[PPT] - Symmetry through Geometry Nalini Joshi @monsoon0 Supported by the PowerPoint Presentation

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The University of Sydney

Nalini Joshi

@monsoon0 Supported by the Australian Research Council

Symmetry through Geometry

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Outline

✑ Symmetry ✑ Geometry ✑ Dynamics

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Photo by Dmitriy Smaglov/Thinkstock slate.com

Symmetry

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P P

Reflection

perpendicular distance from P to the mirror = perpendicular distance to its reflection.

mirror

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Translation

✎ ✎

(a, b) (0, 0) (x, y) (a + x, b + y)

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β = (−1, √ 3) α = (2, 0) (0, 0) 2π/3

An example

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β = (−1, √ 3) sα α = (2, 0)

Place a mirror

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β = (−1, √ 3) sα α = (2, 0) α + β = (1, √ 3)

Reflect

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β = (1, p 3) sα α = (2, 0) α + β = (1, p 3) sβ β = (1, p 3) (α + β) = (1, p 3) α = (2, 0)

Reflect again & again

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α1

α2

are “simple” roots α1 and α2

This is reflection group called A2.

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Roots:
Reflections:
Co-roots:
Weights:

α1, α2, . . . , αn wi(αj) = αj − 2 (αi, αj) (αi, αi) αi ˇ αi = 2 αi (αi, αi) h1, h2, . . . , hn (hi, ˇ αj) = δij

Reflection groups

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α1

α2

s2 s1

α1 + α2 −α1 − α2 −α2 −α1

h1 h2

longest root

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A2(1)

Translation by longest route

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Translation

4 40

T1

1 2 a1 = 0 a0 = 0 a2 = 0

Dynamics on the lattice

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equilateral triangle

a = 0

a1 = 0

a2 = 0

s0, s1, s2

✑ Define

to be

On the lattice

reflections across each edge

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s0(a0, a1, a2) = (−a0, a1 + a0, a2 + a0)

equilateral triangle

a = 0

a1 = 0

a2 = 0

s0, s1, s2

✑ Define

to be

On the lattice

reflections across each edge

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4 40 s1(4)

T1

1 2 1 2 1 2 a1 = 0 a0 = 0 a2 = 0

Translations as reflections

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4 40 s2(s1(4))

T1

1 2 1 2 1 2 1 2 a

1

= 0 a = 0 a2 = 0

Translations as reflections

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4 40 π(s2(s1(4)))

π

T1

1 2 1 2 1 2 1 2 a

1

= a = a2 = 0

Translations as reflections

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Translations

So we have

T1(a0) = a0 + k, T1(a1) = a1 − k, T1(a2) = a2

T1(a0) = π s2 s1(a0) = π s2 (a0 + a1) = π (a0 + a1 + 2a2) = a1 + a2 + 2 a0 = a0 + k

⇒

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Translations again

Define Using

T1(a0) = a0 + 1, T1(a1) = a1 − 1, T1(a2) = a2

un = T n

1 (f1), vn = T n 1 (f0)

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Translations again

Define Using

T1(a0) = a0 + 1, T1(a1) = a1 − 1, T1(a2) = a2

⇒

un = T n

1 (f1), vn = T n 1 (f0)

( un + un+1 = t − vn − a0+n

vn

vn + vn−1 = t − un + a1−n

un

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Translations again

Define Using

T1(a0) = a0 + 1, T1(a1) = a1 − 1, T1(a2) = a2

⇒

This is a discrete Painlevé equation.

un = T n

1 (f1), vn = T n 1 (f0)

( un + un+1 = t − vn − a0+n

vn

vn + vn−1 = t − un + a1−n

un

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How Columbus Discovered America. London: Bancroft & Co. 1961, by Vojtěch Kubašta

https://library.bowdoin.edu/arch/exhibits/popup/practitioners.shtml

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The University of Sydney

Gridlock in India http://thedailynewnation.com/news/57912/a-total-traffic- chaos-at-shahbagh-intersection-in-city-as-vehicles-got-stuck-up-to-make-way- through-the-massive-gridlock-this-photo-was-taken-on-saturday.html

Dynamics

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Motion follows curves

✑ In initial value space ✑ But continuation of the flow fails at singularities of curves and at base points where families of curves all intersect.

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⇢ ˙ u(t) = v(t) ˙ v(t) = − sin

u(t)
Motion of plane pendulum

u(t)

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H(u, v) = v2 2 − cos

u(t)
Phase Space

✑ Level curves of H are curves through initial values. ✑ Phase space ⟺ space

f initial values.

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Another View

⇒ 8 < : ¨ f =

˙ f 2 f − 1 2

f 2 − 1
E

=

˙ f 2 2f + 1 2

⇣ f + 1

f

⌘

f(t) = eiu(t)

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6 4 2 2 4 6 6 4 2 2 4 6

y2 = − x (x − E+) (x − E−) E± = E ± p E2 − 1

Transformed phase plane

✑ Phase curves are given by ✑ This is a pencil of cubic curves. ✑ All curves go through (0,0). ✑ Such a point is called a base point of the pencil.

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Canonical example

dlmf.nist.gov

✑ Base point: (0, 1, 0) ✑ Weierstrass cubic curves

y2 = 4 x3 − g2 x − g3

where is fixed and is free.

g2

g3

✑ Phase space is no longer

compact. In homogeneous

coordinates in the curves are CP2

w v2 = 4u3 − g2uw2 − g3w3

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Base point

6 4 2 2 4 6 6 4 2 2 4 6

Continuation of the flow is not defined through a base point.

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From JJ Duistermaat, Springer Verlag, 2010

Resolving a base point

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An example

y2 = x3

The curve has a singularity at (0,0), which can be resolved.

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Resolution 1

( x1 = x y1 = y

x

⇔ ( x = x1 y = x1 y1

e0

C(1)

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Resolution 1

( x1 = x y1 = y

x

⇔ ( x = x1 y = x1 y1

e0

C(1)

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Resolution 1

( x1 = x y1 = y

x

⇔ ( x = x1 y = x1 y1

f = y2 − x3 = x1

2y1 2 − x3 1

= x1

2

y1

2 − x1

e0

C(1)

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Resolution 1

( x1 = x y1 = y

x

⇔ ( x = x1 y = x1 y1

f = y2 − x3 = x1

2y1 2 − x3 1

= x1

2

y1

2 − x1

e0 = {x1 = 0},

f (1) = y1

2 − x1

e0

C(1)

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Resolution 2

(

x2 = x1

y1

y2 = y1 ⇔ ( x1 = x2 y2 y1 = y2

f (1)(x1, y1) = y2

2 − x2y2

= y2

y2 − x2
e1 = {y2 = 0},

f (2) = y2 − x2

e(1)

C(2)

e1

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e(2) e(1)

1

e2

C(3)

Resolution 3

(

x3 = x2 y3 = y2

x2

⇔ ( x2 = x3 y2 = x3 y3

f (2)(x2, y2) = x3 y3 − x3 = x3

y3 − 1
e2 = {x3 = 0},

f (3) = y3 − 1

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Intersection theory

e(2) e(1)

1

e2

C(3)

2
2
1

✑ Each line has a self- intersection number. ✑ Each blow up reduces the self-intersection number by 1. ✑ The lines of self- intersection -2 play a special role.

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Intersection theory

e(2) e(1)

1

e2

C(3)

2
2
1

✑ Each line has a self- intersection number. ✑ Each blow up reduces the self-intersection number by 1. ✑ The lines of self- intersection -2 play a special role.

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e(2) e(1)

1

e2

C(3)

✑ The curves with self- intersection -2 correspond to nodes of a Dynkin diagram. ✑ This is DuVal (or Mackay) correspondence.

DuVal Correspondence

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e(2) e(1)

1

e2

C(3)

✑ The curves with self- intersection -2 correspond to nodes of a Dynkin diagram. ✑ This is DuVal (or Mackay) correspondence.

DuVal Correspondence

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e(2) e(1)

1

e2

C(3)

✑ The curves with self- intersection -2 correspond to nodes of a Dynkin diagram. ✑ This is DuVal (or Mackay) correspondence. ⇓ A2

DuVal Correspondence

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This A2 is the reflection group we saw earlier.

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Painlevé and discrete Painlevé equations

✑ Equations with initial value space that is compactified and regularised after 9 blow-ups in . ✑ Sakai classified all such equations. ✑ They each have a symmetry group obtained as an

rthogonal complement of

the resolved initial value space inside the Picard lattice.

P2

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Sakai described all such equations.

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Sakai’s Description II

Initial-value spaces of Painlevé equations

Sakai 2001

Ell: A

(1)

Mul: A

(1)

A

(1)

1

A

(1)

2

A

(1)

3

A

(1)

4

A

(1)

5

A

(1)

6

A

(1)

7

A

(1)

8

A

(1)

7

Add: A

(1)

A

(1)

1

A

(1)

2

D

(1)

4

D

(1)

5

D

(1)

6

D

(1)

7

D

(1)

8

E

(1)

6

E

(1)

7

E

(1)

8

Rains 2016

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Sakai’s Description II

Initial-value spaces of Painlevé equations

Sakai 2001

Ell: A

(1)

Mul: A

(1)

A

(1)

1

A

(1)

2

A

(1)

3

A

(1)

4

A

(1)

5

A

(1)

6

A

(1)

7

A

(1)

8

A

(1)

7

Add: A

(1)

A

(1)

1

A

(1)

2

D

(1)

4

D

(1)

5

D

(1)

6

D

(1)

7

D

(1)

8

E

(1)

6

E

(1)

7

E

(1)

8

Rains 2016

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Sakai’s Description I

Symmetry groups of Painlevé equations

Ell: E

(1)

8

Mul: E

(1)

8

E

(1)

7

E

(1)

6

A

(1)

3

A

(1)

4

(A2 + A1)(1) (A1 + A1)(1) A

(1)

1

A

(1)

A

(1)

1

Add: E

(1)

8

E

(1)

7

E

(1)

6

D

(1)

4

A

(1)

3

2A

(1)

1

A

(1)

1

A

(1)

A

(1)

2

A

(1)

1

A

(1)

Sakai 2001 Rains 2016

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Sakai’s Description I

Symmetry groups of Painlevé equations

Ell: E

(1)

8

Mul: E

(1)

8

E

(1)

7

E

(1)

6

A

(1)

3

A

(1)

4

(A2 + A1)(1) (A1 + A1)(1) A

(1)

1

A

(1)

A

(1)

1

Add: E

(1)

8

E

(1)

7

E

(1)

6

D

(1)

4

A

(1)

3

2A

(1)

1

A

(1)

1

A

(1)

A

(1)

2

A

(1)

1

A

(1)

Sakai 2001 Rains 2016

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Elliptic difference equations ✑ Sakai’s elliptic difference equation is at the top of the classification. ✑ Its symmetry group is E8(1).

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At the other end of the classification is the first Painlevé equation, which has initial value space E8(1).

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PI : y00 = 6y2 + x PII : y00 = 2y3 + xy + α PIV : y00 = y02 2y + 3 2y3 + 4xy2 + 2(x2 + α)y + β y Analysis of solutions in limit x → ∞

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Solutions of

Solutions possess movable

poles

General solutions are highly

transcendental functions.

Asymptotic behaviours needed

for applications.

Fornberg & Weideman 2009

PI

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General Solutions

PI:
in system form
has t-dependent Hamiltonian

d dt ✓ w1 w2 ◆ = ✓ w2 6 w2

1 − t

◆ H = w2

2

2 − 2 w3

1 + t w1

wtt = 6w2 − t

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Perturbed Form

In Boutroux’s coordinates:
a perturbation of an elliptic curve as |z| → ∞

E = u2

2

2 − 2u3

1 + u1 ⇒ dE

dz = 1 5z (6E + 4u1) w1 = t1/2 u1(z), w2 = t3/4u2(z), z = 4 5t5/4 ✓ ˙ u1 ˙ u2 ◆ = ✓ u2 6u2

1 − 1

◆ − 1 5z ✓2u1 3u2 ◆

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Perturbed Form

In Boutroux’s coordinates:
a perturbation of an elliptic curve as |z| → ∞

E = u2

2

2 − 2u3

1 + u1 ⇒ dE

dz = 1 5z (6E + 4u1) w1 = t1/2 u1(z), w2 = t3/4u2(z), z = 4 5t5/4 ✓ ˙ u1 ˙ u2 ◆ = ✓ u2 6u2

1 − 1

◆ − 1 5z ✓2u1 3u2 ◆

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Initial-Value Space of PI

There are nine base points:
Only the last one differs from the elliptic case.

b0 : u031 = 0, u032 = 0 b1 : u111 = 0, u112 = 0 b2 : u211 = 0, u212 = 0 b3 : u311 = 4, u312 = 0 b4 : u411 = 4, u412 = 0 b5 : u511 = 0, u512 = 0 b6 : u611 = 0, u612 = 0 b7 : u711 = 32, u712 = 0 b8 : u811 = − 28 (5 z), u812 = 0

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PI

L9 L8(1) L7(2)

Duistermaat & Joshi, 2011

−2

L6(3) L5(4) L4(5) L3(6)

−2 −2 −2 −2

−2

L0(9)

−2

L1(9)

−2

L2(8)

−2

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PI

L9 L8(1) L7(2)

Duistermaat & Joshi, 2011

−2

L6(3) L5(4) L4(5) L3(6)

−2 −2 −2 −2

−2

L0(9)

−2

L1(9)

−2

L2(8)

−2

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PI

L9 L8(1) L7(2)

E8(1)

Duistermaat & Joshi, 2011

−2

L6(3) L5(4) L4(5) L3(6)

−2 −2 −2 −2

−2

L0(9)

−2

L1(9)

−2

L2(8)

−2

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PI

L9 L8(1) L7(2)

E8(1)

Duistermaat & Joshi, 2011

autonomous eqn

−2

L6(3) L5(4) L4(5) L3(6)

−2 −2 −2 −2

−2

L0(9)

−2

L1(9)

−2

L2(8)

−2

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PII

Howes & Joshi, 2014

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PII

Howes & Joshi, 2014

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PII

E7(1)

Howes & Joshi, 2014

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PII

E7(1)

Howes & Joshi, 2014

autonomous eqn

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PIV

L∗ L∗

1

L∗

2

L∗

3

L∗

4

L7(z) L∗

5

L8(z) L∗

6

L9(z)

Joshi & Radnovic, 2015

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PIV

L∗ L∗

1

L∗

2

L∗

3

L∗

4

L7(z) L∗

5

L8(z) L∗

6

L9(z)

Joshi & Radnovic, 2015

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PIV

L∗ L∗

1

L∗

2

L∗

3

L∗

4

L7(z) L∗

5

L8(z) L∗

6

L9(z)

E6(1)

Joshi & Radnovic, 2015

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PIV

L∗ L∗

1

L∗

2

L∗

3

L∗

4

L7(z) L∗

5

L8(z) L∗

6

L9(z)

E6(1)

Joshi & Radnovic, 2015

autonomous eqn

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Global results for PI , PII , PIV

✑ The union of exceptional lines is a repeller for the

flow.

✑ There exists a complex limit set, which is non-empty,

connected and compact.

✑ Every solution of PI , every solution of PII whose limit

set is not {0}, and every non-rational solution of PIV intersects the last exceptional line(s) infinitely many times ⇒ ∃ infinite number of movable poles and movable zeroes.

Duistermaat & J (2011); Howes & J (2014); J & Radnovic (2015, 2016)

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Ingredients of proofs

– Energy and Jacobians of each coordinate chart. – Use each as a measure between the flow and the   exceptional lines. – Use constancy of measure near an exceptional line to estimate the domain in each chart in which the solution is analytic. – Use compactness to show results about the limit set.

E = u2

2

2 − 2u3 1 + u1

Jij = ∂uij1

∂u1 ∂uij2 ∂u2 − ∂uij1 ∂u2 ∂uij2 ∂u1

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A similar approach provides asymptotic behaviours of discrete Painlevé equations.

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qP1

⇒ w w = 1 w − 1 ξ w2

qPI
w = w(q ξ), w = w(ξ), w = w(ξ/q)

y00 = 6 y2 − t

PI: in continuum limit.

7!

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qP1

e2 e5-e7 e4-e8 e8 hv - e1-e3 hu -e2-e4 hv -e4 -e6

P1 × P1

A7(1)

hu-e3-e5

J & Lobb, 2016

e7 e3-e5 e1

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Remark on elliptic difference equations ✑ Sakai’s equation arises by translation from one point to the nearest neighbour on the E8(1) lattice. ✑ Recently, we discovered a new equation from translations to the next- nearest neighbour on the E8(1) lattice.

J. & Nakazono 2017

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Remark on elliptic difference equations ✑ Sakai’s equation arises by translation from one point to the nearest neighbour on the E8(1) lattice. ✑ Recently, we discovered a new equation from translations to the next- nearest neighbour on the E8(1) lattice.

J. & Nakazono 2017

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Summary

New mathematical models of physics pose new

questions for applied mathematics

Global dynamics of solutions of non-linear

equations, whether they are differential or discrete, can be found through geometry.

Geometry provides the only analytic approach

available in for discrete equations.

Tantalising questions about finite properties of

solutions remain open.

C

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Summary

✑Global dynamics of solutions of non-

linear equations, whether they are differential or discrete, can be found through geometry.

✑Geometry provides the only analytic

approach available in for discrete equations.

✑Tantalising questions about finite

properties of solutions remain open and discrete transcendents remain open.

C

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