Creative TelescopingZeilberger’s Algorithm

ByDylan Heeg

Christopher JordanDeanne PieperBrent Serum

Introduction and OverviewComputerized Proofs

and Combinatorial Identities

ByDeanne Pieper

Some proofs are essentially computations. In particular, proofs of combinatorial identities are mostly computational. With computers, it has been possible to devise algorithms to do these computations and prove combinatorial identities.

What is a Combinatorial Identity?

Some examples of combinatorial identity are:

k

kn

nn

kn

kn

2

2

2

Or,

What is a Proper Hypergeometric Function?

kV

iiii

U

iiii

xwkvnu

ckbnaknPknF *

)!(

)!(),(),(

1

1

Where 1.) P(n,k) ~ Polynomial

2.) ai’s, bi’s, ui’s, vi’s, ci’s, wi’s are in z

3.) 0 u, v <

4.) x is indeterminate (i.e. a variable)

Proper Hypergeometric Function Example

)!13()!3(

131),(

kn

knkn

knF

Sister Celine’s Algorithm is used in solving definite hypergeometric summations.

While Gosper’s Algorithm solves indefinite hypergeometric summations.

A Review of Sister Celine’s Algorithm

Celine’s Theorem(The Fundamental Theorem)Suppose F(n,k) is proper hypergeometric. Then, F(n,k) satisfies a k-free recurrence relation of the form

for some I, J that are positive integers

I

i

J

jji ikjnfna

0 0, 0),()(

Moreover, there will be a particular pair (I*, J*) that will work with

1*)deg(1*

*

ss

ss

ss

ss

uaJPI

vbJ

I

i

J

jji ikjnfna

0 0, 0),()(

Holds that every point (n,k) where F(n,k) 0

and all the values of F that occur in it are well defined

A Review of Gosper’s Algorithm

Gosper’s Algorithm

Input:

Output: (if possible)

Steps: i) compute r(n) ii) Perform Gosper Factorization

}{ nt

nZ

)()1(

)()()(

ncnc

nbnanr

Gosper’s Algorithm

iii) Solve Gosper’s Polynomial Recurrence

iv) Return

v) Set

)()()1()1()( ncnxnbnxna

nn tnc

nxnbZ)(

)()1(

0ZZS nn

Zeilberger’s Creative Telescoping Algorithm

Zeilberger’s TheoremLet’s take a look at the theorem that this algorithm is based on.

Let F(n,k) be a proper hypergeometric term. Then F satisfies a nontrivial recurrence of the form

in which is a rational function of n and k.

J

jj knGknGkjnFna

0

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

),(),(

knFknG

Zeilberger’s AlgorithmZeilberger’s Algorithm is a hybrid of Celine’s and Gosper’s Algorithms. They both play roles in the completion of the Creative Telescoping proof and example we will get to shortly….

A General Description….

The problem?

Strategy: find a Zeilberger Recurrence

n

knFnF ),(:

J

jj knGknGkjnFna

0

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

Usual Assumptions

i) F(n,k) Proper Hypergeometric

ii) Rational in (n,k)),(),(

knFknG

The basic mechanicsLet

We’ll be applying Gosper’s algorithm to these tk’s.

),(...),(),()(:0

0

kJnFknFkjnFn aaat j

J

jjk

J

jj

J

jj

k

k

kjnFna

kjnFna

tt

0

01

),()(

)1,()(

Linear Recurrence Relations

A homogenous linear recurrence equation with constant coefficients is one of the form:

We use its Characteristic(auxillary) equation to solve it.

\

0...110

mmm ababa

0)(...)1()( 10 mnXanXanXa m

This equation will produce (possibly repeated or even complex) roots:

We now have two possible cases to consider.

mbbb ,...,, 21

Case 1In this case the general solution to the problem is then given by linear combinations of the form:

Where the C’s are arbitrary constants determined in practice by initial conditions of the terms X(0),…,X(m-1).

nmm

nn bCbCbCnX ...)( 2211

Case 2A root b of multiplicity k will contribute the k terms

to the general solution.

nknn bnnbb 1,...,,

The mechanics of Zeilberger’s Algorithm

General ObservationsProblem: to sum in closed form something of the form

Strategy: Find a Zeilberger recurrence. This is of the form:

k

knFnf ),(:)(

J

jj knGknGkjnFna

0

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

General ObservationsOur unknowns in this recurrence are aj(n), J, and G(n,k).

We assume the following F(n,k) is proper hypergeometric is rational in n, k

),(),(

knFknG

MechanicsDefine

This means

J

jjk kjnFnat

0

),()(:

),(...),1(),( 10 kJnFaknFaknFat Jk

MechanicsNow define

If we multiply both the top and bottom by

J

jj

J

jj

k

k

kjnFna

kjnFna

ttkr

0

01

),()(

)1,()()(

),()1,(

knFknF

MechanicsWe get the equation

Which is equal to r(k)

),()1,(*

),(/),()(

)1,(/)1,()(

0

0

knFknF

knFkjnFna

knFkjnFna

J

jj

J

jj

MechanicsNote that is a rational function

If we write we know

that r1(n,k), r2(n,k) are polynomials

),()1,(

knFknF

),(),(

),()1,(

2

1

knrknr

knFknF

MechanicsSimilarly, we know that s1(n,k), s2(n,k) are polynomials when they are defined

),(),(

),1(),(

2

1

knskns

knFknF

MechanicsAlso note that we can rewriteas

Which is

),(),(

knFkjnF

),(),1(*...*

),2(),1(*

),1(),(

knFknF

kjnFkjnF

kjnFkjnF

1

0 ),1(),1(j

i kijnFkjnF

MechanicsUsing our s1 and s2 functions this can be written as

Which condenses to

),1(),1(*...*

),1(),1(*

),(),(

2

1

2

1

2

1

knskns

kjnskjns

kjnskjns

1

0 2

1

),(),(j

i kijnskijns

MechanicsNow our original can be written as

which is

k

k

tt 1

),(),(*

),(),()(

)1,()1,()(

2

1

0

1

0 2

1

0

1

0 2

1

knrknr

kijnskijnsna

kijnskijnsna

J

j

j

ij

J

j

j

ij

),(),(*

),(),()(

)1,()1,()(

2

1

0

1

0

0

1

0

knrknr

knFkjnFna

knFkjnFna

J

j

j

ij

J

j

j

ij

MechanicsNext we clear denominators. The LCD of this fraction is

Clearing denominators we get

1

0

1

022 ),()1,(

j

i

j

i

kijnskijns

J

i

J

iJ

j

j

i

J

ij

J

j

j

i

J

jij

kijnsknr

kijnsknr

kijnskijnsna

kijnskijnsna

022

021

0

1

0 021

0

1

021

)1,(),(

),(),(*

),(),()(

)1,()1,()(

MechanicsNow, we do some renaming of the parts of this fraction

The left hand side we call The top of the right is r(k) and the bottom of the right is s(k).

J

i

J

iJ

j

j

i

J

ij

J

j

j

i

J

jij

kijnsknr

kijnsknr

kijnskijnsna

kijnskijnsna

022

021

0

1

0 021

0

1

021

)1,(),(

),(),(*

),(),()(

)1,()1,()(

)()1(

0

0

kPkP

MechanicsSo now we have a simple expression fork

k

tt 1

)()1(*

)()(

0

01

kPkP

kskr

tt

k

k

Relation to Gosper’s Algorithm

If we treat as the r(n) from Gosper’s

algorithm we perform the Gosper factorization and obtain

Where for j=0,1,2,…

)()(

kskr

)()1(*

)()(

)()(

1

1

3

2

kPkP

kPkP

kskr

1))(),(gcd( 23 jkPkP

Relation to Gosper’s Algorithm

Substituting this into our previous equation for gives

k

k

tt 1

)()(*

)()()1()1(

3

2

10

101

kPkP

kPkPkPkP

tt

k

k

Relation to Gosper’s Algorithm

If we define

Then we get

This is a Gosper factorization of

)1()1(:)1( 10 kPkPkP)()(:)( 10 kPkPkP

)()1(*

)()(

3

21

kPkP

kPkP

tt

k

k

k

k

tt 1

Relation to Gosper’s Algorithm

We can now try to solve Gosper’s polynomial recurrence

for b(k).If there exists a polynomial solution for b(k) then can be summed

)()()1()1()( 32 kPkbkPkbkP

1n

kkt

Relation to Gosper’s Algorithm

The process of finding b(k) allows us to find the aj ’s. b(k) is never actually used in finding the hypergeometic sum.The b(k) can be used to find G(n,k) although the G(n,k) is also not used.

Finding G(n,k)First, define

This means

(compare with tk = Zk+1 - Zk from Gosper’s algorithm)

J

jjk knGknGjknFnat

0

),()1,(),()(:

),()1,(: knGknGtk

Finding G(n,k)Now let

Since G(n,k) has compact support over k this will just be

1

03

0 )()()1(m

kmmkm Zt

mPmbmPZZtS

mm tmP

mbmPS)(

)()1(3

Finding G(n,k)We can also write

Since G(n,m) has compact support over k, everything cancels out except for G(n,m). Thus

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

),()1,(11

mnGmnGnGnGnGnG

knGknGtSm

k

m

kkm

),( mnGSm

Finding G(n,k)So we can change the variable m back to k and combine our two formulas for Sm to get

ktkPkbkPknG

)()()1(),( 3

Finding J and f(n)In order to fully solve sums of the form

We need to know the unknowns. We will now go through the process of finding J and f(n). The aj(n)’s are found when solving for b(n).

J

jj knGknGkjnFna

0

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

Finding UnknownsFirst, we must assume that has compact support in k for each n Thus if we sum our original equation over all k. We get

),( knG

J

jj jnfna

0

0)()(

Finding UnknownsNext, we must examine several possible cases for J. Case 1: J = 1 This means that Solving for f gives:

So

0)1()()()( 10 nfnanfna

)()(

)()1(

1

0

nana

nfnf

1

010 ))(/)(()0()(

n

j

jajafnf

Finding UnknownsCase 2: J > 1 but are constant.This means we have

Which can be solved.

Ji na 0)}({

)(...)1()()( 10 Jnfanfanfanf J

Finding UnknownsCase 3: J > 1 and are polynomials. Alternatively we can say that f(n) can be expressed as a finite linear combination of hypergeometric terms. When that is the case, we can use something called Petkovšek’s algorithm f(n).

Ji na 0)}({

Finding UnknownsWe can pick a J that is bigger than needed and find the correct answer. However the smaller J is the less work required to solve the problem. There are algorithms that give an upper bound on J but we do not explore them here.

IV. A Hand Example Implementing the Creative Telescoping Algorithm

2

),(

kn

knF

2

0

)(

K

k kn

nf

Using Creative Telescoping

Take:

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

knGknGkjnFnaJ

jj

Creative Telescoping Recurrence

),()1,()(2

0

knGknGk

jnna

J

jj

Then

2

0

)(

kjn

natJ

jjk

Next we guess J=1 and apply Gosper Theory

2

1

2

0

1

kn

akn

atk

2

1

2

01 11

1

kn

ak

natk

2

1

2

0

2

1

2

01

1

11

1

kn

akn

a

kn

ak

na

tt

k

k

2

1

2

0

2

1

2

01

)!1(!)!1(

)!(!!

)!()!1()!1(

)!1()!1(!

knkna

knkna

knkna

knkna

tt

k

k

2)!1()!1( knkLcd

21

20

21

201

)1()!1()1)(1(!)1()!1()1)((!

knaknknaknnaknknna

tt

k

k

22

12

0

221

201

)1()1()1()1()1()(

knaknaknnakna

tt

k

k

21

200 )1()()1( naknakP

21

200 )1()1()( naknakP

2)1()( knkr

2)1()( kks

)()(

)()1(

)()1(

0

0

kskr

kPkP

ktkt

Gosper Factorize

2

2

)1()1(

)()(

k

knkskr

11

)1()1(

2

2

k

kn1)(1 kP

22 )1()( knkP

23 )1()( kkP

Gosper Side Condition

1))(),(( 32 hkPkPGcd For h

,...}2,1,0{

Set )()()( 10 kPkPkP

1)1()1()( 21

200 naknakP

This gives the final Gosper Factorization:

2

2

21

20

21

20

)1()1(

)1()1()1()(

kkn

naknanakna

)()(

)()1(

)()1(

3

2

0

0

kPkP

kPkP

ktkt

Gosper Recurrence

Using the fact that if the Gosper Factorization of

)()1(

)()(

)()1(

kCkC

kBkA

ktkt

Then the Gosper Recurrence is:

)()()1()1()( ncnxnBnxnA

We get a Gosper Recurrence of:

)()()1()1()( 32 kPkbkPkbkP

Using the Method of Undetermined Coefficients for

kkb )(We get:

)1(3 n2

)12(20 na

11 na

Plug these values into the original Zeilberger’s Recurrence

The original recurrence

),()1,()(2

0

knGknGk

jnna

J

jj

),()1,(1

)()(2

1

2

0 knGknGk

nna

kn

na

),()1,(1

)1()12(222

knGknGk

nn

kn

n

Now sum both sides over k

0)1()1()()12(2 nfnnfn

)()12(2)1()1( nfnnfn

)1()()12(2)1(

n

nfnnf

We Know

6)1(3)2(2)0(2)1(

ffff

1)0( f

1)0( f

1

0 1

0

)()()0()(

n

j jajafnf

nn

jjn

j

21

)12(21

0

nn

nf2

)(

BibliographyPetkovsek, Marko, Herbert Wilf, Doron Zeilberger. A=B. Massachusetts: Addison Wesley, Reading, 1968.

We would like to thank everyone for coming and to Dr. D for everything he has done to help!