Find the value of this infinitely nested radical (it appears to obtain multiple values)

Question

Find the value of $$\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\cdots}}}}$$

This is not as simple as it looks for one reason - there are $2$ real solutions to the equation $$x=\sqrt{1-\sqrt{\frac{17}{16}-x}}\implies\begin{cases}x_1=0.5\\x_2\approx 0.073\end{cases}$$

You can see that for yourself on Wolfram Alpha.

How can we know the real value of this infinite radical?

Answer 1

Let $f(x)=\sqrt{1-\sqrt{\frac{17}{16}-x}}$ on $\left[\frac1{16},\frac{17}{16}\right]$:

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$f$ is monotonically increasing on $\left[\frac1{16},\frac{17}{16}\right]$. To be precise, $$ f'(x)=\frac1{4f(x)\left(1-f(x)^2\right)}\tag{1} $$

$B$, $C=\!\!\left(\frac12,\frac12\right)$, and $D=\!\!\left(\frac{17}{16},1\right)$ are the points where $f(x)=\frac89x+\frac1{18}$.

$A$ and $C$ are the points where $f(x)=x$.

Evaluating $f'$ at the two roots of $f(x)=x$, we get $$ f'(A_x)\doteq3.4345\tag{2} $$ and $$ f'\left(\frac12\right)=\frac23\tag{3} $$ This means that $A$ is an unstable fixed point of $f$ and $C$ is a stable fixed point of $f$. In fact, for $x\in\left[B_x,D_x\right]$, we have $$ \frac23\le\frac{f(x)-\frac12}{x-\frac12}\le\frac89\tag{4} $$ Furthermore, for $x\in\left(\frac1{16},B_x\right]$, $$ f'(x)\gt\frac53\tag{5} $$ $(4)$ and $(5)$ say that

1. if $x_0\in\left(A_x,\frac{17}{16}\right]$, then $x_{n+1}=f(x_n)$ will converge to $\frac12$.
2. if $x_0=A_x$, then $x_{n+1}=f(x_n)$ is the constant sequence $x_n=A_x$.
3. if $x_0\in\left[\frac1{16},A_x\right)$, then $x_{n+1}=f(x_n)$ will get smaller until at some point $x_n\lt\frac1{16}$.

where $A_x\doteq0.073182744516454277369$.

Answer 2

Solution technique:

Let $$L = \sqrt{1 - \sqrt{\frac{17}{6} - \sqrt{1 - \sqrt{\frac{17}{6} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }} } $$

The natural method of solution is to observe that

$$L^2 = 1 - \sqrt{\frac{17}{6} - \sqrt{1 - \sqrt{\frac{17}{6} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }}$$

So...

$$ L^2 - 1= -\sqrt{\frac{17}{16} - \sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }}$$

Therefore

$$ (L^2 - 1)^2 = \frac{17}{16} - \sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }$$

Implying:

$$(L^2 - 1)^2 - \frac{17}{16} = -\sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }$$

And therefore we substitute

$$(L^2 - 1)^2 - \frac{17}{16} = -L$$

Now this equation has 4 possible solutions each expression for $L$ behaves algebraically the same as this nested radical.

The reason is because if you recall from algebra there are two possible $\sqrt{x}$ for any number (a positive and negative root). Now we need to de-nest 2 roots before we can exploit our substitution so what ends up happening is that that are really four possible infinite-radicals (depending on the types of roots we alernated with) that could've yielded this $L$. Your job is to then approximate YOUR problem to a good amount and see which of the 4 solutions to

$$(L^2 - 1)^2 - \frac{17}{16} = -L$$

Corresponds and even then, we may have more than one converging value if our nested radical trends in a way analogous to

$$1 - 1 + 1 - 1 + 1 - 1...$$

Whose sum can be seen as 1 or 0 depending on how one chooses to enumerate its values.

Answer 3

Let's study the following sequel: $u_{n+1}=f(u_n)$. Here, $f(u_n)= \sqrt{ 1 - \sqrt{\frac{17}{16}-u_n}}$.

1/ Where is it defined? Well, necessarily $u_n\leq\frac{17}{16}$ and $u_n\geq\frac{17}{16}-1 = \frac{1}{16}$.

2/ What intervals are stable by $f$? Well looking at $f$ pretty much its entire domain: $f([0,\frac{17}{16}])\subset [0,\frac{17}{16}]$

3/ What are the possible fixed points? As you've noted, there are 2 solutions, x=0.5 and $x=0.0731...$

4/ Checking that the points are attractive / repulsive. Basically, you're checking if $|u_{n+2} - u_{n+1}|=|f(u_{n+1}) - f(u_n)| < K |u_{n+1} - u_{n}|$, that is if $f-Id$ is K-lipschitz with K<1. To make things easy, you simply look at your derivative and take the sup on the relevant interval.

What you find is that $|f-Id|'_{x=0.0783} =2.43$: this point is going to repulse the sequel. On the other hand, $|f-Id|'_{x=0.5} =0.33..$ and will be attractive.

5/ In conclusion, either you start with $u_0 = 0.0783$ and $u_n = u_0=0.0783$ or $u_0 \not = 0.0783$ and $\lim_{n\rightarrow +\infty} u_n = 0.5$.

If you want to be really clean, you need a proper study of what intervals are stable by $f$, and take the sup on those, showing that the rest of the domain will bring you there.