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I wonder about a closed form for

$ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] {1 + 4 \sqrt[3] {1 + 5 \sqrt[3] \cdots}}}}} $

To be clear

$$? = \sqrt[3]{ 1 + \color{Red}{1}\sqrt[3]{ 1 + \color{Red}{2} \sqrt[3]{ 1 + \color{Red}{3} \sqrt[3]{\cdots}}}} $$

Where the red coefficients are just the natural numbers.

I tried solving the related equation $ f(x) ^3 = 1 + (x+y) f(x+1) $ for various fixed integer values $y,$ but I failed.

It appears

$$ ? = \sqrt[3] {1 + \sqrt[3] 5} $$

But I am not able to prove it.

See also https://en.m.wikipedia.org/wiki/Nested_radical#Ramanujan.27s_infinite_radicals

user376343
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mick
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    As in $$ ? = \sqrt[3]{ 1 + \sqrt[3]{ 1 + 2 \sqrt[3]{ 1 + 3 \sqrt[3]{\cdots}}}} $$ ? – Ian Miller Aug 30 '16 at 11:32
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    How the sequence 1, 1, 2, 3... being generated? I don't quite follow how it arises from solving the given equation. (Esp. given that it relates $f$ at different points, and I don't know what your given value for $y$ is.) – Alex Meiburg Aug 30 '16 at 11:34
  • Do you mean $\sqrt[3] {1+2\sqrt[3] {1+3\sqrt[3] {1+4\sqrt[3] {\cdots}}}}$? With the extra term in front, it's impossible to guess what the pattern of coefficients is. – lulu Aug 30 '16 at 11:38
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    Perhaps Ramanujan's solution could be adapted to work here. – Arthur Aug 30 '16 at 11:38
  • Ramanujan's solution seems to be solving the functional equation directly using the quadratic formula, and mick explicitly mentions that his problem arose from trying to solve a functional equation. It seems rather we should just borrow the idea of solving that directly. (Cubic formula?) – Alex Meiburg Aug 30 '16 at 11:43
  • how do you know that it converges to a real number? – Konstantinos Gaitanas Aug 30 '16 at 13:03
  • It converges to a real number because the sqrt analogue does. – mick Aug 31 '16 at 11:11
  • The simpler case IS KNOWN – mick Sep 08 '16 at 13:30
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    I don't think it converges to $(1+5^{1/3})^{1/3}\approx 1.394189514203921$. Computing your expression out to $500$ gives $\approx 1.392858038264747$. – rogerl Sep 08 '16 at 14:09
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    Then it better to take $\sqrt[3]{\sqrt[3]{2}+\sqrt[3]{3}}$. But this is also only a speculation. :-) – user90369 Sep 08 '16 at 14:16
  • Really? What is the closed form of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$? Do you have a reference? – Winther Sep 08 '16 at 16:11
  • @Winther are you kidding ? It is famous. And it was already in the links of the comments too ! A famous Ramanujan result. https://en.m.wikipedia.org/wiki/Nested_radical#Ramanujan.27s_infinite_radicals – mick Sep 08 '16 at 16:15
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    Not everybody knows everything. Thanks for the link. – Winther Sep 08 '16 at 16:17
  • I thought you knew everything :) @Winther – mick Sep 08 '16 at 16:18
  • Yeah not to be confused with the nested radical constant ! – mick Sep 08 '16 at 16:19
  • I have tried to adapt Ramanujan's formula with nested square roots to cube roots, but it is not coming out in a useful form that will help me solve the expression. – Biggs Sep 08 '16 at 22:30
  • You might be interested in my comment at http://math.stackexchange.com/a/1940171/1714 I could vaguely imagine, that proceeding the way which I indicate in my very last remarks there one could find a Ramanujan-like closed-form-formula – Gottfried Helms Sep 26 '16 at 08:52

1 Answers1

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Generalizing Ramanujan's Radical, you can get trivial results like this,

$$2=\sqrt[3]{4+1^2\sqrt[3]{10+3^{2}\sqrt[3]{16+5^{2}\sqrt[3]{22+7^{2}\sqrt[3]{28+9^{2}\sqrt[3]{...}}}}}}$$ $$3=\sqrt[3]{7+2^{2}\sqrt[3]{13+4^{2}\sqrt[3]{19+6^{2}\sqrt[3]{25+8^{2}\sqrt[3]{...}}}}}$$ Where the terms, $4,10,16,..$ are in an arithmetic progression with $d=6$. And so are the terms $7,13,19..$

While your stated question looks similar to Ramanujan's, it isn't. A closed form expression is highly unlikely.

Also, the above expressions are from;

$$n+1=\sqrt[3]{1+3n+n^{2}\sqrt[3]{1+3\left(n+2\right)+\left(n+2\right)^{2}\sqrt[3]{1+3\left(n+4\right)+\left(n+4\right)^{2}\sqrt[3]{...}}}}$$

Some approximations for the constant, one of them include;

$$\approx \sqrt{\frac{\sqrt{15}+\sqrt[3]{2}}{\sqrt{7}}}$$

At 6 decimal places, other approximations include; $12^{\frac{2}{15}},\sqrt{\frac{97}{50}},\exp\left(\frac{1}{3}-\frac{1}{500}\right),\sinh^{-1}\left(\frac{17}{9}+.0001\right),\frac{39}{28}$

Math2718
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Mourad
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