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  1. Gosper’s Algorithm By Zachary Vogel

  2. Binomial Coefficients • Binomial Coefficients • The Binomial Theorem

  3. Pascal’s Triangle • Base identity for Pascal’s Triangle • 1 • 1 1 • 1 2 1 • 1 3 3 1 • 1 4 6 4 1 • 5 10 10 5 1 • 1 6 15 20 15 6 1

  4. Binomial Identities • Parallel Summation identity • Negation identity

  5. Binomial Identities • There are volumes of identities with binomial coefficients. • Here is one taken from a book: • Nanjundiah’s identity

  6. Visual of Parallel Summation 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 n=0

  7. Visual of Parallel Summation 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 n=1

  8. Visual of Parallel Summation 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 • 5 10 10 5 1 1 6 15 20 15 6 1 n=2

  9. Visual of Parallel Summation 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 • 5 10 10 5 1 1 6 15 20 15 6 1 n=3

  10. Hypergeometrics • How to find some order in all these identities with binomial coefficients? • Hypergeometric notation can be used to standardize identities.

  11. Notation

  12. Hypergeometric Series

  13. Special Cases • Exponential series • Geometric series

  14. Hypergeometric Terms • General Form of a hypergeometric term

  15. Successive Terms • If you take the ratio of successive terms of a hypergeometric series, the ratio is a rational polynomial of k. • If the ratio of successive terms form a rational function of k, then the series is hypergeometric up to a constant multiple.

  16. Gosper’s Algorithm Overview • Takes a hypergeometric term and sums it indefinitely • Example

  17. Gosper’s Algorithm Overview • The algorithm determines if the sum is a multiple of another hypergeometric term –OR – • It determines that the sum cannot be put in this form.

  18. Gosper’s Algorithm Step 1 • We will assume that some such T(k) exists. • If we get an impossible situation, then no such T(k) exists. • The first step is to work out the term ratio of the summand t(k)

  19. Term Ratio • Write the term ratio for t(k) • where initially p(k) = 1 • We will pull out some factors of q and r into p.

  20. Gosper’s Algorithm Step 1 • We require that q(k) and r(k) must have no factors • such that

  21. Gosper’s Algorithm Step 1 • For if • then divide out the factors from q(k) and r(k) and absorb those terms into p(k) as follows • p(k+1)/p(k) telescopes nicely.

  22. Gosper’s Algorithm Step 2 • Cleverly set • s(k) is some unknown function which will be the focus of the remainder of the algorithm • If we can determine what s(k) is, we can determine the final summation T(k).

  23. Gosper’s Algorithm Step 2 • By applying • we get • We look to solve for s(k).

  24. Unknown function s(k) • In order to determine T(k) we must solve for s(k). This requires a few steps • Determine that s(k) is a rational function of k • Determine that s(k) is a polynomial in k • Determine a bound on the degree of s(k)

  25. s(k) is a rational function of k • By substitution • and • With the left hand side a rational function of k, and p(k) and r(k) are polynomials, s(k) must be a rational function of k

  26. s(k) is a polynomial • Knowing s(k) is a rational function of k, we can write it as the quotient of polynomials • such that f(k) and g(k) have no common factor • we will also assume that g(k) has a root, then find a contradiction • any polynomial without a root is just a constant, so s(k) will, itself, be a polynomial

  27. s(k) is a polynomial • Suppose that g(a) = g(b) = 0, and b-a is a nonnegative integer. (In particular, we might have a = b). • Since • We have

  28. s(k) is a polynomial • Substitute a=k+1, and separately b = k, we get: • Since f and g have no common root, • So either g(a-1)=0, or g(b+1)=0, or both r(a-1) = q(b) = 0. • The last choice is impossible by construction.

  29. s(k) is a polynomial • We now know that g(b+1) or g(a-1) is a root • By repeating this argument with a-1 and b, or a and b+1, we get infinitely many roots for g(k). • Therefore, g(k) has no root, thus is a constant. So s(k) is, itself a polynomial.

  30. If we know a bound to the degree d of s(k), then we can solve it by a system of d+1 linear equations, as given by the equation: Bounding degree of s(k)

  31. By manipulating our previous equations, it can be seen that With Bounding degree of s(k) We also Know change to ≤

  32. Bounding degree of s(k) • Now if Then the degree of the RHS will be Therefore, Otherwise, one of two options will occur i) ii) remove RHS

  33. Solving s(k) • Knowing the degree of s(k), solve • Then simply plug the known s(k) into

  34. Example of Gosper’s Algorithm • To provide an example of Gosper’s algorithm at work we will attempt to solve the negation identity • To begin we set our t(k) to the summand

  35. Negation Identity • Now by setting up a term ratio we will arrive at values for r(k), q(k) and p(k): • This satisfies the conditions on r(k) and q(k), so long as n is non-negative.

  36. Negation Identity • The next step is to determine s(k). • We can bound the degree by calculating R(k) and Q(k)

  37. Negation Identity • Since deg(R(k)) > deg(Q(k)), we have two options: d=0 or d=n. We will try d = 0 first. • So

  38. Solution to negation identity • Now we know our s(k) = -1/n, so we plug in to get T(k):

  39. Solution to negation identity • So Gosper’s algorithm gives the identity