1 / 39

Gosper’s Algorithm

Gosper’s Algorithm. By Zachary Vogel. Binomial Coefficients. Binomial Coefficients The Binomial Theorem. 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. Binomial Identities.

vance-pace
Télécharger la présentation

Gosper’s Algorithm

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related