Experimental Mathematics& Computer Algebra

Bruno Salvy

Saint-Flour, June 21-25, 2018

Part II

Short Summary of Part I

Experimental Mathematics

A 3-step process:

1. Compute a high order approximation

(high precision numerical approx., power series truncated to high order, large number of terms in a sequence,…)

2. Guess/conjecture a general formula

3. Prove it

(with the help of a computer)

(using computer-algebra algorithms)

Basic Computer Algebra

Simplification is undecidable.

Tools for conjectures:

OEIS (integer sequences) ISC (real numbers) Padé-Hermite approximants (relations between power series) LLL (relations between numerical approximations).

Fast computation with large integers, polynomials, power series, matrices,…

Exercises for Part I

Ex 1. Identities for Nice Constants1X

i=0

16�i

8i+ j, j = 1, . . . , 8

<latexit sha1_base64="jOmSrHScwIWBUZgX2HYDYh7P7r0=">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</latexit><latexit sha1_base64="5XftniVulWRjwiL6XNqHiLpA+b4=">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</latexit><latexit sha1_base64="5XftniVulWRjwiL6XNqHiLpA+b4=">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</latexit><latexit sha1_base64="q5KxLpBFplp/oFAhjchS1bJFdzQ=">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</latexit>

lead to formulas for ⇡, ln 2, ln 3, ln 5, arctan 2, arctan 3,

p2 arctan(1/

p2),

p2 ln(1 +

p2).

<latexit sha1_base64="6JNNI6LNQHzzLkVY0I+aCgokkmE=">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</latexit><latexit sha1_base64="qeYsROo+1EYDUwIcyBI5alLS3zE=">AAADC3icbVJNb9NAEN2YrxK+UjhyWbWKVChK7VSoHCO4cKNITVspjqL1epysurs2u+O2luWfACf+CTfElRO/oD+DK1xYxwm0KSNZ8/TejGbfjKNMCou+f9Hybty8dfvO2t32vfsPHj7qrD8+tGluOAx5KlNzHDELUmgYokAJx5kBpiIJR9HJm1o/OgVjRaoPsMhgrNhUi0Rwho6adCDMxAsaSk37Tdpt0kuXmOHIGmEBa9F+MFj2qwW1FewsmWeXROmE7b9Cb9LZ9Hv+POh1ECzA5mAj3P58MSj2J+utT2Gc8lyBRi6ZtaPAz3BcMoOCS6ja3dCCM6unOCtDhHM8EzHOqnLX56pqh7mFjPETNoWRg5opsOPyfL6uinYdFdMkNe7TSOfs5ZaSKWsLFblKxXBmV7Wa/J82yjF5NS6FznIEzZtBSS4pprTePY2FAY6ycIBxI5wVymfMMI7uQu3QgIYznirFdPw8tEkMCcsllnmaVatqwpSQxbLiX7Eb2qXvtJshEoozoPXP0fhMG2K5Vips/bAIqGXaUgtGJFfcHATjsm6srbTdBYPVe10Hh/1e4PeC9+6Ur0kTa+Qp2SBbJCB7ZEDekn0yJJz8ID/JL/Lb++h98b5635pSr7XoeUKuhPf9D+X++i8=</latexit><latexit sha1_base64="qeYsROo+1EYDUwIcyBI5alLS3zE=">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</latexit><latexit sha1_base64="Vxwkcr4ZSFTWayr8CoHE5dx89+s=">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</latexit>

Ex 2. Stanley’s Problem E 2297

Number of monomials in a generic nxn symmetric matrix

Ex 3. Number of Domino Tilings of the Aztec Diamond

Guess & Prove a formula

Part II: Tools for Proofs

A short tour of ``univariate’’ computer algebra

1. Resultants 2. D-finite sequences & series 3. Creative telescoping

1. Resultants

Polynomials as a data-structure

Definition

The Sylvester matrix of A = amxm + · · ·+ a0 2 K[x], (am 6= 0), and of

B = bnxn + · · ·+ b0 2 K[x], (bn 6= 0), is the square matrix of size m+ n

Syl(A,B) =

2

666666666666666664

am am�1 . . . a0

am am�1 . . . a0. . .

. . .. . .

am am�1 . . . a0

bn bn�1 . . . b0

bn bn�1 . . . b0. . .

. . .. . .

bn bn�1 . . . b0

3

777777777777777775

The resultant Res(A,B) of A and B is the determinant of Syl(A,B).

I Definition extends to polynomials with coe�cients in a commutative ring R.

Definition

n rows

m rows

Basic observation

If A = amxm + · · ·+ a0 and B = bnxn + · · ·+ b0, then

2

6666666666664

am am�1 . . . a0. . .

. . .. . .

am am�1 . . . a0

bn bn�1 . . . b0. . .

. . .. . .

bn bn�1 . . . b0

3

7777777777775

⇥

2

6666664

↵m+n�1

...

↵

1

3

7777775=

2

6666666666664

↵n�1A(↵)...

A(↵)

↵m�1B(↵)...

B(↵)

3

7777777777775

Corollary: If A(↵) = B(↵) = 0, then Res (A,B) = 0.

Basic Observation

Example: the discriminant

The discriminant of A is the resultant of A and of its derivative A0.

E.g. for A = ax2 + bx+ c,

Disc(A) = Res (A,A0) = det

2

664

a b c

2a b

2a b

3

775 = �a(b2 � 4ac).

E.g. for A = ax3 + bx+ c,

Disc(A) = Res (A,A0) = det

2

666666664

a 0 b c

a 0 b c

3a 0 b

3a 0 b

3a 0 b

3

777777775

= a2(4b3 + 27ac2).

I The discriminant vanishes when A and A0 have a common root, that is

when A has a multiple root.

Example: the Discriminant

Main properties

• Link with gcd Res (A,B) = 0 if and only if gcd(A,B) is non-constant.

• Elimination property

There exist U, V 2 K[x] not both zero, with deg(U) < n, deg(V ) < m and

such that the following Bezout identity holds:

Res (A,B) = UA+ V B in K \ (A,B).

• Poisson formula

If A = a(x� ↵1) · · · (x� ↵m) and B = b(x� �1) · · · (x� �n), then

Res (A,B) = anbmY

i,j

(↵i � �j) = anY

1im

B(↵i).

• Bezout-Hadamard bound

If A,B 2 K[x, y], then Res y(A,B) is a polynomial in K[x] of degree

degx(A) degy(B) + degx(B) degy(A).

Main Properties

Application: computation with algebraic numbers

Let A =Q

i(x� ↵i) and B =

Qj(x� �j) be polynomials of K[x]. Then

Res x(A(x), B(t� x)) =Y

i,j

(t� (↵i + �j)),

Res x(A(x), B(t+ x)) =Y

i,j

(t� (�j � ↵i)),

Res x(A(x), xdegBB(t/x)) =Y

i,j

(t� ↵i�j),

Res x(A(x), t�B(x)) =Y

i

(t�B(↵i)).

In particular, the set of algebraic numbers is a field.

Proof: Poisson’s formula. E.g., first one:Y

i

B(t� ↵i) =Y

i,j

(t� ↵i � �j).

I The same formulas apply mutatis mutandis to algebraic power series.

Application: Computation with Algebraic Numbers

Exercise for the afternoon:A Nice Number

Guess and prove a simple formula for

sin 2⇡7

sin2 3⇡7

�sin ⇡

7

sin2 2⇡7

+sin 3⇡

7

sin2 ⇡7

.<latexit sha1_base64="RObmBAj89If+4l8HN4zZB2/Jwo8=">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</latexit><latexit sha1_base64="ZdoC/jfHX4qtRvefMO2NyaqhoHI=">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</latexit><latexit sha1_base64="ZdoC/jfHX4qtRvefMO2NyaqhoHI=">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</latexit><latexit sha1_base64="AgxEsncO9xmcoZd8DtmhsIANXCo=">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</latexit>

Rothstein-Trager resultant

Let A,B 2 K[x] with deg(A) < deg(B) and squarefree monic denominator B.

The rational function F = A/B has simple poles only.

If F =X

i

�ix� �i

, then the residue �i of F at the pole �i equals �i =A(�i)

B0(�i).

Theorem. The residues �i of F are roots of the Rothstein-Trager resultant

R(t) = Res x

�B(x), A(x)� t ·B0(x)

�.

Proof. Poisson formula again: R(t) =Y

i

⇣A(�i)� t ·B0(�i)

⌘.

I This special resultant is useful for symbolic integration of rational functions.

Rothstein-Trager Resultant

Application: diagonal Rook paths

Question: A chess Rook can move any number of squares horizontally or

vertically in one step. How many paths can a Rook take from the lower-left

corner square to the upper-right corner square of an N ⇥N chessboard?

Assume that the Rook moves right or up at each step.

1, 2, 14, 106, 838, 6802, 56190, 470010, . . .

Application: Diagonal Rook Paths

Application: diagonal Rook paths

1, 2, 14, 106, 838, 6802, 56190, 470010, . . .

Diag(F ) = [s0]F (s, x/s) =1

2i⇡

IF (s, x/s)

ds

s, where F =

1

1� s

1�s�

t

1�t

.

By the residue theorem, Diag(F ) is a sum of roots of the Rothstein-Trager

resultant

> F:=1/(1-s/(1-s)-t/(1-t)):

> G:=normal(1/s*subs(t=x/s,F)):

> factor(resultant(denom(G),numer(G)-t*diff(denom(G),s),s));

2 2 2

x (-1 + 2 t) (x - 1) (-x + 36 t x + 1 - 4 t )

Answer: Generating series of diagonal Rook paths is1

2

1 +

r1� x

1� 9x

!.

Diagonal Rook Paths II

Application: certified algebraic guessingGuess + Bound = Proof

Theorem. Suppose A 2 K[[x]] is an algebraic series, and that it is a root of a

(unknown) polynomial in K[x, y] of degree at most d in x and at most n in y.

If

nX

i=0

Qi(x)Ai(x) = O(x2dn), then

nX

i=0

Qi(x)Ai(x) = 0.

Proof: Let P 2 K[x, y] be an irreducible polynomial such that

P (x,A(x)) = 0, and degx(P ) d, degy(P ) n.

• By Hadamard, R(x) = Res y(P,Q) 2 K[x] has degree at most 2dn.

• By elimination, R(x) = UP + V Q for U, V 2 K[x, y] with degy(V ) < n.

• Evaluation at y = A(x) yields

R(x) = U(x,A(x))P (x,A(x))| {z }0

+V (x,A(x))Q(x,A(x))| {z }O(x2dn)

= O(x2dn).

• Thus R = 0, that is gcd(P,Q) 6= 1, and thus P |Q, and A is a root of Q.

Application: Certified Algebraic GuessingGuess+Bound=Proof

Systems of two equations and two unknowns

Geometrically, roots of a polynomial f 2 Q[x] correspond to points on a line.

Roots of polynomials A 2 Q[x, y] correspond to plane curves A = 0.

Let now A and B be in Q[x, y]. Then:

• either the curves A = 0 and B = 0 have a common component,

• or they intersect in a finite number of points.

Systems of 2 Equations in 2 Unknowns

Application: Resultants compute projections

Theorem. Let A = amym + · · · and B = bnyn + · · · be polynomials in Q[x][y].

The roots of Res y(A,B) 2 Q[x] are either the abscissas of points in the

intersection A = B = 0, or common roots of am and bn.

Proof. Elimination property: Res (A,B) = UA+ V B.

Thus A(↵,�) = B(↵,�) = 0 implies Res y(A,B)(↵) = 0

Resultants Compute Projections

Application: implicitization of parametric curves

Task: Given a rational parametrization of a curve

x = A(t), y = B(t), A,B 2 K(t),

compute a non-trivial polynomial in x and y vanishing on the curve.

Recipe: take the resultant in t of numerators of x�A(t) and y �B(t).

Example: for the four-leaved clover (a.k.a. quadrifolium) given by

x =4t(1� t2)2

(1 + t2)3, y =

8t2(1� t2)

(1 + t2)3,

Res t((1+t2)3x�4t(1�t2)2, (1+t2)3y�8t2(1�t2)) = 224�(x2 + y2)3 � 4x2y2

�.

Application: Implicitization of Parametric Curves

2. D-finite Series and Sequences

Differential or Recurrence equations as a data-structure

D-finite Series & Sequences

Definition: A power series f(x) 2 K[[x]] is D-finite over K when its derivatives

generate a finite-dimensional vector space over K(x).

A sequence un is D-finite (or P-recursive) over K when its shifts (un, un+1, . . . )

generate a finite-dimensional vector space over K(n).

equation + init conditions = data structure

About 25% of Sloane’s encyclopedia, 60% of Abramowitz & Stegun

Examples: exp, log, sin, cos, sinh, cosh,

arccos, arccosh, arcsin, arcsinh, arctan,

arctanh, arccot, arccoth, arccsc, arccsch,

arcsec, arcsech, pFq (includes Bessel J , Y , I

andK, Airy Ai and Bi and polylogarithms),

Struve, Weber and Anger functions, the

large class of algebraic functions,. . .

D-finite Series & Sequences

Automatic Proof of Identities

> series(sin(x)^2+cos(x)^2-1,x,4);

O(x4)

Why is this a proof?

1. sin and cos satisfy a 2nd order LDE: y’’+y=0; 2. their squares and their sum satisfy a 3rd order LDE; 3. the constant -1 satisfies y’=0; 4. thus sin2+cos2-1 satisfies a LDE of order at most 4; 5. the Cauchy-Lipschitz theorem concludes.

Proofs of non-linear identities by linear algebra!

f satisfies a LDE⟺

f,f’,f’’,… live in a finite-dim. vector space

Mehler’s identity for Hermite polynomials

1X

n=0

Hn(x)Hn(y)un

n!=

exp⇣

4u(xy�u(x2+y2))1�4u2

⌘

p1� 4u2

1. Definition of Hermite polynomials: recurrence of order 2;

2. Product by linear algebra: Hn+k(x)Hn+k(y)/(n+k)!, k∈ℕgenerated over (x,n) by → recurrence of order at most 4;

3. Translate into differential equation.

QHn(x)Hn(y)

n!,Hn+1(x)Hn(y)

n!,Hn(x)Hn+1(y)

n!,Hn+1(x)Hn+1(y)

n!

Guess & prove continued fractions

arctan x =x

1+13x

2

1+415x

2

1+935x

2

1+ · · ·

1. Taylor expansion produces first terms (easy):

2. Guess a formula (easy): an =n2

4n2 � 13. Prove that the CF with these an converges to arctan.

No human intervention needed. gfun[ContFrac]

Automatic Proof of the Guessed CF (1/2)

arctanx?=

x1

+ · · ·+n2

4n2�1x2

1+ · · ·

1. Compute a linear recurrence for 2. Using the initial conditions, find a smaller-order one; 3. Conclude that

Hn;

Hn = O(xn).

Let Hn := Q2n((x

2 + 1)(Pn/Qn)0 � 1).

Lemma. Let be the nth convergent. ThenPn/Qn

limn!1

(x2 + 1)

✓Pn

Qn

◆0� 1

!= 0 ) lim

n!1

Pn

Qn= arctanx.

Automatic Proof of the Guessed CF (2/2)

1. Pn and Qn satisfy un=un-1+anx2un-2 and Qn(0)≠0.

2. Hn is a polynomial in Pn,Qn and their derivatives.

3. All its shifts Hn+k are linear combinations ofP 0n+iQn+j , Pn+iQ0

n+j , Pn+iPn+j , Qn+iQn+j , i and j in {0, 1}

Hn := Q2n((x

2 + 1)(Pn/Qn)0 � 1).

All these steps are easy to automate.

4. → by linear algebra

Hn+4 = Hn+3 + (· · ·x2 + · · ·x4)Hn+2 + · · ·x4Hn+1 + · · ·x8

Hn

5. Using the initial conditions gives

(2n+ 3)2Hn+1 + (n+ 2)2x2Hn = 0.

Algebraic Series can be Computed Fast

P (X,Y (X)) = 0

Px(X,Y (X)) + Py(X,Y (X)) · Y 0(X) = 0

Y 0(X) = (�PxP�1y mod P )(X,Y (X))

Y (X), Y 0(X), Y 00(X), . . . inVectQ(X)(1, Y, Y2, . . . )

a polynomial

finite dimension

Wanted: the first N Taylor coefficients of Y.

→ a LDE by linear algebra

[Abel1827,Cockle1861,Harley1862,Tannery1875]

P irreducible

Note: F sol LDE

⇒ F(Y(X)) sol LDE

(same argument)

An Olympiad Problem

Question: Let (an) be the sequence with a0 = a1 = 1 satisfying the recurrence

(n+ 3)an+1 = (2n+ 3)an + 3nan�1.

Show that all an is an integer for all n.

Computer-aided solution: Let’s compute the first 10 terms of the sequence:

> rec:=(n+3)*a(n+1)-(2*n+3)*a(n)-3*n*a(n-1): ini:=a(0)=1,a(1)=1:

> pro:=gfun:-rectoproc({rec,ini}, a(n), list);

> pro(10);

[1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188]

gfun’s seriestoalgeq command allows to guess that GF is algebraic:

> pol:=gfun:-listtoalgeq(%,y(x))[1];

2 2

1 + (x - 1) y(x) + x y(x)

An Olympiad Problem

Thus it is very likely that y =P

n�0 anxn verifies 1 + (x� 1)y + x2y2 = 0.

By coe�cient extraction, (an) conjecturally verifies the non-linear recurrence

an+2 = an+1 +nX

k=0

ak · an�k. (1)

Clearly (1) implies an 2 N. To prove (1), we proceed the other way around:

we start with P (x, y) = 1 + (x� 1)y + x2y2, and show that it admits a power

series solution whose coe�cients satisfy the same linear recurrence as (an):

> deq:=gfun:-algeqtodiffeq(pol,y(x)):

> recb:=gfun:-diffeqtorec(deq,y(x),b(n));

recb := {(3 + 3 n) b(n) + (2 n + 5) b(n + 1) + (-4 - n) b(n + 2),

b(0) = 1, b(1) = 1}

I In fact,

an =nX

k=0

(�1)n�k

k + 2

✓n

k

◆✓2k + 2

k + 1

◆=

X

j�0

(�1)j✓n+ 1

j

◆✓2n� 3j

n

◆,

(which clearly implies an 2 Z) but how to find algorithmically such a formula?

3. Creative Telescoping

Examples I: hypergeometric summation

•X

k2Z(�1)k

✓a+ b

a+ k

◆✓a+ c

c+ k

◆✓b+ c

b+ k

◆=

(a+ b+ c)!

a!b!c!

• An =nX

k=0

✓n

k

◆2✓n+ k

k

◆2

satisfies the recurrence [Apery78]:

(n+ 1)3An+1 = (34n3 + 51n2 + 27n+ 5)An � n3An�1.

(Neither Cohen nor I had been able to prove this in the intervening two

months [Van der Poorten]).

•nX

k=0

✓n

k

◆2✓n+ k

k

◆2

=nX

k=0

✓n

k

◆✓n+ k

k

◆ kX

j=0

✓n

k

◆3

[Strehl92]

Examples II: Integrals

•Z 1

0

cos(zu)p1� u2

du =

Z +1

1

sin(zu)pu2 � 1

du =⇡

2J0(z);

•Z +1

0xJ1(ax)I1(ax)Y0(x)K0(x) dx = �

ln(1� a4)

2⇡a2[Glasser-Montaldi94];

•1

2⇡i

I (1 + 2xy + 4y2) exp⇣

4x2y2

1+4y2

⌘

yn+1(1 + 4y2)32

dy =Hn(x)

bn/2c![Doetsch30].

Examples III: Diagonals

Definition If f(x1, . . . , xk) =X

i1,i2,...,ik�0

ci1,...,ikxi11 · · ·xik

k2 K[[x1, . . . , xk]], then

its diagonal is Diag(f) =X

n�0

cn,...,nxn2 K[[x]].

• Diagonal k-D rook paths: Diag1

1� x11�x1

� · · ·�xk

1�xk

;

• Hadamard product: F (x)�G(x) =P

nfngnxn = Diag(F (x)G(y));

• Algebraic series [Furstenberg67]: if P (x, S(x)) = 0 and Py(0, 0) 6= 0 then

S(x) = Diag

✓y2

Py(xy, y)

P (xy, y)

◆.

• Apery’s sequence [Dwork80]:

XAnz

n = Diag1

(1� x1)((1� x2)(1� x3)(1� x4)(1� x5)� x1x2x3).

Theorem [Lipshitz88] The diagonal of a rational (or algebraic, or even

D-finite) series is D-finite.

Summation by Creative Telescoping

In :=nX

k=0

✓n

k

◆= 2n.

IF one knows Pascal’s triangle:✓n+ 1

k

◆=

✓n

k

◆+

✓n

k � 1

◆= 2

✓n

k

◆+

✓n

k � 1

◆�

✓n

k

◆,

then summing over k gives

In+1 = 2In.

The initial condition I0 = 1 concludes the proof.

Creative Telescoping for Sums

Fn =X

k

un,k =?

IF one knows A(n, Sn) and B(n, k, Sn, Sk) s.t.

(A(n, Sn) +�kB(n, k, Sn, Sk)) · un,k = 0

(where �k is the di↵erence operator, �k · vn,k = vn,k+1 � vn,k),

then the sum “telescopes”, leading to

A(n, Sn) · Fn = 0.

Zeilberger’s Algorithm [1990]

Input: a hypergeometric term un,k, i.e., un+1,k/un,k and un,k+1/un,k rational

functions in n and k;

Output:

• a linear recurrence (A) satisfied by Fn =P

kun,k

• a certificate (B), s.t. checking the result is easy from

A(n, Sn) · un,k = �kB · un,k.

Example: SIAM flea

1/4

1/4

1/4-ε 1/4+ε

Un,k :=

✓2n

2k

◆✓2k

k

◆✓2n� 2k

n� k

◆✓1

4+ c

◆k ✓1

4� c

◆k 1

42n�2k.

> SumTools[Hypergeometric][Zeilberger](U,n,k,Sn);

[�4n2 + 16n+ 16

�Sn2 +

��4n2 + 32 c2n2 + 96 c2n� 12n+ 72 c2 � 9

�Sn

+ 128 c4n+ 64 c4n2 + 48 c4, ...(BIG certificate)...]

Creative Telescoping for Integrals

I(x) =

Z

⌦u(x, y) dy =?

IF one knows A(x, @x) and B(x, y, @x, @y) s.t.

(A(x, @x) + @yB(x, y, @x, @y)) · u(x, y) = 0,

then the integral “telescopes”, leading to

A(x, @x) · I(x) = 0.

Special Case: Diagonals

Analytically,

Diag(F (x, y)) =1

2⇡i

IF

✓x

y, y

◆dy

y.

On power series,

(A(x, @x) + @yB) ·1

yF

✓x

y, y

◆

| {z }U

= 0 =) A(x, @x) ·DiagF = 0.

Proof:

1. [y�1]U = Diag(f);

2. [y�1]A · U + [y�1]@yB · U = A · [y�1]U .

Extends to more variables: DiagF (x, y, z) obtained from [y�1z�1]U ,

U = 1yzF⇣

x

y, y

z, z⌘, if one finds

(A(x, @x) + @yB(x, y, z, @x, @y, @z) + @zC(x, y, z, @x, @y, @z)) · U = 0.

Provided by Chyzak’s algorithm

Summary of the Exercisesfor this Afternoon

4. A nice number

Guess and prove a simple formula for sin 2⇡7

sin2 3⇡7

�sin ⇡

7

sin2 2⇡7

+sin 3⇡

7

sin2 ⇡7

.<latexit sha1_base64="RObmBAj89If+4l8HN4zZB2/Jwo8=">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</latexit><latexit sha1_base64="ZdoC/jfHX4qtRvefMO2NyaqhoHI=">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</latexit><latexit sha1_base64="ZdoC/jfHX4qtRvefMO2NyaqhoHI=">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</latexit><latexit sha1_base64="AgxEsncO9xmcoZd8DtmhsIANXCo=">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</latexit>

See also the exercises on resultants.

Pb. Kreweras Excursions

· · ·· ·· · ·

��✓�?

•

•

·

·

·

·

·

·

•

·

·

·

·

·

•

•

•

•

·

·

·

·

•

·

•

·

·

·

·

•

•

•

·

·

·

·

•

•

•

•

·

·

·

·

•

•

•

•

·

·

·

·

·

·

·

·

���✓

���✓

���✓

�

?

?

?���✓

���✓

���✓

����✓

���✓

?

?

?�

?�����

?

K(t;x, y) =1X

n=0

0

@X

i,j�0

k(n; i, j)xiyj

1

A tn

<latexit sha1_base64="70//y1Hq8oQHTSOV2p4HH1cKQFE=">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</latexit><latexit sha1_base64="tEdBeVywuwH+n8Bku2S8hvohJDY=">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</latexit><latexit sha1_base64="tEdBeVywuwH+n8Bku2S8hvohJDY=">AAAC8HicbVJNaxRBEO0dv+L6tdGjl8awsCshzMSDQggGvYg5GCGbBHZ2l56emp3OTvcM3TVmh2HO3jx7E6+e/Av+BkHwpL9DEOzZSdBsLGh4/V4V1a+qgywRBl33e8u5dPnK1Wsr19s3bt66faezevfApLnmMOBpkuqjgBlIhIIBCkzgKNPAZJDAYTB7XuuHb0Abkap9LDIYSTZVIhKcoaUmnZe7Pdyarxf9bd/kclKqbbca+0JFWJR+AhH2Gl6sH/tTcKty1lNb9tKfj0UxPq58LaYx9nGsqklnzd1wF0EvAu8UrO3s/n779NeXr3uT1dY7P0x5LkEhT5gxQ8/NcFQyjYInULW7vgHrSE0xLn2EOZ6IEOOqfORyWbX93EDG+IxNYWihYhLMqJwvZlLRrqVCGqXaHoV0wf5bUjJpTCEDmykZxmZZq8n/acMcoyejUqgsR1C8aRTlCcWU1gOmodDAMSksYFwLa4XymGnG0a6h7WtQcMJTKZkKH/omCiFieYJlnmbVshoxKZLiLONvsm3apa+U7SEiijHQ+gc0PtOGOBsrFaZ+WADUMGWoAS2ic272vVFZF9ZW2naD3vK+LoKDzQ3P3fBe21U+I02skPvkAekRjzwmO+QF2SMDwsln8o38ID8d7bx3Pjgfm1SndVpzj5wL59MfOXH1lA==</latexit><latexit sha1_base64="LIFuzUChHaiBSu6qQ5lFVEnlxtI=">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</latexit>

is algebraic.

THE END

(Except for the exercises!)