Pick’s theorem and Riemann sums:a Fourier analytic tale

Giancarlo Travaglini (Universita di Milano-Bicocca)

joint work withL. Brandolini, L. Colzani and S. Robins

Discrepancy Theory and ApplicationsCIRM, 30 Nov. - 1 Dec. 2020

We are going to talk about a few results from the papers

L. Brandolini, L. Colzani, S. Robins, G. Travaglini, Convergence of multipleFourier series and Pick’s theorem.

L. Brandolini, L. Colzani, S. Robins, G. Travaglini, An Euler-Maclaurin formulafor polygonal sums.

Both of them are in arXiv.

Leonardo Colzani, Luca Brandolini, and Sinai Robins in Bergamo

Let P ⊂ R2 be an integral polygon, that is all vertices of P are integer points. Anelegant result proved by G. Pick in 1899 allows us to compute the area |P| of Pin terms of the integer points therein. In order to introduce Pick’s theorem weassociate the following “angle” ωP(k) to each point k ∈ Z2:

ωP(k) = 1 if k is in the interior of P,ωP(k) = 0 if k is outside P,ωP(k) = α/2π if k is a vertex of P with interior

angle (or union of interior angles) α,ωP(k) = 1/2 if k is on a side of P, but is not a

vertex.

P

1/4

1/2

3/8

1/21/21/8

1/21

1/4

1/8

1/41/8

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0

0

That is to say, for a small ε > 0, ωP(k) = |B(k,ε)∩P||B(0,ε)| , where

B (k , ε) ={t ∈ R2 : |t − k | ≤ ε

}is the disc centred at k and having radius ε.

Theorem (Pick)

Let P be an integral polygon. Then∑k∈Z2

ωP (k) = |P| .

The proof consists of the following steps.

1. The statement of Pick’s theorem is additive:∑k∈Z2

ωP (k) =∑k∈Z2

ωA (k) +∑k∈Z2

ωB (k) ,

|P| = |A|+ |B| .

P

1/2

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1/21/8

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1/4

1/8

1/8

1/83/8

1/2

1/2A

B

2. Then polygon triangulation allows us to consider only triangles (this step is abit cumbersome when P is not convex).

3. It is enough to prove the theorem in the case of an aligned integralright-angled triangle, and therefore (by symmetry) of an aligned integralrectangle. In the latter case case the computation is immediate.

R

A

B

R

Actually Pick’s theorem is usually stated in a different form.Let P be an integral polygon (which we now assume to be simple). Let B be thenumber of integer points on the sides of P and I the number of integer points inthe interior of P. Assume that P has L sides, so that the sum of its internalangles is π (L− 2). Then

|P| =∑k∈Z2

ωP (k) = I︸︷︷︸from theinteriorpoints

+1

2ππ (L− 2)︸ ︷︷ ︸

from thevertices

+ (B − L)1

2︸ ︷︷ ︸from the points

on sides, butnot on vertices

+ 0︸︷︷︸from theexteriorpoints

= I +B

2− 1 ,

and (with the above definitions of I and B) we have

Theorem (Second, and more familiar, statement of Pick’s theorem)

Let P be a simple polygon. Then

|P| = I +B

2− 1 .

Georg Pick

This latter statement of Pick’s theoremcan be explained to Primary School kids,and they should be asked to check it ontheir own examples of polygons drawn ona sheet of squared paper.

Pick’s theorem has been hidden for aboutforty years, until it has been popularizedby H. Steinhaus in the Polish edition ofhis book Mathematical Snapshots.

Georg Pick (Vienna 1859 - Theresienstadtconcentration camp 1942) is known forthe above theorem, for the Nevanlinna-Pick theorem in Complex Analysis, andfor having introduced A. Einstein to theworks of Ricci-Curbastro and Levi-Civita.

Among the applications of Pick’s theorem we point out

The computation of the Frobenius number for the case of two coins:Let α, β be positive integers such that gcd(α, β) = 1. Then the equationn = jα + kβ has no non-negative integer solutions if n = αβ − α− β, and hasnon-negative integer solutions for every n > αβ − α− β.

The study of Farey sequences:For every n ∈ N let h/k and h′/k ′ be consecutive fractions in

Fn :=

{h

k: h, k ∈ Z, k 6= 0, 0 ≤ h ≤ k ≤ n, gcd (h, k) = 1

},

e.g.

F7 :

{0

1

1

7

1

6

1

5

1

4

2

7

1

3

2

5

3

7

1

2

4

7

3

5

2

3

5

7

3

4

4

5

5

6

6

7

1

1

}then kh′ − hk ′ = 1.Then Farey sequences can be used to prove Hurwitz’s theorem on the bestconstant in the Dirichlet’s diophantine approximation theorem:For every irrational α there are infinitely many n, p such that

∣∣α− pn

∣∣ < 1√5 n2

.

Here we are going to show a connection between Pick’s theorem and Fourierseries.

The following geometric estimate will be useful.

Theorem (Podkorytov)

Let C ⊂ R2 be a convex body. For Θ := (cos θ, sin θ) and small δ > 0 let

λ(δ, θ) = λC (δ, θ) :=

{t ∈ C : δ + t ·Θ = sup

y∈C(y ·Θ)

}

be the chord perpendicular to Θ and “at distance δ from the boundary” ∂C of C .Then for ρ > 1 we have (here |λC | is the length of λC )

|χC (ρΘ)| ≤ c ρ−1(∣∣λC (ρ−1, θ)

∣∣+∣∣λC (ρ−1, θ + π)

∣∣)

In particular, if C is a disc we have

|χC (ξ)| ≤ c |ξ|−3/2 ,

while for an arbitrary convex body C we have nothing better than

|χC (ξ)| ≤ c |ξ|−1 .

See the figures, where ρ = |ξ|:

ρ-1

ρ-1/2

ρ-11

discconvex body

Podkorytov’s theorem is a consequence of the following 1-dimensional result.Let f be a real non-negative function supported and concave on [−1, 1] (we mayassume f even and f (±1) = 0, for simplicity). Then, for ξ > 1,∣∣∣f (ξ)

∣∣∣ ≤ c |ξ|−1 f (1− 1/ξ)

Indeed integration by parts allows us to replace f (ξ) with

− |ξ|−1∫ 1

0

f ′ (x) sin (2πξx) dx

f(x)

1-1/ξ

1

-f'(x)

-f'(x)sin(2πξx)

+

+

+

-

--

Then observe that −∫ 1

0f ′ (x) sin (2πξx) dx is a Leibniz sum, thereby bounded

by its largest term (the grey area), which is ≈ f (1− 1/ξ).

Now let us go back to Pick’s theorem and Fourier series.As before, let P ⊂ R2 be an integral polygon. We periodize its characteristic

function χP (t) =

{1 if t ∈ P0 if t /∈ P

, thereby introducing the periodic function

λP (t) :=∑k∈Z2

χP (t + k) .

Then λP(m) =∑

k∈Zd∫[− 1

2 ,12 )2 χP(t + k) e−2πim·t dt = χP(m) for all m ∈ Z2.

The following figures shows an integral triangle P and the periodization λP of itscharacteristic function.

In the particular case in the figure, λP is a characteristic function too. In generalit is a superposition of characteristic functions (of polygons).

We need to set a summation method for the Fourier series of λP (t). We choosea small ε > 0 and consider the (normalized) characteristic function of the unitdisc with radius ε: that is ϕε (t) := 1

πε2 χ{|t|≤ε} (t) . Let ϕ := ϕ1. We have

(ϕε ∗ λP) (t) =∑m∈Z2

ϕ (εm) χP (m) e2πim·t .

The above Fourier series converges (for every ε) since ϕ decays of order 3/2 andχP decays of order at least 1. At the origin the series sums to∑

m∈Z2ϕ (εm) χP (m) = (ϕε ∗ λP) (0) =

1

πε2

∫{|t|≤ε}

λP (t) dt ,

0

1/2

1/2-1/2

-1/2

ε

that is (if ε is not too large) the sum of the(normalized) measures of the angles of P at theinteger points (see the figure).In order to prove Pick’s theorem we have to showthat this sum equals the area of P. In otherwords, Pick’s theorem is equivalent to the fol-lowing identity:

|P| =∑m∈Z2

ϕ (εm) χP (m) , that is∑

06=m∈Z2ϕ (εm) χP (m) = 0 .

Proof of∑

06=m∈Z2 ϕ (εm) χP (m) = 0 .

Let {Pj}Lj=1 ⊂ Z2 and {Pj + s (Pj+1 − Pj) : s ∈ [0, 1]}j=1,...,L be the verticesand the sides of P respectively. For each j let nj be the outward unit normal tothe jth-side. The divergence theorem yields

χP (m) =

∫P

e−2πim·t dt =

∫P

div

(m

−2πi |m|2e−2πim·t

)dt

=−1

2πi

L∑j=1

m · nj|m|2

|Pj+1 − Pj |∫ 1

0

e−2πim·(Pj+1−Pj )s ds

=−1

2πi

L∑j=1

m·(Pj+1−Pj )=0

m · nj|m|2

|Pj+1 − Pj | .

Then

∑06=m∈Z2

ϕ (εm) χP (m) =−1

2πi

L∑j=1

|Pj+1 − Pj |

∑06=m∈Z2

m·(Pj+1−Pj )=0

ϕ (εm)m · nj|m|2

= 0 ,

because ϕ (εm) is even and m · nj is odd. This ends the proof of Pick’s theorem.

Pick’s theorem is part of the general problem of studying integer points in convexbounded integral polyhedra, which is motivated by questions such as“How many integer solutions does a given system of linear inequalities have?”.

In terms of Fourier series, let P ⊂ Rd be a bounded polyhedron containing theorigin in its interior, and let

DN (t) =∑

m∈NP

e2πim·t

be the corresponding (polyhedral) Dirichlet kernel. Then the correspondingFourier partial sum of a function f can be written as∑

m∈NP

f (n) e2πim·t = (f ∗ DN) (t) .

Then (by Parseval identity)√card (NP ∩ Zd) = ‖DN‖L2(Td ) = sup

‖f ‖L1(Td )=1

‖f ∗ DN‖L2(Td )

is the L1(Td)→ L2

(Td)

operator norm of the Fourier partial sum operator.

There is no easy extension of Pick’s theorem to several variables. Indeed, thetetrahedron in the figure

(1,1,0)

(0,0,R)

(1,0,0)(0,0,0)

(0,1,0)

has large volume, 4 integer vertices, and no inner integer points. Then thevolume of this polyhedron does not seem to depend on the integer points.

Anyway the higher dimensional case shows several interesting features.As in the planar case, given a d-dimensional convex polyhedron P and an integerpoint k ∈ Zd we define the solid angle

ωP(k) =|B (k , ε) ∩ P||B (0, ε)|

,

where B (k, ε) ={t ∈ Rd : |t − k | ≤ ε

}and ε is small. That is the proportion of

a small ball of radius ε, centered at k , which intersects P.We consider the solid angles sum ∑

k∈ZdωP (k)

and we say that P satisfies Pick’s identity if∑k∈Zd

ωP (k) = Vol (P) .

Remarks

Assume we do not know that Pick’s identity may fail in dimension d ≥ 3, so thatwe try to carry on the argument used in the 2-dimensional proof.Everything works for a while, and again Pick’s identity is equivalent to∑

06=m∈Zdϕ (εm) χP (m) = 0 .

Then the proof no longer works since we may have∑

06=m∈Zd ϕ (εm) χP (m) 6= 0(because of the appearance of even terms).

It is interesting to find relevant families of integral polyhedra which satisfy Pick’sidentity, so that for each one of them its volume is the sum of its solid angles atthe integer points.An example is given by multiple tilings.Let H be a positive integer and let P be a convex bounded integral polyhedron inRd . We say that P multi-tiles (or H-tiles) Rd by translation if (for a.e. t)∑

k∈ZdχP (t + k) = H (1)

We have already observed that the periodic function t 7→∑

k∈Zd χP (t + k) has

Fourier coefficients χP (m) for every m ∈ Zd .Computing the Fourier coefficients on both sides of (??) yields χP (m) = 0 forevery m 6= 0.Then Pick’s identity holds true for the integral convex polyhedra that multi-tileRd .

N. Gravin, S. Robins and D. Shiryaev have proved the following characterisation.

Theorem (Gravin, Robins, Shiryaev)

A convex bounded integral d-dimensional polyhedron multi-tiles Rd if and only ifit is centrally symmetric and has centrally symmetric facets (the codimension-1faces).

Eugene Ehrhrart, self-portrait

While talking about integer points inpolyhedra, we cannot ignore the funda-mental result proved by E. Ehrhart in1962.

Ehrhart (1906-2000) spent most of hislife teaching mathematics in high schoolsin France, and did mathematics researchjust for his personal pleasure; he startedpublishing in mathematics at the age of45, and got his Ph.D 15 years later.

Theorem (Ehrhart)

Let P be an integral closed convex bounded d-dimensional polyhedron. For everyinteger N ≥ 0 let E (N) := card

(NP ∩ Zd

).

Then E (N) is d-degree polynomial:

E (N) = cdNd + cd−1N

d−1 + . . .+ c1N + c0

(termed Ehrhart polynomial) having rational coefficients.

Then c0 = E (0) = 1 and cd = Vol (P) because, for every positive N,

Vol (P) = N−dVol (NP) = N−d(card

(NP ∩ Zd

)+O

(Nd−1))

= N−d(cdN

d + cd−1Nd−1 + . . .+ c1N + c0

)+O

(N−1

)= cd +O

(N−1

).

Moreover Ehrhart proved that cd−1 = |∂P| /2.The meaning of the remaining coefficients is an open problem.Ehrhart’s theorem allows us to compute explicitly card

(NP ∩ Zd

)for a given N.

As an example, let P ⊂ R4 be a closed integral bounded polyhedron, and say thatwe need to know how many integer points are contained in 1000P, that isE (1000). Then we count the number of integer points in P, 2P, 3P, and 4P. Inthis way we quickly get the coefficients of the Ehrhart polynomialE (N) = c4N

4 + c3N3 + c2N

2 + c1N + 1. Then we compute E (1000). In particularwe get cd = Vol (P), as if we had a d-dimensional analog of Pick’s theorem.

The proof of Ehrhart’s theorem is rather intricate and combinatorial in nature.Here we like to show a Fourier analytic argument (not really a proof) which“leads” to Ehrhart’s theorem.

A few figures for the planar case will be useful. Let P be a convex integralpolygon having vertices P1,P2, . . . ,PL. For every positive integer N let

DNP (t) =∑n∈NP

e2πin·t

be the polygonal Dirichlet kernel associated to P. We are going to show that

DNP (t) =

∑sj=1 e

2πiNaj ·t Fj (t)∏rj=1 (e2πivj ·t − 1)

, (2)

where F1,F2, . . . ,Fs are trigonometric polynomials (independent of N) withinteger coefficients, and {v1, v2, . . . , vr} ⊂ Z2. We observe that (??) is ad-dimensional (polyhedral) analog of the well-known identity

N∑n=−N

e2πinx =sin ((2N + 1)πx)

sin (πx).

Indeed let P, v1, v2, v3, v4 be as in the figure on the left, so that DNP (n) = 1 at

the black points and DNP (n) = 0 elsewhere.

v v v v1 2 3 4

NP

Pv

= +1

= -1v1

We translate the black points by v1, we keep another copy of the black pointsstill, and we subtract them. This amounts to computing

DNP (t)(e2πiv1·t − 1

)=∑n∈NP

e2πi(n+v1)·t −∑n∈NP

e2πin·t

The figure on the right represents the Fourier coefficients of the polynomialDNP (t)

(e2πiv1·t − 1

). The interior of NP “has vanished” and a few strips of +1

(black) or −1 (white) remain, but not close to the interior of the side of NPparallel to v1.

By repeating the previous argument with v2 in place of v1 we obtain

DNP (t)(e2πiv1·t − 1

) (e2πiv2·t − 1

),

and again the points close to the side(s) parallel to v2 almost disappear. Twomore steps and we see that the Fourier coefficients of

DNP (t)(e2πiv1·t − 1

) (e2πiv2·t − 1

) (e2πiv3·t − 1

) (e2πiv4·t − 1

),

consist of “clouds” or “ovals” close to the vertices of NP:

=+1

=-1

v2

=+1

=-1

v4

Then

DNP (t) =

∑sj=1 e

2πiNaj ·t Fj (t)∏rj=1 (e2πivj ·t − 1)

,

where the trigonometric polynomials e2πiNaj ·tFj (t) (the “clouds”) do not dependon N and have their non-zero Fourier coefficients inside the above “ovals”.Finally

card(NP ∩ Z2

)= DNP (0) = lim

t→0

∑sj=1 e

2πiNaj ·t Fj (t)∏rj=1 (e2πivj ·t − 1)

.

Computing the limit through L’Hopital’s rule shows that card(NP ∩ Z2

)is a

polynomial in N (then an obvious dimensionality argument shows that its degreemust be equal to the dimension of the Euclidean space).

In Pick’s theorem we have compared the area of a polygon and the (weighted)number of integer points therein. Of course this is a particular case of thecomparison between integrals and Riemann sums.

We are going to discuss an Euler-Maclaurin summation formula for polygons.

We recall the

Theorem (Euler-Maclaurin summation formula)

Let g be a smooth function on [1,+∞) and let 0 < M < N be integers. Then

N∑n=M

g (n) =1

2(g (M) + g (N)) +

∫ N

M

g (x) dx

+k−1∑h=0

1

(h + 1)!ψh+1 (0)

(g (h) (N)− g (h) (M)

)− (−1)k

∫ N

M

1

k!ψk (x) g (k) (x) dx ,

where B0 (x) ,B1 (x) , . . . are the Bernoulli polynomials, defined by

B0 (x) = 1 , B ′k+1 (x) = (k + 1)Bk (x) ,

∫ 1

0

Bk+1 = 0,

and

ψk (x) = Bk ({x}) =∑

06=n∈Z

−k!

(2πin)ke2πinx

We also recall the Poisson summation formula:

Let f ∈ L1(Rd). Then (formally)∑m∈Zd

f (t + m) =∑m∈Zd

f (m) e2πim·t

If t = 0 we have the beautiful identity∑m∈Zd

f (m) =∑m∈Zd

f (m) .

The above formula, as it is written, does not immediately apply to non-smoothfunctions, such as characteristic functions. Without additional assumptions theseries in both sides of this identity may not converge pointwise. And, even whenboth sides converge, they may differ (for example, when we modify f (x) on a setof measure zero). A correct formula can be obtained either assuming regularityconditions for f (x) or using suitable summability methods for the Fourier series.That is why we shall replace f with ϕε ∗ f , where ϕε (x) = ε−1ϕ

(ε−1x

)and ϕ is

a suitable radial function with compact support and integral 1.The Euler-Maclaurin summation formula and the Poisson summation formula arerelated. In particular, Poisson himself has given the following proof of theEuler-Maclaurin summation formula.Poisson’s argument is the starting point for the second part of this talk.

Poisson’s proof of the Euler-Maclaurin summation formula. Let M,N ≥ 0and g (x) smooth. As at the beginning of the proof of Pick’s theorem we have

limε→0+

(ϕε ∗ gχ[M,N]

)(x) =

g (x) if M < x < N12g (x) if x = M or x = N0 otherwise.

Hence Poisson summation formula yields

1

2g (M) +

N−1∑n=M+1

g (n) +1

2g (N) = lim

ε→0+

{∑n∈Z

ϕ (εn) gχ[M,N] (n)

}.

Repeated integration by parts yields

gχ[M,N] (n) = −k−1∑h=0

g (h) (N)− g (h) (M)

(2πin)h+1+

1

(2πin)k

∫ N

M

g (k) (x) e−2πinx dx .

Then

1

2g (M) +

N−1∑n=M+1

g (n) +1

2g (N)

=

∫ N

M

g +k−1∑h=0

1

(h + 1)!ψh+1 (0)

(g (h) (N)− g (h) (M)

)− (−1)k

∫ N

M

1

k!ψkg

(k) .

Now we can write a two-dimensionalEuler-Maclaurin type summation for-mula over an integral polygon P.

Again we consider, for ε > 0 small, theangle

ωP (q) =

∣∣P ∩ {t ∈ R2 : |t − q| < ε}∣∣

|{t ∈ R2 : |t| < ε}|

(which can be defined for every q ∈ R2).

P

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0

Theorem

Let P be an integral polygon and g a smooth function on R2. Then there exist acomputable real sequence δh such that for every choice of positive integers k,Kthere exist a real number R (k ,K ) such that |R (k ,K )| ≤ C (k) and

K−2∑q∈Z2

ωP (q/K ) g (q/K ) =

∫P

g +k∑

h=1

K−2hδh + K−2k−2R (k,K ) .

We are going to sketch the very first part of the proof. As before we have

limε→0+

{∫R2

ε−2ϕ(ε−1s

)(gχP) (t − s) ds

}= ωP (t) g (t) .

Then Poisson summation formula yields

K−2∑n∈Z2

ωP (n/K ) (gχP) (n/K )

= limε→0+

{∑n∈Z2

ϕ (εn) gχP (Kn)

}=

∫P

g + limε→0+

∑06=n∈Z2

ϕ (εn) gχP (Kn)

.

As in Possion’s proof we need to estimate the Fourier transform gχP (ξ).While proving Pick’s theorem we have applied the divergence theorem to write

χP (m) =

∫P

e−2πim·t dt =

∫P

div

(m

−2πi |m|2e−2πim·t

)dt

=−1

2πi

L∑j=1

m·(Pj+1−Pj )=0

m · nj|m|2

|Pj+1 − Pj | .

Now some extra work gives

gχP (ξ) =

∫∂P

(ξ · ν (t)) g (t)

−2πi |ξ|2e−2πiξ·t dt

+

∫∂P

(ξ · ν (t)) (ξ · ∇g (t))

4π2 |ξ|4e−2πiξ·t dt −

∫P

ξTHessg (t) ξ

4π |ξ|4e−2πiξ·t dt .

Let P1,P2, . . . ,PL ∈ Z2 be the vertices of P (write PL+1 = P1). Let νj be theoutward unit vector to the side PjPj+1. For simplicity, we look only at the firstterm in the previous RHS, which equals

L∑j=1

ξ · νj−2πi |ξ|2

|Pj+1 − Pj |∫ 1

0

Gj (t) e−2πiξ·(Pj+(Pj+1−Pj )t) dt =:L∑

j=1

Aj (ξ) ,

where Gj (t) = g (Pj + |Pj+1 − Pj | t). For simplicity, let us write (with suitableorthogonal vectors X and Y )

Aj (Kn) = cn · XK |n|2

∫ 1

0

G (t) e−2πiKY ·nt dt .

If Y · n = 0 we have

Aj (Kn) = cn · XK |n|2

∫ 1

0

G (t) dt ,

so that ∑06=n∈Z2

ϕ (εn)Aj (Kn) = 0

because we are adding an odd sequence.If Y · n 6= 0 integration by parts yields

Aj (Kn) = cn · XK |n|2

∫ 1

0

G (t) e−2πiKY ·nt dt

=k−1∑h=0

chn · X

(G (h) (1)− G (h) (0)

)K h+2 |n|2 (Y · n)h+1

+c ′n · X

K k+1 |n|2 (Y · n)k

∫ 1

0

G (k) (t) e−2πiKY ·nt dt .

Going back to∑Y ·n 6=0

ϕ (εn)Aj (Kn)

=k−1∑h=0

chG (h) (1)− G (h) (0)

K h+2

∑Y ·n 6=0

ϕ (εn)n · X

|n|2 (Y · n)h+1︸ ︷︷ ︸=0 if h is odd

+c ′1

K k+1

∫ 1

0

G (k) (t)

∑Y ·n 6=0

ϕ (εn)n · X

|n|2 (Y · n)ke−2πiKY ·nt

dt

we need to check the convergence of the series.

We observe that ∑n·Y 6=0

n · X|n|2 (Y · n)k

e−2πiKY ·nt

plays a similar role to the periodised Bernoulli polynomials ψk in the 1-variableEuler-Maclaurin summation formula.

The first order approximation case in the previous theorem yields the followingresult, which can be considered as an estimate for the error in a 2-dimensionaltrapezoidal rule.

Corollary

Let P ⊂ R2 be an open integral polygon. Let g (t) be a smooth function on anopen set containing the closure of P. Then, for every K ∈ N,∣∣∣∣∣∣∣∣∫P

g −

1

K 2

∑m∈Z2,

K−1m∈P

g(K−1m

)+

1

2K 2

∑m∈Z2,

K−1m∈∂P

g(K−1m

)∣∣∣∣∣∣∣∣ ≤

C

K 2.

We end by showing how to improve the speed of convergence from K−2 to K−4.

Corollary

Let P and g (t) be as before. Let

S (K ) =1

K 2

∑m∈Z2

ωP

(K−1m

)g(K−1m

).

Then ∣∣∣∣∫P

g −(−1

3S(K ) +

4

3S(2K )

)∣∣∣∣ ≤ C

K 4.

Indeed, by the previous result there exist constants α, β, γ, . . . such that

S (K ) =

∫P

g +α

K 2+

β

K 4+ . . .

so that, for x , y to be determined,

xS (K ) + yS (2K ) = (x + y)

∫P

g +(x +

y

4

) α

K 2+(x +

y

16

) β

K 4+ . . .

The corollary stems from the observation that x = −1/3 and y = 4/3 satisfy{x + y = 1

x + y/4 = 0.

Since

−1

3S (K )+

4

3S (2K )

= −1

3

1

K 2

∑m∈Z2

ωP

(K−1m

)g(K−1m

)+

4

3

1

(2K )2

∑m∈Z2

ωP

((2K )−1 m

)g(

(2K )−1 m)

=1

3K 2

∑m∈Z2�2Z2

ωP

((2K )−1 m

)g(

(2K )−1 m),

it is a bit surprising to see that we have obtained a better approximation by ...deleting 25% of the points.

As an example, only the large points in the figure appear in the above sampling.

(0, 0)

(2, 1)

(1, 2)

In a similar way we can prove∣∣∣∣∫P

g (x) dx −(

1

45S (N)− 20

45S (2N) +

64

45S (4N)

)∣∣∣∣ ≤ C

N6.

And so on...

The above trick goes back to Huygens (1654). In 1676 Newton explainedHuygens’ argument in terms of the power series expansion of the sine function:

sin (x) = x − x3/6 + x5/120− ...,4

3

sin (x)

x− 1

3

sin (2x)

2x= 1− x4/30 + ...

Then our previous corollary (speed of convergence from K−2 to K−4) can beconsidered a proof of the following very old conjecture.

Conjecture

What has been will be again, what has been done will be done again; there isnothing new under the sun. (Ecclesiastes)

THANK YOU!(and hope to meet you soon)