Trigonometry and Nested Radicals

Early last month, I was chatting with one of my officemates about a curious problem I had studied in high school. I hadn't written any of the results down, so much of the discussion involved me rediscovering the results and proving them with much more powerful tools than I once possessed.

Before writing about the problem I had played around with, I want to give a brief motivation. For as long as humans have been doing mathematics, finding values of \(\pi\) has been deemed worthwhile (or every generation has just found it worthwhile to waste time computing digits).

One such way the Greeks (particularly Archmides) computed \(\pi\) was by approximating a circle by a regular polygon and letting the number of sides grow large enough so that the error between the area of the unit circle (\(\pi \cdot 1^2\)) and the area of the polygon would be smaller than some fixed threshold. Usually these thresholds were picked to ensure that the first \(k\) digits were fully accurate (for some appropriate value of \(k\)).

In many introductory Calculus courses, this problem is introduced exactly when the limit is introduced and students are forced to think about the area problem in the regular polygon:

Given \(N\) sides, the area is \(N \cdot T_N\) where \(T_N\) is the area of each individual triangle given by one side of the polygon and the circumcenter.

Call one such triangle \(\Delta ABC\) and let \(BC\) be the side that is also a side of the polygon while the other sides have \(\left|AB\right| = \left|AC\right| = 1\) since the polygon is inscribed in a unit circle. The angle \(\angle BAC = \frac{2\pi}{N}\) since each of the triangles has the same internal angle and there are \(N\) of them. If we can find the perpendicular height \(h\) from \(AB\) to \(C\), the area will be \(\frac{1}{2} h \left|AB\right| = \frac{h}{2}\). But we also know that
\[\sin\left(\angle BAC\right) = \frac{h}{\left|AC\right|} \Rightarrow h = \sin\left(\frac{2\pi}{N}\right).\] Combining all of these, we can approximate \(\pi\) with the area:
\[\pi \approx \frac{N}{2} \sin\left(\frac{2\pi}{N}\right) = \pi \frac{\sin\left(\frac{2\pi}{N}\right)}{\frac{2 \pi}{N}}. \] As I've shown my Math 1A students, we see that
\[\lim_{N \to \infty} \pi \frac{\sin\left(\frac{2\pi}{N}\right)}{\frac{2 \pi}{N}} = \pi \lim_{x \to 0} \frac{\sin(x)}{x} = \pi\] so these are indeed good approximations.

Theory is Nice, But I Thought We Were Computing Something

Unfortunately for us (and Archimedes), computing \(\sin\left(\frac{2\pi}{N}\right)\) is not quite as simple as dividing by \(N\), so often special values of \(N\) were chosen. In fact, starting from \(N\) and then using \(2N\), the areas could be computed via a special way of averaging the previous areas. Lucky for us, such a method is equivalent to the trusty half angle identities (courtesy of Abraham De Moivre). To keep track of these polygons with a power of two as the number of sides, we call \(A_n = \frac{2^n}{2} \sin\left(\frac{2\pi}{2^n}\right)\).

Starting out with the simplest polygon, the square with \(N = 2^2\) sides, we have
\[A_2 = 2 \sin\left(\frac{\pi}{2}\right) = 2.\] Jumping to the octagon (no not that "The Octagon"), we have
\[A_3 = 4 \sin\left(\frac{\pi}{4}\right) = 4 \frac{\sqrt{2}}{2} = 2 \sqrt{2}.\] So far, the toughest thing we've had to deal with is a \(45^{\circ}\) angle and haven't yet had to lean on Abraham (him, not him) for help. The hexadecagon wants to change that:
\[A_4 = 8 \sin\left(\frac{\pi}{8}\right) = 8 \sqrt{\frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}} = 8 \sqrt{\frac{2 - \sqrt{2}}{4}} = 4 \sqrt{2 - \sqrt{2}}.\]

To really drill home the point (and motivate my next post) we'll compute this for the \(32\)-gon (past the point where polygons have worthwhile names):

\[A_5 = 16 \sin\left(\frac{\pi}{16}\right) = 16 \sqrt{\frac{1 - \cos\left(\frac{\pi}{8}\right)}{2}}.\] Before, we could rely on the fact that we know that a \(45-45-90\) triangle looked like, but now, we come across \(\cos\left(\frac{\pi}{8}\right)\), a value which we haven't seen before. Luckily, Abraham has help here as well:
\[\cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{1}{2} \sqrt{2 + \sqrt{2}}\] which lets us compute
\[A_5 = 16 \sqrt{\frac{1 - \frac{1}{2} \sqrt{2 + \sqrt{2}}}{2}} = 8 \sqrt{2 - \sqrt{2 + \sqrt{2}}}.\]

So why have I put you through all this? If we wave our hands like a magician, we can see this pattern continues and for the general \(n\)

\[A_n = 2^{n - 2} \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{\cdots + \sqrt{2}}}}}\]

where there are \(n - 3\) nested radicals with the \(\oplus\) sign and only one minus sign at the beginning.

This motivates us to study two questions, what is the limiting behavior of such a nested radical:

\[\sqrt{2 + s_1 \sqrt{2 + s_2 \sqrt{ \cdots }}}\] as the signs \(s_1, s_2, \ldots\) takes values in \(\left\{-1, 1\right\}\). Recasting in terms of the discussion above, we want to know how close we are to \(\pi\) as we increase the number of sides.

When I was in high school, I just loved to nerd out on any and all math problems, so I studied this just for fun. Having heard about the unfathomable brain of Ramanujan and the fun work he had done with infinitely nested radicals, I wanted to examine which sequences of signs \((s_1, s_2, \ldots)\) produced an infinite radical that converged and what the convergence behavior was.

I'm fairly certain my original questions came from an Illinois Council of Teachers of Mathematics (ICTM) contest problem along the lines of

\[\text{Find the value of the infinite nested radical } \sqrt{2 + \sqrt{2 + \cdots}}\] or maybe the slightly more difficult \[\text{Find the value of the infinite nested radical } \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \cdots}}}}.\] Armed with my TI-83, I set out to do some hardcore programming and figure it out. It took me around a month of off-and-on tinkering. This second time around as a mathematical grown-up, it took me the first half of a plane ride from SFO to Dallas.

In the next few weeks/months, I hope to write a few blog posts, including math, proofs and some real code on what answers I came up with and what other questions I have.

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Continued fraction expansions of irrational square roots

I had no idea (until this Thursday, July 16) that I had never seen a proof of the fact that the continued fraction expansion of \(\sqrt{D}\) is periodic whenever \(D\) is not a perfect square. But have no fear, I found out about something called a reduced quadratic irrational and now have a proof. Here we go.

Definition: An irrational root \(\alpha\) of a quadratic equation with integer coefficients is called reduced if \(\alpha > 1\) and its conjugate \(\tilde{\alpha}\) satisfies \(-1 < \tilde{\alpha} < 0\). \(\Box\)

Solutions (since assumed real) of such quadratics can be written as $$\alpha = \frac{\sqrt{D} + P}{Q}$$ where \(D, P, Q \in \mathbf{Z}\) and \(D, Q > 0\). It is also possible (though not required) to ensure that \(Q\) divides \(D - P^2\). This is actually a necessary assumption for some of the stuff I do, is mentioned here and generally frustrated the heck out of me, so that. As an example for some enlightenment, notice $$\alpha = \frac{2 + \sqrt{7}}{4}$$ is reduced but \(4\) does not divide \(7 - 2^2\). However, if we write this as \(\frac{8 + \sqrt{112}}{16}\), we have our desired condition.

Definition: We say a reduced quadratic irrational \(\alpha\) is associated to \(D\) if we can write $$\alpha = \frac{P + \sqrt{D}}{Q}$$ and \(Q\) divides \(D - P^2\). \(\Box\)


Lemma 1: Transforming a reduced irrational root \(\alpha\) associated to \(D\) into its integer part and fractional part via $$\alpha = \lfloor \alpha \rfloor + \frac{1}{\alpha'},$$ the resulting quadratic irrational \(\alpha'\) is reduced and associated to \(D\) as well. (This is what one does during continued fraction expansion, and as I did with \(\sqrt{2}\) during my last post.)

Proof: Letting $$\alpha = \frac{\sqrt{D} + P}{Q}$$ and \(X = \lfloor \alpha \rfloor\), we have $$\frac{1}{\alpha'} = \frac{\sqrt{D} - (QX - P)}{Q}.$$

• Since \(\sqrt{D}\) is irrational, we must have \(\frac{1}{\alpha'} > 0\) and since \(\frac{1}{\alpha'}\) is the fractional part we know $$0 < \frac{1}{\alpha'} < 1 \Rightarrow \alpha' > 1.$$ 
• Transforming $$\alpha' = \frac{Q}{\sqrt{D} - (QX - P)} \cdot \frac{\sqrt{D} + (QX - P)}{\sqrt{D} + (QX - P)} = \frac{\sqrt{D} + (QX - P)}{\frac{1}{Q}\left(D - (QX - P)^2\right)},$$ we have \(P' = QX - P\) and \(Q' = \frac{1}{Q}\left(D - (QX - P)^2\right)\) and need to show \(Q' \in \mathbf{Z}\). But \(D - (QX - P)^2 \equiv D - P^2 \bmod{Q}\) and since \(\alpha\) is associated to \(D\), \(Q\) must divide this quantity, hence \(Q'\) is an integer.
• Since \(X = \lfloor\frac{\sqrt{D} + P}{Q}\rfloor\) is an integer and \(\alpha\) is irrational, we know \(X < \frac{\sqrt{D} + P}{Q}\) hence \(P' = QX - P < \sqrt{D}\) forcing \(\tilde{\alpha}' < 0\).
• Since \(\alpha > 1\) we know \(X \geq 1 \Leftrightarrow 0 \leq X - 1\). Thus \begin{align*}\tilde{\alpha} = \frac{P - \sqrt{D}}{Q} &< 0 \leq X - 1 \\ \Rightarrow Q &< \sqrt{D} + (QX - P) \\ \Rightarrow Q(\sqrt{D} - (QX - P)) &< D - (QX - P)^2 \\ \Rightarrow -\tilde{\alpha}' = \frac{\sqrt{D} - (QX - P)}{\frac{1}{Q}\left(D - (QX - P)^2\right)} &< 1 \end{align*} hence \(\tilde{\alpha}' > -1\) and \(\alpha'\) is reduced.
• Since \(Q' = \frac{1}{Q}\left(D - (P')^2\right)\), we know $$D - (P')^2 \equiv Q Q' \equiv 0 \bmod{Q'}$$ hence \(\alpha'\) is associated to \(D\).

Thus \(\alpha'\) is both reduced and associated to \(D\). \(\Box\)


Lemma 2: There are finitely many reduced quadratic irrationals associated to a fixed \(D\).

Proof: As above write an arbitrary reduced irrational as \(\alpha = \frac{\sqrt{D} + P}{Q}\).  Since \(\alpha > 1\) and \(\tilde{\alpha} > -1\), we know \(\alpha + \tilde{\alpha} = \frac{2P}{Q} > 0\) hence with the assumption \(Q > 0\) we have \(P > 0\). Since \(\tilde{\alpha} < 0\) we also have \(P < \sqrt{D}\). Also, since \(\alpha > 1\) by assumption we have \(Q < P + \sqrt{D} < 2\sqrt{D}\) thus there are finitely many choices for both \(P\) and \(Q\), forcing finitely many reduced quadratic irrationals associated to a fixed \(D\) (this amount is strictly bounded above by \(2D\)). \(\Box\)

Claim: The continued fraction expansion of \(\sqrt{D}\) is periodic whenever \(D\) is not a perfect square.

Proof: We'll use Lemma 1 to establish a series of reduced quadratic irrationals associated to \(D\) and then use Lemma 2 to assert this series must repeat (hence be periodic) due to the finite number of such irrationals.

Write \(a_0 = \lfloor \sqrt{D} \rfloor\) and \(\sqrt{D} = a_0 + \frac{1}{\alpha_0}\). From here, we will prove

• \(\alpha_0\) is a reduced quadratic irrational associated to \(D\).
• By defining \(a_{i+1} = \lfloor \alpha_i \rfloor\) and \(\alpha_i = a_{i + 1} + \frac{1}{\alpha_{i + 1}}\), \(\alpha_{i + 1}\) is also a reduced quadratic irrational associated to \(D\) (assuming all \(\alpha\) up until \(i\) are as well).

Since \(\frac{1}{\alpha_0}\) is the fractional part of the irrational \(\sqrt{D}\), we have $$0 < \frac{1}{\alpha_0} < 1 \Rightarrow \alpha_0 > 1.$$ By simple algebra, we have $$\alpha_0 = \frac{a_0 + \sqrt{D}}{D - a_0^2}, \qquad \tilde{\alpha_0} = \frac{a_0 - \sqrt{D}}{D - a_0^2}.$$ Since \(a_0\) is the floor, we know \(a_0 - \sqrt{D} < 0 \Rightarrow \tilde{\alpha_0} < 0\). Since \(D \in \mathbf{Z} \Rightarrow \sqrt{D} > 1\) and \(\sqrt{D} > a_0\), we have $$1 < \sqrt{D} + a_0 \Rightarrow \sqrt{D} - a_0 < D - a_0^2 \Rightarrow a_0 - \sqrt{D} > -(D - a_0^2) \Rightarrow \tilde{\alpha_0} > -1.$$ Thus \(\alpha_0\) is a reduced quadratic irrational. Since \(P_0 = a_0\) and \(Q_0 = D - a_0^2 = D - P_0^2\), \(Q_0\) clearly divides \(D - P_0^2\) so \(\alpha_0\) is associated to \(D\) as well.

Following the recurrence defined, since each \(\alpha_i\) is a reduced quadratic irrational, each \(a_i \geq 1\). Also, by Lemma 1, each \(\alpha_{i + 1}\) is reduced and associated to \(D\) since \(\alpha_0\) is. By Lemma 2, we only have finitely many choices for these, hence there must be some smallest \(k\) for which \(\alpha_k = \alpha_0\). Since \(\alpha_{i + 1}\) is determined completely by \(\alpha_i\) we will then have \(\alpha_{k + j} = \alpha_j\) for all \(j > 0\), hence the \(\alpha_i\) are periodic. Similarly, as the \(a_i\) for \(i > 0\) are determined completely by \(\alpha_{i - 1}\), the \(a_i\) must be periodic as well, forcing the continued fraction expansion $$\sqrt{D} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \ddots}}$$ to be periodic.\(\Box\)

Update: I posted this on
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