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Creative Telescoping

Creative Telescoping. Zeilberger’s Algorithm By Dylan Heeg Christopher Jordan Deanne Pieper Brent Serum. Introduction and Overview. Computerized Proofs and Combinatorial Identities By Deanne Pieper.

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Creative Telescoping

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  1. Creative Telescoping Zeilberger’s Algorithm By Dylan Heeg Christopher Jordan Deanne Pieper Brent Serum

  2. Introduction and Overview Computerized Proofs and Combinatorial Identities By Deanne Pieper

  3. 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.

  4. What is a Combinatorial Identity? Some examples of combinatorial identity are: Or,

  5. What is a Proper Hypergeometric Function? 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)

  6. Proper Hypergeometric Function Example

  7. Sister Celine’s Algorithm is used in solving definite hypergeometric summations. While Gosper’s Algorithm solves indefinite hypergeometric summations.

  8. A Review of Sister Celine’s Algorithm

  9. 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

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

  11. Holds that every point (n,k) where F(n,k) 0 and all the values of F that occur in it are well defined

  12. A Review of Gosper’s Algorithm

  13. Gosper’s Algorithm Input: Output: (if possible) Steps: i) compute r(n) ii) Perform Gosper Factorization

  14. Gosper’s Algorithm iii) Solve Gosper’s Polynomial Recurrence iv) Return v) Set

  15. Zeilberger’s Creative Telescoping Algorithm

  16. Zeilberger’s Theorem • Let’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.

  17. Zeilberger’s Algorithm • Zeilberger’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….

  18. A General Description…. • The problem? Strategy: find a Zeilberger Recurrence

  19. Usual Assumptions • i) F(n,k) Proper Hypergeometric • ii) Rational in (n,k)

  20. The basic mechanics • Let • We’ll be applying Gosper’s algorithm to these tk’s.

  21. 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.

  22. This equation will produce (possibly repeated or even complex) roots: • We now have two possible cases to consider.

  23. Case 1 • In 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).

  24. Case 2 • A root b of multiplicity k will contribute the k terms to the general solution.

  25. The mechanics of Zeilberger’s Algorithm

  26. General Observations • Problem: to sum in closed form something of the form • Strategy: Find a Zeilberger recurrence. This is of the form:

  27. General Observations • Our unknowns in this recurrence are aj(n), J, and G(n,k). • We assume the following • F(n,k) is proper hypergeometric • is rational inn, k

  28. Mechanics • Define • This means

  29. Mechanics • Now define • If we multiply both the top and bottom by

  30. Mechanics • We get the equation • Which is equal to r(k)

  31. Mechanics • Note that is a rational function • If we write we know that r1(n,k), r2(n,k) are polynomials

  32. Mechanics • Similarly, we know that s1(n,k), s2(n,k) are polynomials when they are defined

  33. Mechanics • Also note that we can rewriteas • Which is

  34. Mechanics • Using our s1 and s2 functions this can be written as • Which condenses to

  35. Mechanics • Now our original can be written as which is

  36. Mechanics • Next we clear denominators. The LCD of this fraction is • Clearing denominators we get

  37. Mechanics • Now, we do some renaming of the parts of this fraction • The left hand side we callThe top of the right is r(k) and the bottom of the right is s(k).

  38. Mechanics • So now we have a simple expression for

  39. 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,…

  40. Relation to Gosper’s Algorithm • Substituting this into our previous equation for gives

  41. Relation to Gosper’s Algorithm • If we define • Then we get • This is a Gosper factorization of

  42. 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

  43. 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.

  44. Finding G(n,k) • First, define • This means (compare with tk = Zk+1 - Zkfrom Gosper’s algorithm)

  45. Finding G(n,k) • Now let • Since G(n,k) has compact support over kthis will just be

  46. 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

  47. Finding G(n,k) • So we can change the variable mback to kand combine our two formulas for Smto get

  48. 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). Theaj(n)’s are found when solving for b(n).

  49. Finding Unknowns • First, we must assume that has compact support in kfor each n • Thus if we sum our original equation over all k. We get

  50. Finding Unknowns • Next, we must examine several possible cases for J. • Case 1: J = 1 • This means that • Solving for f gives: • So

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