Gelfand’s Question and Poncelet’sClosure Theorem

Bachelor thesis in MathematicsAugust 28, 2013

Student: Jaap EisingFirst supervisor: Prof. Dr. J. TopSecond supervisor: Prof. Dr. H. Waalkens

Abstract

In this thesis the underlying systems of Poncelet’s Closure theorem and Gelfand’squestion are shown to be closely related to rotations on a circle. Also, solutionsto Gelfand’s questions are given: It is shown that in Gelfand’s system no rowsof equal digits can appear, and that the row ‘23456789’ can not reappear afterthe first row.

1

Introduction

Part of the beauty of mathematics is in discovering that two very dissimilar prob-lems have, in fact, the same underlying mechanisms. This paper will describethe relations between two problems that have this property. These problemsare Poncelet’s closure theorem and Gelfand’s question. The first of these is atheorem about tangents to conic sections, and the second is a series of questionsabout the first digit of powers of integers. The relations between both problemsis not a new discovery, it is in fact the subject of a paper by J. L. King, whichcan be found in [1]. This paper was the starting point of this present work. Theaim of this thesis is to make the relations given by King more explicit, and toactually answer the questions posed by Gelfand.

Poncelet’s Closure Theorem

To state Poncelet’s theorem we will first require the notion of a Poncelet trans-verse.

Definition 1. For any two smooth conic sections C1 and C2 in the real pro-jective plane (RP2), starting position x0 ∈ C1 and starting line segment l0 =[x0 x1] tangent to C2, the Poncelet transverse is the union of line segmentsli = [xi xi+1] tangent to C2.

A Poncelet transverse closes up if for some n, xn = x0, thus forming ann-gon. This leads us to stating Poncelet’s Theorem.

Theorem 1. If a Poncelet transverse of C1 and C2 closes up in n steps fora certain starting point that is not in C1 ∩ C2, it closes up in n steps for allstarting points x ∈ C1.

**plaatje**

Gelfand’s question

To properly define the system from which Gelfand’s question arises, we requirea couple

Definition 2. For each x ∈ R>0 we can take ai ∈ {0, 1, ...9} such that x =∞∑n=k

an10−n, with ak 6= 0. This is called a decimal expansion of x.

Because 0.999... = 1.00... this decimal expansion is not always unique. If inthese cases we only allow the commonly preferred expansion ending in ai = 0for i > N , we obtain a unique expansion for each x ∈ R>0.

This allows us to define a ‘take the first digit’-operator on the multiplicationgroup (R>0, ·, 1), which is given by:

〈〈x〉〉 : R>0 → {1, 2, ..., 9}x 7→ ak

An example of this is 〈〈10〉〉 = 1, or 〈〈π〉〉 = 3

2

This operator can be made explicit by viewing it as a two step process: Firstdivide by the highest power of ten, to shift the decimal point to the first digit,and then floor the result. This gives us a direct expression

〈〈x〉〉 = bx · 10−blog10 xcc.

Gelfand’s question arises from a system defined by taking the first decimaldigits of nth powers of integers. At the first step the numbers 2, 3, ..., 9 are con-sidered. Then, for step two the first decimal digits of 22, 32, ...92 are considered,and continuing for cubes and nth powers. This can be shown in a matrix definedby (yi,j) = 〈〈(yi,1)j〉〉, which is shown beneath here for the first 10 values.

1 2 3 4 5 6 7 8 92 4 9 1 2 3 4 6 83 8 2 6 1 2 3 5 74 1 8 2 6 1 2 4 65 3 2 1 3 7 1 3 56 6 7 4 1 4 1 2 57 1 2 1 7 2 8 2 48 2 6 6 3 1 5 1 49 5 2 2 2 1 4 1 310 1 5 1 9 6 2 1 3

An extension of this table up to a hundred rows can be found in appendixC on page 21. Regarding this table, Gelfand’s question is actually a series ofquestions, most of which are related to the frequencies and ergodicity in thissystem.

1. Will a ‘9’ ever occur in the column of 2n?

2. Will the row ‘23456789’ ever occur again? If so, will it have a frequency?

3. Will a row of the same numbers appear?

4. Will the decimal expansion of an 8-digit prime ever occur?

The first of these is rather easy to solve: Yes it does, for the first time at n = 53.The fourth one is less obvious, but knowing that 23456789 and 21443183 areprime numbers, gives us rows 1 and 11 as examples.

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1 Rotations on circles

One of the central mechanisms in this thesis will be rotations on a circle. Theoperation on the circle group can be described in many ways, rotations over anangle or multiplication on the complex plane. A description analogue to rotationis by parametrizing the circle by the interval K=[0,1), using ths isomorphismt 7→ (cos 2πt, sin 2πt). On this interval we can define the operation ⊕ as additionmodulo 1. This makes a rotation of 2πα equal to adding α modulo one.

An n-torus is the Cartesian product of n circles, and thus can be viewed asa hypercube [0,1)× . . .×[0,1). Therefore n simultaneous rotations on circles canbe viewed as rotations on a torus. An example is rotating by 2π

3 on one circle,while rotating by π on another. This can be seen as the mapping(

xy

)7→(x⊕ 1

3y ⊕ 1

2

)on K×K, which ends up at the starting position after 6 iterations.

According to a theorem by L.Kronecker[3] the orbit of simultaneous rotationsover 2πx1, . . . , 2πxn on an n-torus is uniformly distributed over the closure ofthe orbit. This is the entire torus if and only if 1, x1, . . . , xn are rationallyindependent.

Let E be an elliptic curve over R defined by: η2 = f(λ) with f(λ) a thirddegree polynomial with exactly one root, α. Then the group E(R) is isomorphicto the circle group,K = [0, 1), by the isomorphism ϕ:

E(R) → K(η, λ) 7→ ϕ(η, λ)

ϕ(η, λ) =

{1− g(λ) if η < 0

g(λ) if η ≥ 0, g(λ) =

∞∫λ

dt√f(t)

2∞∫α

dt√f(t)

A proof of this statement can be found in [2].This isomorphism can be visualized by putting the point (α, 0) at the left-

most point of a circle and then ‘folding’ the elliptic curve around it, putting thepoint at infinity (O) at the rightmost point of the circle.

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2 Gelfand’s question and rotations on a circle

In this section the relation between Gelfand’s system and rotations on a circleare shown. This relation is then used to answer Gelfand’s questions.

2.1 Isomorphism to Rotations

As the first digits of x and 10·x are the same We can define the equivalence rela-tion ∼ on R>0 as x ∼ y ⇔ (∃n ∈ Z such that x = y · 10n). As 10Z is a subgroupof R>0, we can define the factorgroup A = R>0/10Z as the set of all equiv-alence classes x̂ = {y ∈ R>0|y ∼ x}. This construction is similar to modulararithmetic, except using the multiplication operator instead of addition.

If we view the interval K = [0, 1), with addition modulo 1 (⊕), as a circlewe can construct an isomorphism between A and K by beginning with the(surjective) mapping:

ψ : R>0 −→ K = [0, 1)x 7−→ log10 x (mod 1)

(1)

This is a group homomorphism because

ψ(xy) = log10 xy (mod 1) = log10 x⊕ log10 y

Using the first isomorphism theorem and the fact that 10Z is the kernel of ψ Kis isomorphic A.

Due to the constructions of both the ‘take the first digit’-operator and thegroup A we know x ∼ y ⇒ 〈〈x〉〉 = 〈〈y〉〉. As k ∈ K implies that 10k ∈ [1, 10) wecan simplify 〈〈10k〉〉 to simply b10kc.

R>0x 7→xn−−−−→ R>0

x 7→ x̂ ↓ x7→〈〈x〉〉 ↘A x̂ 7→x̂n−−−−→ A x̂ 7→〈〈x〉〉−−−−−→ {1, 2, ..., 9}

x̂ 7→ ψ(x) ↓ k 7→ 1̂0k ↑ k 7→b10kc ↗K k 7→n·kmod1−−−−−−−−→ K

This means that multiplication in A corresponds to rotation (addition modulo1) on a circle, therefore making Gelfand’s system correspond to rotations on ann-torus. This can be visualized as rotating numbered points around a circle, wewill show this for n =1, 2, 3, 4:

2.2 Results on Gelfand’s Question

Using the isomorphism from the last section, calculations on Gelfand’s Systemcan be easily executed, as for each entry only the log10 x (mod 1), has to bestored. We can do this in an ‘underlying’ array for Gelfand’s system. We callthe nth element of the column starting with log10 α: xα,n

xα,n = n log10 α (mod 1)= xα,n−1 + log10 α (mod 1);

xα,1 = log10 α (mod 1).

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The array generated by simultaneously taking 2,3,. . . ,9 for α is directly re-lated to Gelfand’s system by taking

yα,n = 〈〈αn〉〉 = b10xα,nc⇒ yα,n = k ⇔ log10 k ≤ xα,n < log10 k + 1.

It is important to take caution when performing numerical calculations usingthis system. An error ε in xα,1 will proliferate to an error of n · ε in xα,n. As thexα,i ∈ [0, ) this error can quickly cause problems. The questions asked aboutGelfand’s system were:

1. Will a ‘9’ ever occur in the collumn of 2n?

2. Will the row ‘23456789’ ever occur again? If so, will it have a frequency?

3. Will a row of the same numbers appear?

4. Will the decimal expansion of an 8-digit prime ever occur?

2.2.1 Question 1

Question 1 can be easily answered by using Kronecker’s Theorem (in section 1)and noting that log10 2 and 1 are rationally independent. This implies that theorbit of rotation over an angle 2π log10 2 on the circle is uniformly distributed,

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which implies that the probability that x2,n is in a certain interval is equal tothe relative length of the interval.

y2,n = 9 ⇐⇒ log10 9 ≤ x2,n < 1

Therefore the probability of y2,n = 9 is equal to 1−log10 9 ≈ 0.0458. This meansthat for close to 4.6 % of the integers n > 0, 〈〈2n〉〉 = 9. Checking with Matlabshowed that in the first 1,000 values 45 start with a 9, and in the first 1,000,000values, 45757 start with a 9. This suggests that indeed:

limn→∞

{#k|y2,k = 9, k < n}n

= 1− log10 9.

Generalizing this result by taking α ∈ [2, 3, . . . , 9], easily shows that theserotations are also uniformly distributed. Among other things, this means thatthe columns of the system independently satisfy the famous Benford’s Law[4].

2.2.2 Questions 2,3 and 4

The other three questions are about the relations between the different columns.Each column is uniformly distributed on the circle, but this does not imply thatthe system is uniformly distributed on the 8-torus. For example, if we considerthe columns starting with 2 and 4, some simple calculations reveal that theseare not independent.

y2,n = 2 =⇒ log10 2 ≤ x2,n < log10 3=⇒ 2 log10 2 ≤ 2x2,n < 2 log10 3=⇒ log10 4 ≤ 2n log10 2 < log10 9=⇒ log10 4 ≤ n log10 4 < log10 9=⇒ 4 ≤ y4,n < 9

This is caused by the rational relation log10 4 = 2 log10 2. Not only the log-arithms 4 and 2 have a rational relations.

22 = 42 · 5 = 102 · 3 = 6

23 = 832 = 9

⇒

log10 4 = 2 log10 2log10 5 = − log10 2log10 6 = log10 2 · log10 3log10 8 = 3 log10 2log10 9 = 2 log10 3

How these rotations behave on a torus can be seen in appendix A on page18.

Suppose α, β, γ ∈ {2, 3, ..., 9}, and γ = αβ. We know αn = (yα,n + ε) · 10k,with ε smaller than 1. The same is true for β and γ. Having ε = 0 meansαn = yα,n ·10k. Using the unique prime factorization of both sides makes it clearthat for n > 3 this can not happen, so we consider the first 3 rows separately.

γn = αnβn

= (yα,n + ε1)10k1(yβ,n + ε2)10k2 .

This gives us two conditions for γn:

γn ∈ 10(k1+k2) ·(yα,n · yβ,n, (yα,n + 1) · (yβ,n + 1)

)And

γn ∈ 10k3 · (yγ,n, yγ,n + 1)

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This implies that checking if yα,n, yβ,n and yγ,n can simultaneously appearin rows α, β and γ respectively comes down to checking whether:

(yα,n · yβ,n, (yα,n + 1) · (yβ,n + 1)) ∩ 10(k3−k1−k2) · (yγ,n, yγ,n + 1) 6= ∅

The size of k3 is also bounded.

k1 + k2 ≤ k3 < k1 + k2 + log10((yα,n + 1) · (yβ,n + 1))0 ≤ k3 − k2 − k1 < log10(yα,n + 1) + log10(yβ,n + 1))

Therefore checking whether a triplet yα,n, yβ,n, yγ,n can appear as a row inthe system can be checked using a simple algorithm.

2.2.3 Example

If, for example, we take α = 2, β = 3, and γ = 6 and assume that the valuesy2,n and y3,n are 3 and 4 respectively, we can check what values for y6,n can beattained.

0 ≤ k3 − k2 − k1 < log10(5) + log10(4)) < 2

Therefore we have to check two possible cases.

[12, 20) ∩ [y6,n, y6,n + 1) 6= ∅OR

[12, 20) ∩ 10 · [y6,n, y6,n + 1) 6= ∅

This condition leaves us with only y6,n = 1.

2.2.4 Answers

If row [y2,n, . . . , y9,n] is attainable, this means a set of [2n, . . . , 9n] exists in theintersections corresponding to all rational relations. As these are intersectionsof open sets, we know that there exists an open interval in R8

>0 that maps to[y2,n, . . . , y9,n]

If we view the logarithms of this, we get an open interval in K8. Using thetheorem by L. Kronecker[3], the probability that for a certain n the simultaneousrotations will be in an interval is relative to the size of the interval. As a non-empty open set always has length greater then zero, this implies that everyattainable value, will also be attained.

Suppose 〈〈2n〉〉 = 2 and 〈〈5n〉〉 = 5. This means that there is an ε ∈ [0, 1)

such that 2n ∼ 2 + ε. As 5̂n is the inverse of 2̂n in A, we know 5n ∼ 102+ε . Due

to the construction of the equivalence relation, we know 〈〈5n〉〉 = 〈〈 102+ε 〉〉 = 5.

This means that 102+ε ≥ 5, which can only happen if ε = 0. Therefore 2n ∼ 2, or

in other words, 2n−1 = 10k, which can only happen if n = 1. A result of this isthat the row ‘23456789’ can only appear at row number 1.

A row of equal numbers can only appear if all xα,n are in the same interval[log10 k, log10 k + 1) for some k and n. Using the condition n log10 4 (mod 1) =2n log10 2 (mod 1), we know that this interval has to be close to 0, leavingonly k = 1 and k = 9. Now using the condition n log10 5 (mod 1) = −n log10 2

8

(mod 1) shows that n log10 2 (mod 1) and n log10 5 (mod 1) can not both be in[0, log10 2) or [ log10 9, 1) unless both are zero. Therefore, only row zero can haveall the same numbers (the row with the first digits of α0).

Generalizing this for all possibilities, a Matlab program (appendix B) waswritten to check for all 98 = 43046721 possible 8-tuples whether they were at-tainable in Gelfand’s system. Only 153999 different 8-tuples were attainable forn > 3. As this already contained row 2 and 3, only row 1 needs to be added.This gives us a total of 154000 different rows, one of which can only appear inthe first row.

The last question about prime numbers was solved using Matlab, by inter-secting the set of possible values with the set of 8-digit prime numbers. Thisnets us 9890 prime numbers that can (and will) appear.

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3 Poncelet’s Theorem and rotations on a circle

Poncelet’s Theorem can be seen as a statement about a system. After all,for each pair (xi, li), we can find (xi+1, li+1). Using elementary operations asrotations, translations and scaling, all of which preserve tangents, we can reduceeach conic section C1 in RP2 to the unit circle. Therefore one can consider thePoncelet system as finding the (ordered) points on this circle such that twoconsecutive points define a line tangent to another conic section.

3.1 Concentric Circles

An interesting example of a pair of conic sections would be two concentric circles.As we can ’shrink’ the outer circle to the unit circle, the only parameter of thesystem is the ratio of the radii of C2 and C1, which we will call k. We canparametrize the outer circle, C1 by

C1 :=

{x = 1−s2

1+s2

y = 2x1+s2

, s ∈ R

Using rotational symmetry, it is evident that each chord tangent to C2 will havethe same length. Therefore we can always take (1, 0) as the starting point x0.Every chord from x0 to x1 is in the form:

T (τ) = x0 + τ (x1 − x0) , τ ∈ [0, 1]

To be tangent to C2 means that there is only one point on the chord withdistance k from the origin. Using elementary geometry, this point has to be themiddle of the chord, the point with τ = 1

2 .

|T ( 12 )| =

√(12 + 1

21−s211+s21

)2+(

12

2s11+s21

)2= 1

2

√1 + 2

1−s211+s21

+(

1−s211+s21

)2+(

2s11+s21

)2= 1√

2

√1 +

1−s211+s21

=√

11+s21

= k

Therefore, the upper tangent to C2 from x0 ends for s1 =√

1k2 − 1, which makes

x1 =

1− 1

k2 +1

1+1k2−1

2

√1k2−1

1+1k2−1

=

(2k2 − 1

2k√

1− k2

)

Applying the rotational symmetry built in in this system allows take thisnew point as a starting point and applying the same step again, therefore takingan equal step. This shows that the Poncelet system for this example is the sameas rotating on the circle

Using this, a link between Gelfand’s problem and Poncelet’s Theorem canbe found: In section 2 Gelfand’s system for starting value 10α, the behaviour of

10

〈〈(10α)n〉〉 is shown to correspond to repeated addition modulo one of α on K,or a rotation of 2πα on a circle.

Using the results above, if we take k = | cos(πα)|, taking one step ends upat:

x1 =

(2 cos2(πα)− 1

2 cos(πα)√

1− cos2(πα)

)=

(cos(2πα)sin(2πα)

)Taking α = log10(2) shows the relation between the first column of Gelfand’ssystem and a specific case of Poncelet’s system.

3.2 A more general case

If we start with two conics, and apply an isomorphism, to change C1 into theunit circle, the system will be isomorphic to the system with:

C1 : x2 + y2 − 1 = 0C2 : Ax2 +Bxy + Cy2 +Dx+ Ey + F = fC2

(x, y) = 0

In this thesis the fact that Poncelet’s system is isomorphic to rotations on acircle, is shown, not proven. Therefore a far more specific case is used. Thisis done because using the general case, with it’s 8 variables (6 for the conicsection and 2 for x0, l0), will be horribly incomprehensible. Almost all cases canshown to be isomorphic to rotations using nearly the same procedure as follows:finding an elliptic curve describing the conditions for xi+1, li+1 and then usingthe isomorphism from section 1. This is illustrated by the following example.

The system that is used is again a system with 2 circles, but C2 is shiftedfrom the centre this time.

C1 : x2 + y2 = 1C2 : (x− k)2 + (y − l)2 = l2.

The radius of C2 is chosen to be l to guarantee a point x = (1, 0) ∈ C1 for whichwe can easily find a tangent to C2, namely l0 : y = 0.

For a line l : y = cx+d to be tangent to circle C2 we require the intersectionof l and C2 to contain only one point. As all coefficients of the problem are

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real, this means that the expression describing the intersection needs to have adouble root.

x2 − 2kx+ k2 + (cx+ d)2 − 2l(cx+ d)= (1 + c2)x2 + (2dc− 2k − 2lc)x+ k2 + d2 − 2ld

As this expression is quadratic in x we can find this double root by calculatingthe discriminant.

(l2 − k2)c2 − d2 + 2lkc− 2kcd+ 2ld = 0

Using this relation we can parametrize c and d by a single variable t that linksboth variables by d = tc.

0 = (l2 − k2)c2 − (tc)2 + 2lkc− 2ktc2 + 2ltc= (−t2 − 2kt+ l2 − k2)c2 + (2lt+ 2kl)c

c(t) =2l(k + t)

(k + t)2 − l2, d(t) =

2lt(k + t)

(k + t)2 − l2

The case that c = 0 can be seen as an extension of c(t) if we allow t =∞, as onthe projective plane.

If we pick a point xi = (x(s), y(s)) =(

2s1+s2 ,

1−s21+s2

)∈ C1 we want to be able

to find a line tangent to C2, after which we can easily find the second intersectionwith C1, to find the point xi+1. After this a new tangent can be found byrepeating the process, thus forming the Poncelet system.

Having a tangent through xi amounts to y(s) = c(t)x(s) + d(t), thus:

0 =2l(k + t)

(k + t)2 − l2· 2s

1 + s2+

2lt(k + t)

(k + t)2 − l2− 1− s2

1 + s2

Multiplying both sides by ((k + t)2 − l2)(1 + s2) results in:

0 = 4ls(k + t) + 2lt(k + t)(1 + s2) + (s2 − 1)((k + t)2 − l2)= (2l(1 + s2) + s2 − 1)t2 + 2(2ls+ lk(1 + s2) + k(s2 − 1))t

+4lsk + (s2 − 1)(k2 − l2)= a2(s)t2 + a1(s)t+ a0(s)

We can now complete the square, in the way it is done while deriving thequadratic formula.(a2(s)t+ 1

2a1(s))2

= a2(s)2t2 + a2(s)a1(s)t+ 14a1(s)2

= 14a1(s)2 + a2(s)

(a2(s)t2 + a1(s)t+ a0(s)

)− a2(s)a0(s)

= 14a1(s)2 − a2(s)a0(s)

= (1 + s2)(s2k2 + s2 + 2ls2 − 4sk + k2 − 2l + 1)l2

If we then pick ρ = 1l

(a2(s)t+ 1

2a1(s))

as a variable instead of t, we obtainan equation in ρ and s, namely:

ρ2 = (1 + s2)(s2k2 + s2 + 2ls2 − 4sk + k2 − 2l + 1).

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3.2.1 Isomorphism to an elliptic curve

We defined the problem to allow a certain solution, x = (1, 0), l : y = 0, forwhich we can find a possible point on the curve in the (ρ, s)-plane

x = (1, 0) ⇒ s = 1c, d = 0 ⇒ t = −ks = 1, t = −k ⇒ ρ = 1

l

(−a2(1)k + 1

2a1(1))

= 1l (−4kl + (2l + 2lk))

= −2(k − 1)

As we now have a point in the ρ, s-plane that lies on the genus one curve definedearlier, we know this curve is isomorphic to an elliptic curve, which was foundby using MAGMA. This elliptic curve is given by:

Elliptic Curve defined by y^2 + (2*k^2 - 4*k - 2*l + 2)/(k^2 - 2*k + 1)*x*y +

(2*k^3*l - 4*k^2*l + 2*k*l - 2*l^3)/(k^6 - 6*k^5 + 15*k^4 - 20*k^3 + 15*k^2

- 6*k + 1)*y = x^3 + (-2*k^4 + 6*k^3 + 2*k^2*l - 8*k^2 - 4*k*l + 6*k + 2*l^2

+ 2*l - 2)/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1)*x^2 + (k^6 - 2*k^5*l - 4*k^5 -

k^4*l^2 + 8*k^4*l + 6*k^4 + 2*k^3*l^2 - 12*k^3*l - 4*k^3 + 2*k^2*l^3 -

2*k^2*l^2 + 8*k^2*l + k^2 - 4*k*l^3 + 2*k*l^2 - 2*k*l + l^4 + 2*l^3 -

l^2)/(k^8 - 8*k^7 + 28*k^6 - 56*k^5 + 70*k^4 - 56*k^3 + 28*k^2 - 8*k + 1)*x

Using the isomorphism ψ defined by:

Mapping from: CrvHyp: C to CrvEll: E

with equations :

(-k^4 + 3*k^3 - k^2*l - 4*k^2 + 2*k*l + 3*k + l^2 - l - 1)/(k^4 - 4*k^3 + 6*k^2

- 4*k + 1)*$.1^3 + (k^4 - k^3 + k^2*l - 2*k*l - k - 3*l^2 + l + 1)/(k^4 -

4*k^3 + 6*k^2 - 4*k + 1)*$.1^2*$.3 - 1/(k - 1)*$.1*$.2 + (-k^4 + k^3 + k^2*l

- 2*k*l + k + 3*l^2 + l - 1)/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1)*$.1*$.3^2 +

1/(k - 1)*$.2*$.3 + (k^4 - 3*k^3 - k^2*l + 4*k^2 + 2*k*l - 3*k - l^2 - l +

1)/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1)*$.3^3

(2*k^4 - 6*k^3 + 2*k^2*l + 8*k^2 - 4*k*l - 6*k - 2*l^2 + 2*l + 2)/(k^4 - 4*k^3 +

6*k^2 - 4*k + 1)*$.1^3 + (-4*k^3 + 8*k^2 - 4*k + 4*l^2)/(k^4 - 4*k^3 + 6*k^2

- 4*k + 1)*$.1^2*$.3 + 2/(k - 1)*$.1*$.2 + (2*k^4 - 6*k^3 - 2*k^2*l + 8*k^2

+ 4*k*l - 6*k - 2*l^2 - 2*l + 2)/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1)*$.1*$.3^2

-$.1^3 + 3*$.1^2*$.3 - 3*$.1*$.3^2 + $.3^3

and inverse

$.2*$.3

(4*k - 4)*$.1^3*$.3 + (-4*k^4 + 12*k^3 + 4*k^2*l - 16*k^2 - 8*k*l + 12*k + 4*l^2

+ 4*l - 4)/(k^3 - 3*k^2 + 3*k - 1)*$.1^2*$.3^2 + (-4*k^2 + 8*k + 4*l - 4)/(k

- 1)*$.1*$.2*$.3^2 + (-2*k + 2)*$.2^2*$.3^2

2*$.1*$.3 + $.2*$.3

This elliptic curve can be further simplified by bringing it into Weierstrassform,

y2 +Axy +By = x3 + Cx2 +Dx(y + 1

2Ax+ 12B)2 = x3 + Cx2 +Dx+ ( 1

2Ax+ 12B)2

We can now substitute the square with α2, and shift the x-coordinate to cancelout the term with x2.

α2 = x3 + (C + 14A

2)x2 + (D + 12AB)x+ 1

4B2

= (β − 13 (C + 1

4A2))3 + (C + 1

4A2)(β − 1

3 (C + 14A

2))2

+(D + 12AB)(ξ − 1

3 (C + 14A

2)) + 14B

2

= β3 +(− 1

3C2 +D − 1

48A4 + 1

2AB −16CA

2)β

+ 1864A

6 − 16ABC −

13DC + 1

18C2A2 + 1

72CA4

− 124A

3B + 227C

3 + 14B

2 − 112DA

2

13

Substituting A,B,C and D allows us to return to the old variables k and l.

α2 = β3 − 1

3

(k4 − k2 + 1− 3l2)β

(k − 1)4− 1

27

(k2 + 1)(−9l2 + 2k4 − 5k2 + 2)

(k − 1)6

If we don’t take k = 1, we can simplify this further by scaling the variables:

η =α

(k − 1)3, ξ =

β

(k − 1)2

Applying these substitutions and multiplying by (k − 1)6, yields:

η2 = ξ3 − 13 (k4 − k2 + 1− 3l2)ξ − 1

27 (k2 + 1)(−9l2 + 2k4 − 5k2 + 2)

This has a single solution (η2 = 0 for ξ = − 13 (1 + k2)) in RP2 if and only if

l > | 12 (k2 − 1)|. If this condition holds the other two roots are imaginary. Thelast transformation will be shifting the curve by ξ = (λ− 1

3 (1 + k2)) such thatthis solution will be at the origin.

η2 = λ3 − (1 + k2)λ2 + (k2 + l2)λ

3.2.2 Isomorphism to a circle

If the curve has a single root, we can use section 1to show that the group ofreal points on this curve isomorphic to the rotation group on the circle by theisomorphism:

ϕ(η, λ) =

{1− g(λ) if η < 0

g(λ) if η ≥ 0, g(λ) =

∞∫λ

dt√t3−(1+k2)t2+(k2+l2)t

2∞∫0

dt√t3−(1+k2)t2+(k2+l2)t

The condition on k and l to have a single root is l > | 12 (k2 − 1)|.

xi = (a, b) ∈ C1

li = cx+ d through xi, tangent to C2

↓ a = 2s1+s2 , b = 1−s2

1+s2

c(t) = 2l(k+t)(k+t)2−l2 , d(t) = 2lt(k+t)

(k+t)2−l2

(s, t)↓ ρ = (2t+ k)(1 + s2) + 1

l (k + t)(s2 − 1) + 2s(ρ, s)↓ ψ

(y, x)

↓ η = y(k−1)3 + (k−1)2−l

(k−1)5 x+ kl(k−1)2−l3(k−1)9

ξ = x(k−1)2 −

13k4−2k3+2k2−2k−3l2+1

(k−1)6

(η, ξ)↓ λ = ξ + 1

3 (1 + k2)(η, λ)↓ ϕK

14

3.2.3 Explicit example

Finding the next point (xi+1, li+1) from (xi, li) is a two step process:

• Find the other point on the intersection of C1 and li, call this xi+1

• Find the other tangent to C2 through xi+1, call this li+1

In this example we will take k = 2 and l = 2, which satisfies the conditionto have one root of the elliptic curve.

By construction we can take (x0, l0) =((1, 0), y = 0

). This means s = 1 and

t = −2, which leads to ρ = −2. The isomorphism ψ maps this point to O, thepoint at infinity of the (y, x)-plane. Transforming to (η, λ) will keep the pointat O.

An easy calculation reveals that x1 = (−1, 0), this makes s = −1. To have l1through this point, means d1(t) = c1(t), therefore having t = −2 or t = 1. Theline with t = −2 is l0, so l1 corresponds to having t = 1. Having s and t leadsto ρ = 6. Again using ψ and transforming yields:

(η, λ) = (2, 1)

Continuing these steps, we can calculate:

(x2, l2) =((− 119

169 ,120169

), y = − 5560

18921x−80159

)⇒ (s, t) =

(−717 ,

−238139

)⇒ (ρ, s) =

(− 754

289 ,−717

)⇒ (y, x) =

(3564 ,−

1516

)⇒ (η, λ) =

(− 161

49 ,4916

)For these points we can calculate the ϕ which reveals that adding on the

elliptic curve is indeed isomorphic to addition modulo 1

(x0, l0) ∼ ϕ(O) = 0(x1, l1) ∼ ϕ(2, 1) ≈ 0.3822382958(x2, l2) ∼ ϕ

(− 161

49 ,4916

)≈ 0.7644765858

Note that the rows increase by the same number both times. This illustratesthat indeed Poncelet’s system is very closely related to rotations on a circle.

3.2.4 Rotation number for free k and l

By construction we can take (x0, l0) =((1, 0), y = 0

). In section 3.2.1, we found

the corresponding (ρ, s) to be (−2(k − 1), 1). Using the isomorphism ψ, thecorresponding point on the elliptic curve on the (y, x)-plane, is the point O,which corresponds with the point O in the (η, λ)-plane.

Using the algorithm, it is easy to find x1 = (−1, 0). The line l1 : y = c1x+d1is tangent to C2 and, because it goes through x1 = (−1, 0), we know d1 = c1.As both are functions of t, we can use this to calculate t:

c1(t) = d1(t)⇒ 2l(k + t)

(k + t)2 − l2=

2lt(k + t)

(k + t)2 − l2⇒ t = −k or t = 1

15

Figure 1: Plot of ϕ((O)), ϕ(2, 1) and ϕ(− 161

49 ,4916

)(counterclockwise from the

rightmost point) on a circle.

The first of these options is l0, therefore the second corresponds to l1. Thismeans we have an (s, t) = (−1, 1) for (x1, l1), implying: (ρ, s) = (2(k + 1),−1).Using the transformations to the (η, λ)-plane:

(η, λ) =

(l

(k − 1)6,k6 − 4k5 + 7k4 − 8k3 + 6k2 − 4k + 3

3(k − 1)4

)The last section made it very convincing that ϕ(η, λ) will be the rotation

number of the system. After calculating this it should be easy to deduce thebehaviour of the system: if this number is rational, the transverse will close upin a finite amount of steps, if it is irrational it will never.

Conclusion

In conclusion, both the systems for Poncelet’s Closure Theorem and Gelfand’squestion were shown to be closely related to the circle group. Furthermore,using these results, all of Gelfand’s questions were answered.

References

1. King, J. L. ”Three Problems in Search of a Measure.” Amer. Math.Monthly 101, 609-628, 1994.

2. E. Bakker, J. S. Chahal & J. Top “Albime triangles and Guy’s ellipticcurve”, preprint, 2013

3. http://www.math.lsa.umich.edu/∼rauch/558/Kronecker.pdf

16

4. F. Benford ”The law of anomalous numbers”, Proceedings of the AmericanPhilosophical Society, Vol. 78, No. 4, March 31, 1938.

17

A Plots of Gelfand’s system on an 8-torus

Gelfand’s system was found to be isomorphic to fixed rotations on an 8-torus,which can be viewed as addition modulo one on an 8-cube. If we take x = log10 2,y = log10 3, and z = log10 7, the entire shape in 8 dimensions will be the closureof the orbit defined by:

τn = τn−1 ⊕ (x, y, 2x,−x, xy, z, 3x, 2y)τ0 = (0, 0, 0, 0, 0, 0, 0, 0)

As we can not visualize an 8-cube, we plot the intersections obtained byonly looking at 3 starting values, which already hints at the complexity of theattainable shape in the 8-cube.

The plots are made by taking the first 1000 iterations (additions moduloone of the logarithms), for starting values α, β, γ, and plotting them in a 3-dimensional cube.

Figure 2: This illustrates that the relation between 2 and 4 is the equivalent tothe one between 3 and 9.

Figure 3: The left plot illustrates that any two of 2, 3, 6 are independent, butall together they are dependant. The right plot suggests the independence of2, 3, 7.

18

Figure 4: The left plot illustrates the strong dependence between 2, 4, 5: Only aline can be attained. The right plot illustrates the dependence of 2, 6, 9, whichshows that not only the relations stated in 2.2.2 determine the shape.

The last two plots attempt to show some more of the complexity, by takingthe colour of the points as a function of a fourth rotation.

Figure 5: The left image has the colour as a function of the rotation over log10 8.To show the structure this creates, the right image has it’s colour defined byrotation over log10 3, which is clearly not as structured.

19

Figure 6: A similar case: The left image has the colour as a function of therotation over log10 3, the right log10 7

B Matlab Code

This Matlab code uses the relations between the logarithms to remove rows thatcan no be attained in Gelfand’s system.

k=allcomb(1:9,1:9,1:9,1:9,1:9,1:9,1:9,1:9);

for n=1:(9^8)

if ((k(n,3)>= (k(n,1)+1)^2 ) || (k(n,3)+1<=k(n,1)^2) )

&& ((10*k(n,3)>= (k(n,1)+1)^2 ) || (10*(k(n,3)+1)<=k(n,1)^2) )

k(n,:)=[0,0,0,0,0,0,0,0];

end

if ((k(n,7)>= (k(n,1)+1)^3 ) || (k(n,7)+1<=k(n,1)^3) )

&& ((10*k(n,7)>= (k(n,1)+1)^3 ) || (10*(k(n,7)+1)<=k(n,1)^3) )

&& ((100*k(n,7)>= (k(n,1)+1)^3 ) || (100*(k(n,7)+1)<=k(n,1)^3) )

k(n,:)=[0,0,0,0,0,0,0,0];

end

if ((k(n,8)>= (k(n,2)+1)^2 ) || (k(n,8)+1<=k(n,2)^2) )

&& ((10*k(n,8)>= (k(n,2)+1)^2 ) || (10*(k(n,8)+1)<=k(n,2)^2) )

k(n,:)=[0,0,0,0,0,0,0,0];

end

if ((k(n,5)>= (k(n,1)+1)*(k(n,2)+1) ) || (k(n,5)+1<=k(n,1)*k(n,2) ))

&& ((10*k(n,5)>= (k(n,1)+1)*(k(n,2)+1) ) || (10*(k(n,5)+1)<= k(n,1)*k(n,2)))

k(n,:)=[0,0,0,0,0,0,0,0];

end

if (10<=(k(n,4)*k(n,1)) || 10>=((k(n,4)+1)*(k(n,1)+1)))

k(n,:)=[0,0,0,0,0,0,0,0];

end

end

X2=k(any(k,2),:) ;

clearvars k n

20

C Gelfand Table1 2 3 4 5 6 7 8 92 4 9 1 2 3 4 6 83 8 2 6 1 2 3 5 74 1 8 2 6 1 2 4 65 3 2 1 3 7 1 3 56 6 7 4 1 4 1 2 57 1 2 1 7 2 8 2 48 2 6 6 3 1 5 1 49 5 1 2 1 1 4 1 310 1 5 1 9 6 2 1 311 2 1 4 4 3 1 8 312 4 5 1 2 2 1 6 213 8 1 6 1 1 9 5 214 1 4 2 6 7 6 4 215 3 1 1 3 4 4 3 216 6 4 4 1 2 3 2 117 1 1 1 7 1 2 2 118 2 3 6 3 1 1 1 119 5 1 2 1 6 1 1 120 1 3 1 9 3 7 1 121 2 1 4 4 2 5 9 122 4 3 1 2 1 3 7 923 8 9 7 1 7 2 5 824 1 2 2 5 4 1 4 725 3 8 1 2 2 1 3 726 6 2 4 1 1 9 3 627 1 7 1 7 1 6 2 528 2 2 7 3 6 4 1 529 5 6 2 1 3 3 1 430 1 2 1 9 2 2 1 431 2 6 4 4 1 1 9 332 4 1 1 2 7 1 7 333 8 5 7 1 4 7 6 334 1 1 2 5 2 5 5 235 3 5 1 2 1 3 4 236 6 1 4 1 1 2 3 237 1 4 1 7 6 1 2 238 2 1 7 3 3 1 2 139 5 4 3 1 2 9 1 140 1 1 1 9 1 6 1 141 2 3 4 4 8 4 1 142 4 1 1 2 4 3 8 143 8 3 7 1 2 2 6 144 1 9 3 5 1 1 5 945 3 2 1 2 1 1 4 846 7 8 4 1 6 7 3 747 1 2 1 7 3 5 2 748 2 7 7 3 2 3 2 649 5 2 3 1 1 2 1 550 1 7 1 8 8 1 1 5

51 2 2 5 4 4 1 1 452 4 6 2 2 2 8 9 453 9 1 8 1 1 6 7 354 1 5 3 5 1 4 5 355 3 1 1 2 6 3 4 356 7 5 5 1 3 2 3 257 1 1 2 6 2 1 2 258 2 4 8 3 1 1 2 259 5 1 3 1 8 7 1 160 1 4 1 8 4 5 1 161 2 1 5 4 2 3 1 162 4 3 2 2 1 2 9 163 9 1 8 1 1 1 7 164 1 3 3 5 6 1 6 165 3 1 1 2 3 8 5 166 7 3 5 1 2 5 4 967 1 9 2 6 1 4 3 868 2 2 8 3 8 2 2 769 5 8 3 1 4 2 2 670 1 2 1 8 2 1 1 671 2 7 5 4 1 1 1 572 4 2 2 2 1 7 1 573 9 6 8 1 6 4 8 474 1 2 3 5 3 3 6 475 3 6 1 2 2 2 5 376 7 1 5 1 1 1 4 377 1 5 2 6 8 1 3 278 3 1 9 3 4 8 2 279 6 4 3 1 2 5 2 280 1 1 1 8 1 4 1 281 2 4 5 4 1 2 1 182 4 1 2 2 6 1 1 183 9 3 9 1 3 1 9 184 1 1 3 5 2 9 7 185 3 3 1 2 1 6 5 186 7 1 5 1 8 4 4 187 1 3 2 6 5 3 3 188 3 9 9 3 3 2 2 989 6 2 3 1 1 1 2 890 1 8 1 8 1 1 1 791 2 2 6 4 6 8 1 692 4 7 2 2 3 5 1 693 9 2 9 1 2 3 9 594 1 7 3 5 1 2 7 495 3 2 1 2 8 1 6 496 7 6 6 1 5 1 4 497 1 1 2 6 3 9 3 398 3 5 1 3 1 6 3 399 6 1 4 1 1 4 2 2100 1 5 1 7 6 3 2 2

21