Post on 09-Feb-2016
description
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!