# Solving the infinite radical $\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+…}}}}$

$$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$

This is a modification on the well-known Ramanujan infinite radical, $$\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}}$$, except it cannot be solved by the conventional method -- the functional equation $$F(x)^2=ax+(n+a)^2+xF(x+n)$$, since setting $$n=1$$ with $$a=0$$ requires having $$(n+a)^2=1$$, not $$6$$.

Here are some alternative methods I've tried:

• The functional equation we have instead for this infinite radical is $$F(x)^2=6+xF(x+1)$$. I've tried to solve this, but unfortunately it's easy to demonstrate that $$F(x)$$ cannot be a simple linear function $$F(x)=ax+b$$. I've tried some slightly more complicated versions -- the equation for a hyperbola, etc. -- but nothing seems to work.
• I've tried factoring stuff out from the radical to bring it to a more tenable form. Perhaps not a satisfactorily rigorous approach, I thought of factoring out $$\sqrt{6^{N/2}}$$ where $$N\to\infty$$, which allows us to transform the radical into $$6^{-N/2}\sqrt{6^{N+1}+\sqrt{6^{2N+1}+2\sqrt{6^{4N+1}+\cdots}}}$$, which can be treated as having each term a power of $$6^{N/2}$$ in the limit. For a radical of the form $$\sqrt{\alpha^2+\sqrt{\alpha^4+2\sqrt{\alpha^8+\cdots}}}$$ we have the functional equation $$F(x)^2=\alpha^{2^x}+xF(x+1)$$, or upon letting $$F(x)=\alpha^{2^x}p(x)$$, you get $$p(x)^2-xp(x+1)=\alpha^{-2^x}$$, but I'm stuck there.
• Similarly, I tried factoring out some arbitrary $$N$$ then factoring out a term from each radical inside such that the coefficients go from being $$1,2,3,\cdots$$ to a constant $$1/N,1/N,1/N...$$, transforming the radical into $$N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac{24}{N^2}+\frac1N\sqrt{\frac{864}{N^2}+\frac1N\sqrt{\frac{1990656}{N^2}+\cdots}}}}}$$ where the added terms go as $$k_1=6$$, $$k_{n+1}=\frac{n^2}6k_n^2$$. But how might one proceed?
• I considered differentiating the function $$G(x)=\sqrt{x+\sqrt{x+2\sqrt{x+3\sqrt{x+\cdots}}}}$$. But all I got was an equally weird differential equation:

$$\frac{df}{dx}=\frac{1+\frac{1+\frac{1+\frac{{\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}}}{\frac23\frac{\left(\frac{\left(f(x)^2-x\right)^2-x}{2}\right)^2-x}{3}}}{\frac22\frac{\left(f(x)^2-x\right)^2-x}{2}}}{\frac21\left(f(x)^2-x\right)}}{2f(x)}$$

Any ideas as to how I might proceed?/Any alternative (hopefully less tedious, but regardless) methods that might work?

I created a small program to play with this. The exact answer (perhaps as an infinite series) may contain $$\sqrt{6}+1/2+...$$ somewhere in it, because as you increase the number replacing 6, the radical approaches $$\sqrt{x}+1/2$$. Of course, this term just comes from the binomial series for $$\sqrt{6+\sqrt{6}}$$.

I also got nothing on the inverse symbolic calculator.

Here's another possible approach: one may consider the sequence of polynomials:

$$P_1:x^2-6=x$$ $$P_2:\left(\frac{x^2-6}2\right)^2-6=x$$ $$P_3:\left(\frac{\left(\frac{x^2-6}2\right)^2-6}3\right)^2-6=x$$

Formed by taking recurrent approximations to the infinite radical. The limit of $$P_n$$ as $$n\to\infty$$ is the root of some function with a power series expansion that can perhaps be calculated in this form. But what is the power series expansion?

Note that the polynomial gets very complicated very quick. E.g. here's $$P_5$$:

$$\frac{x^{32}}{2751882854400}-\frac{x^{30}}{28665446400}+\frac{43x^{28}}{28665446400}-\frac{91x^{26}}{2388787200}+\frac{121x^{24}}{191102976}-\frac{53x^{22}}{7372800}+\frac{11167x^{20}}{199065600}-\frac{4817x^{18}}{16588800}+\frac{57659x^{16}}{66355200}-\frac{x^{14}}{1382400}-\frac{9491x^{12}}{1382400}+\frac{367x^{10}}{12800}-\frac{2443x^8}{46080}+\frac{179x^6}{9600}+\frac{2233x^4}{9600}-\frac{71x^2}{160}-x-\frac{33359}{6400}=0$$

• why wouly you expect anything pretty to happen ? – mercio Jul 3 '18 at 13:00
• What is the limit value ? With estimates I get about $\,\sqrt{6}+\frac{1}{\sqrt{2}}\,$ but perhaps that's too much ? Or not enough ? (I have no programm with me to calculate it.) – user90369 Jul 3 '18 at 13:54
• Bravo for typing the expression for $df/dx$ in $\LaTeX$ though :) – TheSimpliFire Jul 3 '18 at 14:15
• @user90369 I ran a quick program to check it out (and the value it gives is right to all the given decimal places) -- unfortunately, yours goes wrong at the third decimal place. It's definitely not $\sqrt{10}$, though. – Abhimanyu Pallavi Sudhir Jul 3 '18 at 16:14
• @AbhimanyuPallaviSudhir : Thanks a lot for checking it, very kind of you. – user90369 Jul 3 '18 at 17:43

## 2 Answers

Currently not correct answer; but I keep it for the record (and hopefully whenever I manage to make any progress on it)

Let $$G:=\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$ Then define $$F:=G^2-6=\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}$$ which is easier to work with. Following on from this method, we can immediately match up $n$ and $x$. They are $n=1$ and $x=2$ (as can be observed in the radical).

Finally, we find $a$. The value of $6$ corresponds to $ax+(n+a)^2=2a+(1+a)^2$ so we solve $$6=a^2+4a+1\implies(a-1)(a+5)=0\implies a=1,-5.$$

The result is given as $$F=x+n+a=3+a$$ and since $F$ is clearly non-negative, we have that $a=1$ so $$G=\sqrt{6+F}=\sqrt{6+3+1}=\color{red}{\sqrt{10}}.$$

• Mistyped the last line maybe ? $F=1$ is not realistic and $6+1$ is not $10$. Should be $\sqrt{6+3+1}=\sqrt{10}$. – Oscar Lanzi Jul 3 '18 at 15:14
• @OscarLanzi Thank you. I have corrected the typo. – TheSimpliFire Jul 3 '18 at 15:15
• This is evidently incorrect -- to use that expression, you need $ax+(n+a)^2=6$, $a(x+n)+(n+a)^2=6$, $a(x+2n)+(n+a)^2=6$, etc. -- only the first of these is true when you set $a=1$ (or any non-zero number). The infinite radical you are evaluating is $\sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9...} } } }$ ... This is why the question is tougher than Ramanujan's original problem, $a$ must equal zero for the term "6" to remain constant, but that isn't possible. – Abhimanyu Pallavi Sudhir Jul 3 '18 at 16:04
• Well spotted. The value of $\sqrt{10}$ can thus be an upper bound. I'll see what I can make out of this problem this week. – TheSimpliFire Jul 3 '18 at 16:33

Hint.

Considering the function

$$f(x) = \sqrt{6+(x+1)\sqrt{6+(x+2)\sqrt{6+(x+3)\sqrt{6+(x+4)(\cdots)}}}}$$

we have the recurrence

$$f(x) = \sqrt{6+(x+1)f(x+1)}$$

or squaring

$$f^2(x) = 6 + (x+1) f(x+1)$$

Those kind of equations have an almost linear behavior so making

$$f(x) = a x + b$$

and substituting into the recurrence relationship we have

$$a^2 x^2+2 a b x-a x^2-2 a x-a+b^2-b x-b-6 = 0$$

Considering that we are interested on values near $x = 0$ we follow with

$$2 a b x-2 a x-a+b^2-b x-b-6 = 0$$

thus obtaining

$$\left\{ \begin{array}{rcl} b^2-b-a-6=0 \\ 2 b a-2 a-b=0 \\ \end{array} \right.$$

obtaining the feasible values

$$a = 0.733360\\ b = 3.142604\\$$

so the guess for $f(0)$ is

$$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}\approx 3.142604$$

NOTE

This value is a little smaller than the real value $\approx 3.15433$

• I can't see how this helps any. If this is simply meant to be a suggestion, using the comments is probably more appropriate. – Simply Beautiful Art Jul 3 '18 at 13:31
• I know you can do this, but this is just an approximation, and isn't particularly more useful than just cutting off the nesting at $6+100\sqrt{6}$, or something like that. – Abhimanyu Pallavi Sudhir Jul 3 '18 at 16:05
• Not quite true. If the sequence were $\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{9+\cdots}}}}$ (Ramanujan) this method should give the exact answer. – Cesareo Jul 3 '18 at 16:29
• Yes, I know, but there you would actually have a linear function. Here it's only an approximation. – Abhimanyu Pallavi Sudhir Jul 3 '18 at 16:33
• And as an approximation was considered. – Cesareo Jul 3 '18 at 16:36