# Is there a pattern to the roots of continued-fraction equations with depth $n$?

I've found what looks like the start of a pattern, but I don't have the tools or background to know if one really exists here. Does anyone know how to analyze this mathematical behavior and continue the pattern?

Consider the continued fraction $$f_n(x) = [0; x, x, x, ...]$$, a zero followed by $$n$$ $$x$$s. This represents the continued fraction

$$f_n(x) = 0 + \frac{1}{x + \frac{1}{x + \frac{1}{x + ...}}}$$

with $$n$$ layers of depth. Thus

• for $$n = 0$$ we have $$f_0(x) = 0$$,
• for $$n = 1$$ we have $$f_1(x) = 0 + \frac{1}{x} = \frac{1}{x}$$,
• for $$n = 2$$ we have $$f_2(x) = 0 + \frac{1}{x + \frac{1}{x}} = \frac{x}{x^2 + 1}$$,
• for $$n = 3$$ we have $$f_3(x) = 0 + \frac{1}{x + \frac{1}{x + \frac{1}{x}}} = \frac{x^2 + 1}{x^3 + 2 x}$$,

and so on.

If we include a single complex point at infinity and claim that $$\frac{1}{∞}$$ is zero, $$f_n(x)$$ has $$n$$ roots, all of which lie on the imaginary axis. All roots are symmetrical about the real axis. These roots seem to follow patterns related to the Beraha Constants, $$B(n) = 2 + 2 \cos\left(\frac{2 π}{n}\right)$$, at least for the first several $$n$$s.

The sets of roots for $$n$$ from 1 to 10, in terms of $$B(n)$$ where I can figure out the connection, are as follows:

• $$n$$ = 1: ∞
• $$n$$ = 2: ∞, $$i \sqrt{B( 2)}$$
• $$n$$ = 3: ∞, $$±i \sqrt{B( 3)}$$
• $$n$$ = 4: ∞, $$±i \sqrt{B( 4)}$$, $$i (B( 4) - 2)$$
• $$n$$ = 5: ∞, $$±i \sqrt{B( 5)}$$, $$±i (B( 5) - 2)$$
• $$n$$ = 6: ∞, $$±i \sqrt{B( 6)}$$, $$±i (B( 6) - 2)$$, 0
• $$n$$ = 7: ∞, $$±i \sqrt{B( 7)}$$, $$±i (B( 7) - 2)$$, $$±i \frac{1}{1 + 2 \cos\left(\frac{2 π}{7}\right)}$$
• $$n$$ = 8: ∞, $$±i \sqrt{B( 8)}$$, $$±i (B( 8) - 2)$$, $$±i \sqrt{2 - \sqrt{2}}$$, 0
• $$n$$ = 9: ∞, $$±i \sqrt{B( 9)}$$, $$±i (B( 9) - 2)$$, $$±i, ±2 i \sin\left(\frac{π}{18}\right)$$
• $$n$$ = 10: ∞, $$±i \sqrt{B(10)}$$, $$±i (B(10) - 2)$$, $$±i \sqrt{\frac{5 - \sqrt{5}}{2}}$$, $$±i \sqrt{\frac{3 - \sqrt{5}}{2}}$$, 0

The first set of roots for each value of $$n$$ greater than 1 has a magnitude of $$\sqrt{B(n)}$$. The second set for $$n$$ greater than 3 has a magnitude of $$B(n) - 2$$. The third set for $$n$$ greater than 5 and the fourth set for $$n$$ greater than 7 seem to be following a similar structure, starting at zero and then rising towards an upper limit with a shallower slope each time, but I haven't been able to figure out a relation to the Beraha Constants for them, and I haven't been able to see any meta-pattern in the relations with only two instances of such a a relation to build off of.

So my question is this: is there a pattern to the roots beyond $$\sqrt{B(n)}$$ and $$B(n) - 2$$? Some way to know the roots of $$f_n(x)$$ for arbitrary values of $$n$$ without needing to calculate a continued fraction dozens of layers deep (equivalent to a function of degree $$n$$, which quickly becomes unmanageable)?

At minimum I'd like an equation $$g(B(n))$$ that gives the values $$0$$, $$\frac{1}{1 + 2 \cos\left(\frac{2 π}{7}\right)}$$, $$\sqrt{2 - \sqrt{2}}$$, $$1$$, and $$\sqrt{\frac{5 - \sqrt{5}}{2}}$$ for $$n$$ from 6 to 10, and an equation $$h(B(n))$$ that gives the values $$0$$, $$2 \sin\left(\frac{π}{18}\right)$$, and $$\sqrt{\frac{3 - \sqrt{5}}{2}}$$ for $$n$$ from 8 to 10, plus an explanation of how they can be derived, since I haven't been able to figure that out myself. Ideally, the procedure for finding those equations would extend to additional roots for higher values of $$n$$, allowing arbitrary roots to be found.

Based on the answer from @dxiv, the magnitude of the roots can more correctly be written in the form $$R(n, k) = 2 \cos\left(\frac{k π}{n}\right)$$, where $$n$$ is the depth of the continued fraction function and $$k$$ is the index of the root counting from largest (one) to smallest ($$\lfloor\frac{n}{2}\rfloor$$). This just happens to be equal to $$\sqrt{B(n)}$$ for $$k = 1$$ and equal to $$B(n) - 2$$ for $$k = 2$$. Thus the roots for all values of $$n$$ up to ten, ordered from largest to smallest, are as follows:

• $$n$$ = 1: ∞
• $$n$$ = 2: ∞, 0
• $$n$$ = 3: ∞, $$±2 i \cos\left(\frac{π}{ 3}\right)$$
• $$n$$ = 4: ∞, $$±2 i \cos\left(\frac{π}{ 4}\right)$$, 0
• $$n$$ = 5: ∞, $$±2 i \cos\left(\frac{π}{ 5}\right)$$, $$±2 i \cos\left(\frac{2 π}{ 5}\right)$$
• $$n$$ = 6: ∞, $$±2 i \cos\left(\frac{π}{ 6}\right)$$, $$±2 i \cos\left(\frac{2 π}{ 6}\right)$$, 0
• $$n$$ = 7: ∞, $$±2 i \cos\left(\frac{π}{ 7}\right)$$, $$±2 i \cos\left(\frac{2 π}{ 7}\right)$$, $$±2 i \cos\left(\frac{3 π}{ 7}\right)$$
• $$n$$ = 8: ∞, $$±2 i \cos\left(\frac{π}{ 8}\right)$$, $$±2 i \cos\left(\frac{2 π}{ 8}\right)$$, $$±2 i \cos\left(\frac{3 π}{ 8}\right)$$, 0
• $$n$$ = 9: ∞, $$±2 i \cos\left(\frac{π}{ 9}\right)$$, $$±2 i \cos\left(\frac{2 π}{ 9}\right)$$, $$±2 i \cos\left(\frac{3 π}{ 9}\right)$$, $$±2 i \cos\left(\frac{4 π}{ 9}\right)$$
• $$n$$ = 10: ∞, $$±2 i \cos\left(\frac{π}{10}\right)$$, $$±2 i \cos\left(\frac{2 π}{10}\right)$$, $$±2 i \cos\left(\frac{3 π}{10}\right)$$, $$±2 i \cos\left(\frac{4 π}{10}\right)$$, 0

Let $$\,f_n = \frac{p_n}{q_n}\,$$ then for $$\,n \ge 1\,$$:

$$\frac{p_{n+1}}{q_{n+1}}= f_{n+1} = \frac{1}{x+f_n}=\frac{1}{x+\frac{p_n}{q_n}}=\frac{q_n}{xq_n+p_n} \;\;\implies\;\;\\ \begin{cases}p_{n+1}=q_n \\ q_{n+1}=xq_n+p_n=xq_n+q_{n-1}\end{cases}$$

It follows that $$\,p_{n+1}\,$$ satisfies $$\,p_{n+1}=xp_n+p_{n-1}\,$$ with $$\,p_0(x)=0, \,p_1(x)=1, \,p_2(x)=x\,$$.

Let $$\,p_n(x)=i^n \,r_n\left(\frac{x}{2i}\right)\,$$, then after substituting and canceling a factor of $$\,i^{n+1}\,$$:

$$r_{n+1}\left(\frac{x}{2i}\right) = 2\,\frac{x}{2i} r_n\left(\frac{x}{2i}\right) - r_{n-1}\left(\frac{x}{2i}\right) \;\;\implies\;\; \\r_{n+1}(z) = 2z\, r_n(z)-r_{n-1}(z)$$

This is the same recurrence satisfied by the Chebyshev polynomials. Moreover:

\begin{align} r_1(z) &= i^{-1}\,p_1(2iz) = -i \cdot 1 \\ r_2(z) &= i^{-2}\,p_2(2iz) = -i \cdot 2z \end{align}

Aside from an index shift of $$\,1\,$$, these match the initial conditions for the Chebyshev polynomials of the second kind $$\,U_0(z)=1\,$$, $$\,U_1(z)=2z\,$$ scaled by a factor of $$\,-i\,$$, so $$\,r_{n+1}(z)=-i\,U_n(z)\,$$.

The roots of $$\,r_{n}(z)\,$$ are therefore the roots of $$\,U_{n-1}(z)\,$$, known to be $$\,z_k=\cos \frac{k\pi}{n} \;\big|_{k = 1,2,\dots,n-1}\,$$ and the roots of $$\,p_{n}(x)\,$$ are $$\,x_k = 2i\,z_k\,$$, which are also the zeros of $$\,f_{n}\,$$.

[ EDIT ] $$\;$$ The relation with Beraha constants follows from trig identities between their definition $$B(n) = 2\left(1 + \cos \frac{2 \pi}{n}\right)$$ and the expression for the roots $$\,x_k = 2i \, \cos \frac{k\pi}{n} \;\big|_{k = 1,2,\dots,n-1}\,$$.

• $$x_1^2 = x_{n-1}^2 = -4 \cos^2 \frac{\pi}{n} = -2\left(\cos \frac{2\pi}{n}+1\right) = -B(n)$$

• $$x_2 = -x_{n-2} = 2i \,\cos\frac{2 \pi}{n} = i \,\big(B(n) - 2\big)$$