I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:

What is the maximum value of $$F_\infty=\sqrt{\frac{x}{x+\sqrt{\dfrac{x^2}{x-\sqrt{\dfrac{x^3}{x+ \sqrt{ \dfrac{x^4}{x-\cdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?

Some observations follow.

  • Let $F_k$ be the nested radical up to $x^k$. For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_\infty$ since the powers increase each time - we can no longer write $F_\infty$ as a function of itself to be solved.

  • Here is a plot of $F_{15}$.

  • What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_\infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $\Bbb R$.

  • Furthermore, I also conjecture that $$\max F_\infty=\phi-1,$$ where $\phi$ is the golden ratio. This seems right as $\max F_{15}=0.6179$ from the plot.

EDIT: The problem can be reduced to proving that for $x\in(0,1]$, $$\frac d{dx}\sqrt{\frac{x^3}{x+\sqrt{\dfrac{x^4}{x-\sqrt{\dfrac{x^5}{x+ \sqrt{ \dfrac{x^6}{x+\cdots}}}}}}}}<1.$$


1 Answer 1


Zachary's and OP's combined solutions

If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_\infty$ at $x=1$ will give the maximum, which will be an infinite fraction:

$$F_\infty (1) = \sqrt{ \frac{1}{1+\sqrt{ \frac{1}{1- \sqrt{\cdots} }} }} = \sqrt{ \frac{1}{1+\sqrt{ \frac{1}{1- F_\infty (1) }} }}\implies F_\infty (1)^2 \left(1+\sqrt{ \frac{1}{1- F_\infty (1) }} \right) = 1$$ so $$(F_\infty(1)^2-1)^2=\frac{F_\infty(1)^4}{1-F_\infty(1)}\implies F_\infty(1)^5-2F_\infty(1)^3+2F_\infty(1)^2+F_\infty(1)-1=0.$$ Factoring out some roots, we get $$(F_\infty(1)^2+F_\infty(1)-1)(F_\infty(1)^3-F_\infty(1)^2+1)=0$$ It can be verified from W|A, for example, that the only positive real solution is at $F_\infty(1)=\phi-1$ derived from the first quadratic factor.

Proof attempt for the domain:

While I'm not sure how this holds up for $F_{\infty}(x)$, we can show that $\forall x>1$ $\exists y \mid \forall n \geq y, F_n(x) \notin \mathbb{R}$.

Now, if we ever get a negative denominator, the end result will be non-real. This is because addition, subtraction, and division between non-reals and non-zero reals will stay non-real, and the square root of a non-real will also be non-real.

Now, when $n>2$ is odd, $\sqrt{x^n}<x$, and thus $F_n(x) \notin \mathbb{R}$. Thus, we must concern ourselves with even $n$.

So, in $F_{2k}(x)$, consider:

$$ \frac{\cdots}{x - \sqrt{ \frac{x^{2k-1}}{x+\sqrt{x^{2k}}} }}$$ $$ x < \sqrt{ \frac{x^{2k-1}}{x+x^k} } \implies F_{2k}(x) \notin \mathbb{R} $$

Now, we will show that the "bottom" of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:

$$ \sqrt{ \frac{x^{2k-1}}{x+\sqrt{x^{2k}}} } < \sqrt{ \frac{x^{2k+1}}{x+\sqrt{x^{2(k+1)}}} } $$ $$ \frac{x^{2k-1}}{x+x^k} < \frac{x^{2k+1}}{x+x^{k+1}} $$ $$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$

The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that's messier in my opinion.

Using this result, it follows that:

$$ x < \sqrt{ \frac{x^{2k-1}}{x+x^k} } \implies x < \sqrt{ \frac{x^{2k+1}}{x+x^{k+1}} } \implies F_{2(k+1)}(x) \notin \mathbb{R} $$

Now, to prove the "bottom" will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $|x| > 1$. Since $x$ is non-negative, the domain is $[0,1]$.

So, for any $x$ there's an $y$ for which $F_n(x)$ is non-real for finite $n$ greater than $y$. However, I'm not certain this is rigorously extends to the infinite case.

Extra notes: I've rigged up a computer program to calculate when $F_n(x)$ becomes non-real. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+\frac{1}{2^k}$.

$$\small\begin{array}{c|c} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\\hline n_{\text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774\end{array}$$

As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go non-real. I've tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $\frac{1}{x-1}$.

Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)

Here I will attempt to prove the monotonicity of $F_\infty$. First, let us introduce some definitions. $$F:=F_{\infty},\quad G:=\sqrt{\frac{x^2}{x-\sqrt{\frac{x^3}{x+\sqrt{\frac{x^4}{x- \sqrt{ \frac{x^5}{x+\cdots}}}}}}}},\quad H:=\sqrt{\frac{x^3}{x+\sqrt{\frac{x^4}{x-\sqrt{\frac{x^5}{x+ \sqrt{ \frac{x^6}{x+\cdots}}}}}}}}$$ Since $F=\sqrt{\dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F'=\frac1{2F}\cdot\frac{1(x+G)-x(1+G')}{(x+G)^2}>0\impliedby G-xG'>0$$ as $(x+G)^2$ and $F$ are clearly non-negative.

Now this is implied by $$G'=\frac1{2G}\cdot\left(1+\frac{H'x^2-H}{(x-H)^2}\right)<\frac Gx$$ and since $G=\sqrt{\dfrac{x^2}{x-H}}$ (note that $H<x$), we get \begin{align}2G^2>x+x\frac{H'x^2-H}{(x-H)^2}&\impliedby\frac{2x}{x-H}>1+\frac{H'x^2-H}{(x-H)^2}\\&\impliedby 2x^2-2Hx>x^2-2Hx+H^2+H'x^2-H^2\\&\impliedby x^2>H'x^2\impliedby H'<1\end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H'$. Of course, the latter two are only close approximations to the real distribution of $H$.

enter image description here

  • 1
    $\begingroup$ The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet. $\endgroup$ Commented Jan 5, 2019 at 18:18
  • $\begingroup$ Alrighty, added a legitimate proof, as long as my understanding of infinities is correct. $\endgroup$ Commented Jan 7, 2019 at 0:43
  • 1
    $\begingroup$ Well done! It's not too rigorous, but I can understand your method. I'll make some edits later. I have made an attempt into showing its monotonicity, but I can't get any farther than $H'<1$. Any help appreciated! $\endgroup$
    – TheSimpliFire
    Commented Jan 11, 2019 at 20:51
  • $\begingroup$ I'm revisiting this again and I'm wondering whether $F_\infty$ is defined on $x=0$. I can't see L'Hopital working immediately, so if it isn't then the domain would have to be $(0,1]$. $\endgroup$
    – TheSimpliFire
    Commented Feb 10, 2019 at 10:21

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