Cfrgtkky

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

This is a modification on the well-known Ramanujan infinite radical, $sqrt{1+sqrt{1+2sqrt{1+3sqrt{1+cdots}}}}$, 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 $Ntoinfty$, which allows us to transform the radical into $6^{-N/2}sqrt{6^{N+1}+sqrt{6^{2N+1}+2sqrt{6^{4N+1}+cdots}}}$, which can be treated as having each term a power of $6^{N/2}$ in the limit. For a radical of the form $sqrt{alpha^2+sqrt{alpha^4+2sqrt{alpha^8+cdots}}}$ 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 $Nsqrt{frac6{N^2}+frac1Nsqrt{frac6{N^2}+frac1Nsqrt{frac{24}{N^2}+frac1Nsqrt{frac{864}{N^2}+frac1Nsqrt{frac{1990656}{N^2}+cdots}}}}}$ 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 $G(x)=sqrt{x+sqrt{x+2sqrt{x+3sqrt{x+cdots}}}}$. But all I got was an equally weird differential equation:

$$frac{df}{dx}=frac{1+frac{1+frac{1+frac{{mathinner{mkern2muraise1pthbox{.}mkern2mu raise4pthbox{.}mkern2muraise7pthbox{.}mkern1mu}}}{frac23frac{left(frac{left(f(x)^2-xright)^2-x}{2}right)^2-x}{3}}}{frac22frac{left(f(x)^2-xright)^2-x}{2}}}{frac21left(f(x)^2-xright)}}{2f(x)}$$

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:

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

Formed by taking recurrent approximations to the infinite radical. The limit of $P_n$ as $ntoinfty$ 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 $P_5$:

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

See What is the region of convergence of $x_n=left(frac{x_{n-1}}{n}right)^2-a$, where $a$ is a constant?

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

Let $$G:=sqrt{6+sqrt{6+2sqrt{6+3sqrt{6+cdots}}}}$$ Then define $$F:=G^2-6=sqrt{6+2sqrt{6+3sqrt{6+cdots}}}$$ 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 $$6=a^2+4a+1implies(a-1)(a+5)=0implies a=1,-5.$$

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

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

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

NOTE

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