About expressions of the form $2=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+...}}}}}}$

Question

In an attempt to associate a unique sequence of nested radicals with a generic real number $r>\frac{1+\sqrt{5}}{2}$ $$r=\sqrt{a_0+\sqrt{a_1+\sqrt{a_2+...}}},$$ being $a_n$ positive integers, I have implemented the following algorithm:

1. look for the smallest solution $x_{-}$ of the equation $x^2-(2r^2+1)x+r^4=0$;
2. take the largest integer $m$ smaller than $x_{-}$ such that $m<r(r-1)$;
3. set $a_0=m$;
4. update $r$ to the value $r^2-m$;
5. return to step 1 to calculate $a_1$ and so on.

I think this procedure assures the increasing monotony of the coefficients $a_n$.

Step 2 is necessary: in fact, in the "worst case", we have to obtain at least $a_k=a_n$ for each $k\gt n$.

For example, $$1.75793...=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}}}$$ $$\pi=\sqrt{6+\sqrt{11+\sqrt{11+\sqrt{18+\sqrt{19+\sqrt{25+...}}}}}}$$

Testing the code with $r=2$, I have obtained

$$2=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+...}}}}}}$$

So, defined $P(n)=n^2-n-1$, we can write

$$2=\sqrt{P(2)+\sqrt{P(3)+\sqrt{P(4)+...}}}$$

Actually, the previous argument can be generalised in

$$n=\sqrt{P(n)+\sqrt{P(n+1)+\sqrt{P(n+2)+...}}}$$

for each natural $n\ge0$ (squaring both sides you obtain the expression for the natural $n+1$).

Is my previous reasoning correct? How to formalise it in a rigorous way? Is there any relation with continued fractions theory?

Many thanks.

---

Addendum

Multiplying the last identity by itself, it is not difficult to prove the following further one:

$$n^2=\sqrt{P(n)Q(n)+\sqrt{P(n+1)Q(n+1)+\sqrt{P(n+2)Q(n+2)+...}}}$$

where $Q(n)=n^2+n+1$.

What is quite curious for me is that both relations seem to remain valid even if we replace $n$ with a generic complex number (and, in particular, a positive real one).

For example, $$\sqrt{3}=\sqrt{2-\sqrt{3} +\sqrt{2+\sqrt{3}+\sqrt{4+3\sqrt{3} +\sqrt{8+5\sqrt{3}+...}}}}$$ $$3=\sqrt{5-2\sqrt{3} +\sqrt{21+12\sqrt{3}+\sqrt{85+50\sqrt{3} +\sqrt{233+136\sqrt{3}+...}}}}$$ $$2+i=\sqrt{3i+\sqrt{4+5i+\sqrt{10+7i +\sqrt{18+9i+...}}}}$$

Answer 1

Let $f_k(r) = \sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \dots \sqrt{a_{k-1}}}}}$ be the approximation we get by nesting a total of $k$ radicals, rather than infinitely many. The recurrence relation can be described as $f_k(r) = \sqrt{m + f_{k-1}(r^2-m)}$ together with the description of how $m$ is chosen.

This ensures that $f_k(r) < r$ for all $k$. We can prove this by induction on $k$. For $k=1$, $f_1(r) = \sqrt{a_0} < \sqrt{r(r-1)} < r$, and if we have $f_k(r) < r$ for all $r$, then $$f_{k+1}(r) = \sqrt{m + f_k(r^2-m)} < \sqrt{m + r^2-m} = r$$ for all $r$.

Now we have a sequence $f_1(r), f_2(r), f_3(r), \dots$ that's increasing and bounded above by $r$, so it converges to something. This is where we need to look into our choice of $m$ in steps 1-2, so that we know it converges to the right thing.

---

Let's try to imagine what might be enough, for this. (Here, the lower bound $\frac{1+\sqrt5}{2}$ will come up a lot, so I'll just call it $\varphi$.)

Suppose we prove that $f_k(r) \ge r - \delta_k$ for all $r > \varphi$. Then $$f_{k+1}(r) = \sqrt{m + f_k(r^2-m)} > \sqrt{m + (r^2-m-\delta_k)} = \sqrt{r^2-\delta_k}.$$ Then set $\delta_{k+1} = \sup\{r - \sqrt{r^2-\delta_k} : r > \varphi\}$, and we can conclude that $f_{k+1}(r) \ge r - \delta_{k+1}$. What is this $\delta_{k+1}$? Well, $r - \sqrt{r^2-\delta_k}$ is decreasing in $r$, so actually our definition of $\delta_{k+1}$ boils down to $\delta_{k+1} = \varphi - \sqrt{\varphi^2 - \delta_k}.$ Assuming this is even defined (i.e. that $\delta_k \le \varphi^2$) then we have $\delta_{k+1} \le \delta_k/\varphi$, so the sequence $\delta_k, \delta_{k+1}, \dots$ will decrease exponentially, and eventually go to $0$, which is what we want.

All that's necessary, then, is to make sure that we have a base case for this induction: that we can find some $\delta_1 \le \varphi^2$ such that $f_1(r) \ge r - \delta_1$ for all $r > \varphi$.

I admit I don't understand why step 1 of the algorithm is there, so let's just say we choose $m$ to be the greatest integer strictly less than $r^2-r$. Then $f_1(r) \ge \sqrt{r^2-r-1}$. Re-using an idea, the expression $r - \sqrt{r^2-r-1}$ is decreasing in $r$, so its worst case is at the boundary: for $r = \varphi$, $r-\sqrt{r^2-r-1}=\varphi$, and therefore also for all $r > \varphi$, we have $r-\sqrt{r^2-r-1} \ge \varphi$.

This ensures that $f_1(r) \ge r - \varphi$ for all $r>\varphi$, so we can set $\delta_1 = \varphi$.

Everything we've done so far guarantees that $$ r - \varphi^{2-k} \le f_k(r) \le r $$ and therefore $\lim_{k \to \infty} f_k(r) = r$ by the squeeze theorem.

Answer 2

Not another answer but a quite interesting generalization, I suppose.

Let's consider, in the complex domain, a "suitable" polynomial $f(z)$ (see Misha's comment below). Under conditions yet to be determined, we can write the following identity

$$z=\sqrt[k]{z^k-f(z)+\sqrt[k]{(f(z))^k-f^2(z)+\dots\sqrt[k]{(f^m(z))^k-f^{m+1}(z)+\dots}}}$$

where $f^m(z)$ stands for the polynomial $f(z)$ composed $m$ times with itself.

For example, if we choose $f(z)=z^2+1$ and $k=3$, we can write

$$2=\sqrt[3]{3+\sqrt[3]{99+\sqrt[3]{16899+\dots}}}$$

because

$$f(2)=5, f(5)=26$$ $$2^3-f(2)=3$$ $$5^3-f(f(2))=125-26=99$$ $$(f(f(2)))^3-f(f(f(2)))=26^3-f(26)=17576-677=16899$$

Another example: $f(z)=2z+1$ and $k=2$

$$3=\sqrt{2+\sqrt{34+\sqrt{194+\sqrt{898+\dots}}}}$$

because

$$f(3)=7, f(7)=15, f(15)=31$$ $$3^2-f(3)=9-7=2$$ $$7^2-f(7)=49-15=34$$ $$15^2-f(15)=225-31=194$$ $$31^2-f(31)=961-63=898$$