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Gosper’s Algorithm

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Gosper’s Algorithm

By Zachary Vogel

- Binomial Coefficients
- The Binomial Theorem

- 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

- Parallel Summation identity
- Negation identity

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

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

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

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

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

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

- Exponential series
- Geometric series

- General Form of a hypergeometric term

- 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.

- Takes a hypergeometric term and sums it indefinitely
- Example

- 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.

- 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)

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

- We require that q(k) and r(k) must have no factors
- such that

- 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.

- 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).

- By applying
- we get
- We look to solve for 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)

- 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

- 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

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

- 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.

- 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.

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:

By manipulating our previous equations, it can be seen that

With

We also Know

change to ≤

- Now if
Then the degree of the RHS will be

Therefore,

Otherwise, one of two options will occur

i)

ii)

remove RHS

- Knowing the degree of s(k), solve
- Then simply plug the known s(k) into

- 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

- 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.

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

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

- Now we know our s(k) = -1/n, so we plug in to get T(k):

- So Gosper’s algorithm gives the identity