# Evenly distributing points on a sphere

By Martin Roberts

How to distribute points on the surface of a sphere as evenly as possibly is an incredibly important problem in maths, science and computing, and mapping the Fibonacci lattice onto the surface of a sphere via equal-area projection is an extremely fast and effective approximate method to achieve this. I show that with only minor modifications it can be made even better.

Introduction

The problem of how to evenly distribute points on a sphere has a very long history and is one of the most studied problems in the mathematical literature associated with spherical geometry. It is of critical importance in many areas of mathematics, physics, chemistry including  numerical analysis, approximation theory, coding theory, crystallography, electrostatics, computer graphics, viral morphology to name just a few.

Unfortunately, with the exception of a few special cases (namely the platonic solids) it is not possible to exactly equally distribute points on the sphere. Furthermore, the solution to this problem is critically dependent on the criteria used to judge the uniformity. There are many criteria in use, and they  include:

• Packing and covering
• Convex hulls, Voronoi cells and Delaunay  triangles,
• Riesz $s$-energy kernels
• Cubature and Determinants

Repeating this point as it is crucial: there is usually no single optimal solution  to this question, because an optimal solution based on one criteria is often not an optimal point distribution for another.  For example, in this post, we will also find that optimising for packing does not necessarily produce an optimal convex hull, and vice-versa.

For sake of brevity, this post focuses on just two of these: the minimum packing distance and convex hull / Delaunay mesh measures (volume and area).

Section 1 will show how we can modify the canonical Fibonacci lattice to consistently produce a better packing distribution.

Section 2 will show how we can modify the canonical Fibonacci lattice that produces better convex hull measures (volume and surface area).

Section 1. Optimizing Packing Distance

This is often referred to as “Tammes’s problem” due to the botanist Tammes, who searched for an explanation of the surface structure of pollen grains. The packing criterion asks us to maximize the smallest neighboring distance among the $N$ points. That is,

$$d_N = \min_{i \neq j} | x_i – x_j |$$

This value decreases at a rate $~ 1/\sqrt{N}$, so it is useful to define the normalized distance, and also the asymptotic limit of the normalized distance as

$$d^*_N = \sqrt{N} d_N ,\quad \quad d^* = \lim_{N \rightarrow \infty} d^*_N$$

The Fibonacci Lattice

One very elegant solution is modeled after nodes appearing in nature such as the seed distribution on the head of a sunflower or a pine cone, a phenomenon known as spiral phyllotaxis. Coxeter demonstrated these arrangements are fundamentally related to the Fibonacci sequence, $F_k =\{1, 1, 2, 3, 5, 8, 13, …\}$ and the golden ratio $\phi = (1+\sqrt{5})/2$.

There are two similar definitions of the spherical Fibonacci lattice point set in the literature. The original one is strictly only defined for $N$ equal to one of the terms of the Fibonacci sequence, $F_m$ and is very well studied in number theory.

$$t_i = \left( \frac{i}{F_m}, \frac{i F_{m-1}}{F_m} \right) \quad \textrm{for }\; 0 \leq i \leq N-1$$

Whilst the second one generalises this to arbitrary $N$, and is used more frequently in computing:

$$t_i = \left( \frac{i}{N}, \frac{i }{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N \tag{1}$$

where

$$\phi = \frac{1+\sqrt{5}}{2} = \lim_{n \rightarrow \infty} \left( \frac{F_{n+1}}{F_n} \right)$$

These points sets map to the unit square $[0, 1]^2$ which can then be mapped to the sphere by the cylindrical equal area projection:

$$(x,y) \rightarrow (\theta, \phi) : \quad \left( \cos^{-1}(2x-1) – \pi/2, 2\pi y \right)$$

$$(\theta,\phi) \rightarrow (x,y,z) : \quad \left (\cos\theta \cos\phi, \cos \theta \sin \phi, \sin \theta \right)$$

You can find numerous versions of python code for this at “Evenly distributing points on a sphere“.

Even though spherical Fibonacci point sets are not the globally best distribution of samples on a sphere, (because their solutions do not coincide with the platonic solids for $n=4,6,8,12,20$), they yield excellent sampling properties and are extremely simple to construct in contrast to other more sophisticated spherical sampling schemes.

As the mapping from the unit square to the surface of the sphere is done via an area-preserving projection, one can expect that if the original points are evenly distributed then they will also quite well evenly distributed on the sphere’s surface. However, this does not mean that it is provably optimal.

This mapping suffers from two distinct but inter-related problems.

The first is that this mapping is area-preserving, not distance preserving. Given that in our case, our objective constraint is maximizing the minimum pairwise distance separation between points, then it is not guaranteed that such distance-based constraints and relationships will hold after the projection.

The second, and from a pragmatic point of view possibly the trickiest to resolve, is that the mapping has a two singularity points at each pole. Consider two points very close to the pole but 180 degrees different in longitude. On the unit square, (and also on the cylindrical projection) they would correspond to two points that are quite distant from each other, and yet when mapped to the surface of the sphere they could be joined be a very small great arc going over the north pole. It is this particular issue that makes many of the spiral mappings sub-optimal.

The spherical fibonacci spiral produced from equation 1, results in a value of $d_N^*$ for all $N$, and so $d^* = 2$.

Lattice 1

The more common version – especially in computing – which produces a better $d^*=3.09$, is:

$$t_i = \left( \frac{i+1/2}{N}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N-1 \tag{2}$$

It places the points in the midpoints of the intervals (aka the midpoint rule in gausssian quadrature), and so for $n=100$, the values of the first coordinate would simply be:

$$\{ \frac{0.5}{100},\frac{1.5}{100},\frac{2.5}{100},\ldots, \frac{97.5}{100},\frac{98.5}{100},\frac{99.5}{100} \}$$

Lattice 2.

A key insight to further improving on Equation 2, is to realize that the $d^*_N$ always corresponds to the distance between the points $t_0$ and $t_3$, which are at the poles. Thus, to improve $d_N$ the points near the poles should be positioned farther apart.

If we define the following distribution:

$$t_i(\varepsilon) = \left( \frac{i+1/2+ \varepsilon}{N+2\varepsilon}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N-1$$

The $d^*_N$-curves for various values. $\varepsilon=0$ (blue);  $\varepsilon=\frac{1}{2}$ (orange); $\varepsilon=\frac{3}{2}$ (green); and $\varepsilon=\frac{5}{2}$ (red).  One can see that $\varepsilon = \frac{5}{2}$ produces results close to the asymptotic results. That is, for $N>20$, compared to the canonical spherical Fibonacci lattice,  the following simple expression produces substantially better results of $d^* = 3.29$:

$$t_i = \left( \frac{i+3}{N+5}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N-1 \tag{3}$$

Thus for $n=100$, the values of the first coordinate would simply be:

$$\{ \frac{3}{105},\frac{4}{105},\frac{5}{105},\ldots, \frac{100}{105},\frac{101}{105},\frac{102}{105} \}$$

Lattice 3.

As stated earlier, one of the greatest challenges of the distributing points evenly on the sphere is that the optimal distribution critically depends on which objective function you use. It turns out that local measures such as $d_N^*$ are at times very “unforgiving” inasmuch as a single point in a suboptimal position can catastrophically reduce measure of the entire point distribution.

In our case, regardless of how large $N$ is, the $D_N^*$ is typically determined by the four points closest to each pole, especially $t_0$ and $t_3$. However, what is also known about this lattice is that the largest Voronoi polygon is at the pole. Thus, in trying to maximize $d_N$ by separating the initial polar points in the sequence, actually makes the void at the pole even larger!  Thus, we present an alternative to lattice 2 which is generally more preferable, as it does not exhibit such a large void near the poles.

It is almost identical to lattice 2 but with two differences. Firstly, it uses $\varepsilon = 11/2$ for $1 \leq i \leq n-2$. Secondly, in addition to these $n-2$ points, the first and final point is defined to be at each pole. That is,

$$t_0=(0,0); \; t_n = (1,0); \quad t_i = \left( \frac{i+6}{N+11}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 < i < N-1 \tag{3}$$

The amazing thing about this method of construction is that although it was motivated by desiring to minimize the gap at the poles, it  actually has the best $d_N$ and $d^*$ value of  all methods, with $d^* = 3.31$!

Thus for $n=100$, the values of the first coordinate would simply be:

$$\{ 0; \; \frac{6}{111},\frac{7}{111},\frac{8}{1111},\ldots, \frac{103}{111},\frac{104}{111},\frac{105}{111} ; \; 1\}$$

Comparison

For large $N$ this value of $d^*$ compares extremely well compared to other methods, such as geodesic domes, which are based on triangulated projections from the faces of platonic solids to the surface of the sphere. Not surprsingly, the best quality geodesic domes are those based on the icosahedron or the dodecahedron.

For various $N$ point geodesic domes based on the dodecahedron.

\begin{array}{|c|cccccccccc|} \hline N & 12 & 42 & 92 & 162 & 252& 362 & 492 & 642 & 812 & 1002  \\ \hline d^*  & 3.64 & 3.54 & 3.34 & 3.22 & 3.15 & 3.09 & 3.06 & 3.03 & 3.00 & 2.99 \\ \hline

\end{array}

And for various $N$ point geodesic domes based on the icosahedron.

\begin{array}{|c|ccccccc|} \hline N & 20 & 32 & 122 & 272 & 482 & 752 & 1082\\ \hline d^* & 3.19 & 3.63 & 3.16 & 2.99 & 2.90 & 2.84 & 2.81 \\ \hline

\end{array}

Also, the truncated icosahedron, which is the shape of a $C_{60}$ buckminster fullerene only corresponds to $d^* = 3.125$.

Thus for $N>100$, lattices based on Eqn 3 are better than any geodesic polyhedra.

As per the first reference,  some of the state-of-the-art methods which are typically complex and require recursive and/or dynamic programming are as follows.

\begin{array}{|lr|} \hline \text{Lattice 1} & 3.09\\ \hline \text{Max Determinant} & 3.19 \\ \hline \text{Lattice 2} & 3.28 \\ \hline \text{Lattice 3} & 3.31 \\ \hline \text{Zonal Equal Area} & 3.32 \\ \hline \text{Coulomb} & 3.37 \\ \hline \text{Log Energy} & 3.37\\ \hline

\end{array}

Section 1 Summary

Lattice 3, as per equation 3, is a modification of the canonical Fibonacci lattice that produces a significantly better packing point distribution. That is,

$$t_0 = (0,0); \; t_{N-1} = (0,1); \; \; t_i = \left( \frac{i+6}{N+11}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N-1$$

Section 2. Optimising the Convex hull (Delaunay mesh)

Although the previous section optimized for $d^*_N$, unfortunately these modifications actually make other measures worse, such as the volume of the convex hull (Delaunay mesh). This section shows how to evenly distribute points on a sphere in a manner that optimizes (maximizes) the a more global measure such as the volume of the convex hull.

Let us define, $C_N$ as the convex hull of the $N$ points,

$$\epsilon_N = N \left( \frac{4\pi }{3} – \textrm{Vol}(C_N) \right)$$

where the normalization factor of $N$ is included, as the absolute discrepancy decreases at a rate $~ 1/N$.

The behavior of $\epsilon_N$ for varying $N$ can be seen in Figure 3 (blue).

The key to improving the volume discrepancy is to note that although the use of $\phi$, the golden ratio intuitively makes sense as $N \rightarrow \infty$, it does not necessarily follow that it is the best value for finite $N$. In science terminology, we could say that need to consider finite-term correction effects.

Thus, let us generalize equation 1 as follows:

$$t_i = \left( \frac{i+1/2}{N}, \frac{i}{g(n)} \right) \quad \textrm{for }\; 0 \leq i \leq N-1 \tag{4}$$

First we define an auxillary value,

$$k = \left\lfloor \textrm{log}_{\phi}(\frac{n}{1.5}) \right\rfloor = \left\lfloor \frac{\ln (n/1.5)}{\ln \phi } \right\rfloor$$

where $\lfloor x \rfloor$ is the floor function.

Now define $g(n)$ as follows:

$$g(n) = \begin{cases} 3-\phi, & \text{if k is even} \\ \phi, & \text{if k is odd} \end{cases} \tag{5}$$

Figure 3 shows that this substantially improves the volume discrepancy for half the values of $N$.

The underlying reason why this works is based on the less well known fact that all numbers $x$ that satisfy the special Mobieus transformation are equivalent to $\phi$ in terms of irrationality.

$$x = \frac{a\phi+b}{c\phi+d}, \quad \textrm{for all integers} \; \; a,b,c,d \; \textrm{such that } |ad-bc|=1$$

And thus the connection why $\phi$ and $3-\phi$ work together so well , is that

$$\frac{1}{\phi} = \frac{\phi+1}{2\phi+1}, \quad \quad \frac{1}{3-\phi }= \frac{2\phi+1}{1\phi+1}$$

For the remaining half, we first define an auxiliary sequence $A_N$ that is variant of the Fibonacci sequence

$$A_1 =1, \; A_2 = 4; \; A_{n+2}= A_{n+1}+A_n \; \textrm{for } n = 1,2,3,…$$

That is,

$$A_N = 1,4,5,9,14,23,37,60,97,157,254,411,…$$

The convergents of this sequence all have elegant continued fractions, and in the limit converge to $\phi$. For example,

$$t _5/t_4 = 1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4}}}}$$.

We now fully generalize $g(n)$ as follows:

$$g(n) = \begin{cases} 3-\phi, & \text{if k is even} \\ A_{j+1}/A_j , & \text{if k is odd, where j= (k+7)/2} \end{cases} \tag{6}$$

The following table is a summary of the value of $g(n)$ for various $n$.

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline N & 4-6 & 7-10 & 11-16& 17-26& 27-43& 44-70& 71-114 & 115-184 & 185-300\\ \hline g(n)  &3-\phi & \frac{23}{14} & 3-\phi & \frac{37}{23} & 3-\phi & \frac{60}{37} & 3-\phi & \frac{97}{60} & 3-\phi  \\ \hline

\end{array}

Figure 4 shows that, in relation to convex hull volume, this new distribution is better than the canonical lattice for all values of $n$.

Figure 4. The discrepancy between the volume of the convex hull of points and the volume of a unit sphere. Note that smaller is better. This shows that the newly proposed method (green) produces consistently better distribution than the canonical Fibonacci lattice (blue).

Of interest, this distribution also slightly but consistently reduces the discrepancy between the surface area of the convex hull and the surface area of the unit sphere. This is shown in figure 6.

Section 2 Summary

The lattice as per equation 6, is a modification of the canonical Fibonacci lattice that produces a significantly better point distribution as measured by the volume and surface area of the convex hull (Delaunay Mesh).

Key References

• A Comparison of poopular point confugrations on S^2″: https://arxiv.org/pdf/1607.04590.pdf
• http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere
• https://perswww.kuleuven.be/~u0017946/publications/Papers97/art97a-Saff-Kuijlaars-MI/Saff-Kuijlaars-MathIntel97.pdf
• https://projecteuclid.org/download/pdf_1/euclid.em/1067634731
• https://maths-people.anu.edu.au/~leopardi/Leopardi-Sphere-PhD-Thesis.pdf
• https://www.irit.fr/~David.Vanderhaeghe/M2IGAI-CO/2016-g1/docs/spherical_fibonacci_mapping.pdf
• https://arxiv.org/pdf/1407.8282.pdf
• https://maths-people.anu.edu.au/~leopardi/Macquarie-sphere-talk.pdf

#### Page 2

Most of us know about the golden ratio, $$\phi = \frac{1+\sqrt{5}}{2}$$ and how it can be considered the most irrational number, but most people never ask the question, “What is the second and third most irrational number?” In this post, I discuss discuss some special irrational numbers, and why they can be considered the 2nd and 3rd most irrational numbers.

Finding good rational approximations to reals.

In number theory, the field of Diophantine approximation deals with the approximation of real numbers by rational numbers. It is named after Diophantus of Alexandria. The problem on how well a real number can be approximated by rational numbers is an extremely well studied problem .

Let us, say that the rational number $\frac{p}{q}$ is a good approximation of a real number $x$ , only if the absolute value of the difference between $a/b$ and $α$ can not be improved with any other rational rational number with a smaller denominator. That is,

$$| x-\frac{p}{q} | < | x-\frac{p’}{q’} | \quad \text{for every } p’,q’ \; \text{such that } 0< q'<q \tag{1}$$

Often, a very similar definition which is generally more amenable to study because it does not include fractions, is used

$$| p-q x | < |p’- q’ x | \quad \text{for every } p’,q’ \; \text{such that } 0< q'<q \tag{2}$$

Continued fractions can be used to compute the best rational approximations of a real number. This is because the successive convergents of the continued fraction $x$ produce the best approximations of $x$ possible, (as per the second definition above).

For example, for the constant $\pi= 3.14159265358…$, the regular continued fraction is:

$$\pi = 3 + \cfrac{1}{7+\cfrac{1}{15+\cfrac{1}{1+\cfrac{1}{292+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}}}}$$

Because the numerators for all regular continued fractions are always 1, we can succinctly describe this continued fraction simply by listing the denominators (and the initial whole number)

$$\pi = [3;7,15,1,292,1,1,1,2,1,3,1,\ldots]$$

These produce the following progressively accurate rational approximations of $\pi$.

$$\{ 3, \frac{22}{7}, \frac{333}{106},\frac{355}{113},\frac{103993}{33102}, \frac{104348}{33215}, \frac{208341}{66317}, \frac{312689}{99532},\frac{833719}{265381}, \ldots \}$$

From this, it is clear why $\frac{22}{7}$ is frequently used as an approximation of $\pi$, and why rationals more accurate than $\frac{355}{113}$ are almost never used.

How accurate are these approximations?

The obvious measure of the accuracy of a rational approximation of a real number $x$ by a rational number $p/q$, is $\epsilon = |x-p/q|$. However, this quantity can always be made arbitrarily small by increasing the absolute values of $p$ and $q$. The previous section showed how we can efficiently find $p,q$ that produce successively more and more accurate rational approximations to $x$.

Therefore, the accuracy of the approximation is usually estimated by comparing this quantity to some the size of the denominator $q$, typically a negative power of it.

The following table shows the same approximations of $\pi$ along with their error $\epsilon$ when multiplied by $q$ and by $q^2$.

$$\begin{matrix} p/q & \epsilon & \epsilon q & \epsilon q^2\\ \hline \frac{3}{1} & 1 \times 10^{-1} & 1 \times 10^{-1} & 0.142 \\ \hline \frac{22}{7} & 1 \times 10^{-3} & 9 \times 10^{-3} & 0.062 \\\hline \frac{333}{106} & 8\times 10^{-5} & 9 \times 10^{-3} & 0.935 \\\hline \frac{355}{113} & 3\times 10^{-7} & 3 \times 10^{-5} & 0.003 \\ \hline \frac{103993}{33102} & 6\times 10^{-10} & 2 \times 10^{-5} & 0.633 \\ \hline \frac{104348}{33215} & 3\times 10^{-10} & 1 \times 10^{-5} & 0.366 \\ \hline \frac{208341}{66317} & 1\times 10^{-10} & 8 \times 10^{-6} & 0.538 \\ \hline \frac{312689}{99532} & 3\times 10^{-11} & 3 \times 10^{-6} & 0.289 \\ \hline \frac{833719}{265381} & 9\times 10^{-12} & 2 \times 10^{-6} & 0.614 \\ \hline \ldots & \ldots & \ldots & \ldots \\ \hline \end{matrix}$$

From this one can see that the values in the columns $\epsilon$, as well as $\epsilon q$ become arbitrarily small, but those in the final $\epsilon q^2$ column , don’t.  Said another way, for good approximations, the error term will go down in proportion to square of the denominator.

Thus, it seems that by using the error term $\epsilon q^2$ we can more meaningfully compare how well each of these approximate  $\pi$. That is, of those listed, $355/133$ is by far best, followed by $22/7$, then $3/1$ then $312689/99532$, etc,…

Dirichlet’s Theorem

Noting that for an irrational number, the equivalent continued fractions are infinite, it means that we can produce an infinite number of convergent rational approximations. Looking at the table above, we see that for the first 10 convergents, $\epsilon q^2 < 1$. The question then naturally arises, “How often is $\epsilon q^2$?”

Dirichlet proved that for any irrational $x$, there are an infinite number of rational approximations such that $\epsilon < 1/q^2$. That is, there are an infinite number of fractions $p/q$ such that :

$$\left| x- \frac{p}{q} \right | < \frac{1}{q^2} \tag{3}$$

Furthermore, if $p/q$ are regular continued fraction convergents, then $\epsilon q^2$ will always be less than 1.

Although this property has  ‘long been known from the theory of continued fractions’, it is named after Dirichlet after he provided a very elegant proof of it in 1838.

Hurwitz’s Theorem

In 1891, Hurwitz showed that Dirichlet’s theorem could be improved. Furthermore, he proved that it was a strict upper bound. That is, it could not be improved any further.  That is, there are an infinite number of fractions $p/q$ such that :

$$\left| x- \frac{p}{q} \right | < \frac{1}{\sqrt{5} q^2} \tag{4}$$

Hurwitz showed that the reason that this is a strict upper bound and can not be improved is because if we consider the real number $x = (1+\sqrt{5})/2$, commonly called the golden ratio, and usually denoted $\phi$, then for $C> \sqrt{5}$, there are are only a finite number of rational numbers $p,q$ such that:

$$\left| \phi – \frac{p}{q} \right | < \frac{1}{C q^2} \tag{5}$$

The continued fraction representation of $\phi$ is simply $[1;1,1,1,1,1,1,1,1,\ldots]$ and its convergent fractions are:

$$\{ \frac{1}{1}, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}, \frac{21}{13}, \frac{34}{21},\ldots \}$$

The error terms $\sqrt{5}q^2 \epsilon$ for these convergents are,

$$1.38,0.854,0.1.056,0.978,1.008,0.996,1.001,0.9995,1.0001,0.999994,1.000002,0.999999,\ldots$$

We see that the error terms tend to 1, which is an indication that $\sqrt{5}q^2$ is the bound.

Equivalence Relations

It is very well known that the golden ratio $\phi$, can be considered the most irrational number. From the commentary above, we can see that this is most notably because it provides the limiting bound for Hurwitz’s theorem.  Technically speaking, the fact that the golden ratio has the continued fraction comprising only of ones, is a consequence rather than a rigorous reason why it is the most irrational number.

One lesser known fact is that there are infinitely many other real numbers that can be considered equally irrational, because they force strictness on the upper bound of the Hurwitz theorem. For example, all of the following numbers are equally irrational and thus could claim the title of ‘the most irrational number’:

$$\{ \frac{1}{ \phi+3}, \; \frac{ \phi-1}{\phi}, \; \frac{ \phi+1}{2 \phi+1} ,\; \frac{3 \phi+2}{4 \phi+3}, \; \frac{ \phi+2}{ \phi+3} ,\; \frac{3 \phi-2}{2 \phi-1},\; \frac{ \phi-4}{\phi-3},\; \ldots \}$$

There are an infinite number of them, and they are all considered equivalent, as they are all of the form

$$\frac{a \phi+b}{c \phi+d}, \quad \textrm{for integers } a,b,c,d \;\; \textrm{such that } ad-bc=\pm 1. \tag{6}$$

The continued fractions for these values are:

$$\frac{1}{\phi+3} \simeq 0.216542 = [0; 4,1,1,1,1,1,1,1,\ldots]$$

$$\frac{\phi-1}{\phi} \simeq 0.216542 = [0; 2,1,1,1,1,1,1,1,\ldots]$$

$$\frac{\phi+1}{2\phi+1} \simeq 0.619034 = [0; 1,1,1,1,1,1,1,1,\ldots]$$

$$\frac{3\phi+2}{4\phi+3} \simeq 0.216542 = [0; 1,2,1,1,1,1,1,1,\ldots]$$

$$\frac{\phi+2}{\phi+3} \simeq 0.783458 = [0; 1,3,1,1,1,1,1,1,\ldots]$$

$$\frac{3\phi-2}{2\phi-1} \simeq 1.276390 = [1; 3,1,1,1,1,1,1,1,\ldots]$$

Serrat showed that with the exception of a finite initial sequence,  equivalent fractions have identical continued fraction expansions.

So although all of the numbers in this class are equally irrational, it is reasonable to select from all of those, the number whose continued fraction is the simplest to be the canonical number in this class. Namely, $\phi = [1;1,1,1,1,1,\;ldots]$.

The second most irrational number

If we consider all the real numbers excluding those of the form described in Eqn 6, then it turns out that Hurwitz’s theorem may be made even stronger still. That is, if $x$ is not equivalent to $\phi = (1+\sqrt{5})/2$, then there are an infinite number of fractions $p/q$ such that :

$$\left| x- \frac{p}{q} \right | < \frac{1}{\sqrt{8} q^2} \tag{4}$$

Again, this inequality is strong, because for any number equivalent to $x = \sqrt{2}$, the value of $C=\sqrt{8}$ can not be improved.

Thus the second most irrational class of numbers are those that are equivalent to $x=\sqrt{2}$.

The continued fraction for $\sqrt{2}$ is $[1;2,2,2,2,2,2,\ldots]$,

therefore using Serrat’s theorem, the number $1+\sqrt{2} = [2;2,2,2,2,….]$ is equivalently irrational.

Thus, we can say that the second most irrational number is $1+\sqrt{2}$.

The third most irrational number

Repeating this line of argument again, if we exclude all the real numbers that are equivalent to $\phi$, or $1+\sqrt{2}$, then Hurwitz’s theorem can be made even stronger.

That is, if $x$ is not equivalent to $\phi = (1+\sqrt{5})/2$ or $1+\sqrt{2}$, then there are an infinite number of fractions $p/q$ such that :

$$\left| x- \frac{p}{q} \right | < \frac{1}{C q^2} \quad \textrm{where } C=\sqrt{221}/5. \tag{4}$$

Again, this inequality is strong, because for any number equivalent to $x = (9+\sqrt{221})/10$, the value of $C=\sqrt{221}/5$ can not be improved.

Thus the second most irrational class of numbers are those that are equivalent to $x=\sqrt{2}$.

The continued fraction for $(9+\sqrt{221})/10$ is $[2;2,1,1,2,2,1,1,2,1,1,2,2,1,1,2,\ldots]$,

Thus we can say that the third most irrational number is $(9+\sqrt{221})/10$.

In a similar fashion, we can say:

• the fourth most irrational number is $(23+\sqrt{1517})/26 = [2; 2,1,1,1,1,2,2,1,1,1,1,2 ,…]$
• the fifth most irrational number is $(5+\sqrt{7565})/58 = [1; 1,1,2,2,2,2,1,1,2,2,2,2,…]$
• and so on…

Markov Spectrum

Recalling equation 5, the first class of irrationals are those that enforce the upper bound when $C=\sqrt{5}$, and those relating to the second class correspond to the bound where $C=\sqrt{8}$, etc.

The question arises, ‘what is the pattern?’. The answer is these numbers  form the Lagrange / Markov spectrum, and are all of the form

$$L_n = \sqrt{9-\frac{4}{m_n^2}},$$ where $m_n$ is the n-th term of the Markov sequence.  The first few terms of the Markov sequence are $\{ 1,2,5,13,29,34,89,169,…\}$.

Thus, $L_n = \{ \sqrt{5}, \sqrt{8}, \sqrt{221}/5, \sqrt{1517}/13, … \}$

References

Much of this post was based on a wonderful paper “The hyperbolic geometry of Marjov’s theorem on Diophanitne approximation and quadratic forms” by Boris Springborn.