ is the generating function of sequence A274850 and $$\sqrt{G_0(1/4)} = \sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-3}+\sqrt{...}}}}$$ but it is unlikely that this has a closed form value. More generally, let $$ f(x,q) = x + \sqrt{f(xq,q)}.$$ We can expand it in a power series $$ f(x,q) = 1 + x\frac{2}{2-q} - x^2\frac{q^2}{(2-q)^2(2-q^2)} + x^3\frac{2q^3}{(2-q)^3(2-q^2)(2-q^3)} + \dots$$ and now $G_0(x) = f(2x,2x).$

Somos
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I haven't answered my exact question, but I have managed to compute the values of some similar "crazy iterated square roots" in which the coefficients inside are in geometric progression. For example, $$\sqrt{a^2+\sqrt{4a^2+\sqrt{16a^2+\sqrt{64a^2+...}}}}=a+1$$ and $$\sqrt{1+\sqrt{a^2+a\sqrt{a^4+a^2\sqrt{a^6+a^3\sqrt{a^8+...}}}}}=\frac{a+\sqrt{a^2+4}}{2}$$ Additionally, we have the following involving the Fibonacci numbers, with $F_0=F_1=1$ (not exactly geometric progression, but whatever): $$\sqrt{1+F_1{-1+F_2\sqrt{1+F_3\sqrt{-1+...}}}}=2$$

Franklin Pezzuti Dyer
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